Skills to Develop

- Simplify expressions using the Quotient Property of Exponents
- Simplify expressions with zero exponents
- Simplify expressions using the Quotient to a Power Property
- Simplify expressions by applying several properties
- Divide monomials

be prepared!

Before you get started, take this readiness quiz.

- Simplify: (dfrac{8}{24}). If you missed the problem, review Example 4.3.1.
- Simplify:(2m
^{3})^{5}. If you missed the problem, review Example 10.3.13. - Simplify: (dfrac{12x}{12y}). If you missed the problem, review Example 4.3.5.

## Simplify Expressions Using the Quotient Property of Exponents

Earlier in this chapter, we developed the properties of exponents for multiplication. We summarize these properties here.

Summary of Exponent Properties for Multiplication

If a, b are real numbers and m, n are whole numbers, then

Product Property | a^{m} • a^{n} = a^{m + n} |

Power Property | (a^{m})^{n} = a^{m • n} |

Product to a Power | (ab)^{m} = a^{m}b^{m} |

Now we will look at the exponent properties for division. A quick memory refresher may help before we get started. In Fractions you learned that fractions may be simplified by dividing out common factors from the numerator and denominator using the Equivalent Fractions Property. This property will also help us work with algebraic fractions—which are also quotients.

Definition: Equivalent Fractions Property

If a, b, c are whole numbers where b ≠ 0, c ≠ 0, then

[dfrac{a}{b} = dfrac{a cdot c}{b cdot c}quad andquad dfrac{a cdot c}{b cdot c} = dfrac{a}{b}]

As before, we'll try to discover a property by looking at some examples.

Consider | $$dfrac{x^{5}}{x^{2}}$$ | and | $$dfrac{x^{2}}{x^{3}}$$ |

What do they mean? | $$dfrac{x cdot x cdot x cdot x cdot x}{x cdot x}$$ | $$dfrac{x cdot x}{x cdot x cdot x}$$ | |

Use the Equivalent Fractions Property | $$dfrac{cancel{x} cdot cancel{x} cdot x cdot x cdot x}{cancel{x} cdot cancel{x} cdot 1}$$ | $$dfrac{cancel{x} cdot cancel{x} cdot 1}{cancel{x} cdot cancel{x} cdot x}$$ | |

Simplify. | $$x^{3}$$ | $$dfrac{1}{x}$$ |

Notice that in each case the bases were the same and we subtracted the exponents.

- When the larger exponent was in the numerator, we were left with factors in the numerator and 1 in the denominator, which we simplified.
- When the larger exponent was in the denominator, we were left with factors in the denominator, and 1 in the numerator, which could not be simplified.

We write:

[egin{split} dfrac{x^{5}}{x^{2}} qquad &quad dfrac{x^{2}}{x^{3}} x^{5-2} qquad &; dfrac{1}{x^{3-2}} x^{3} qquad quad &quad dfrac{1}{x} end{split}]

Definition: Quotient Property of Exponents

If a is a real number, a ≠ 0, and m, n are whole numbers, then

[dfrac{a^{m}}{a^{n}} = a^{m-n}, ; m>n quad and quad dfrac{a^{m}}{a^{n}} = dfrac{1}{a^{n-m}},; n>m]

A couple of examples with numbers may help to verify this property.

[egin{split} dfrac{3^{4}}{3^{2}} &stackrel{?}{=} 3^{4-2} qquad ; dfrac{5^{2}}{5^{3}} stackrel{?}{=} dfrac{1}{5^{3-2}} dfrac{81}{9} &stackrel{?}{=} 3^{2} qquad ; ; dfrac{25}{125} stackrel{?}{=} dfrac{1}{5^{1}} 9 &= 9; checkmark qquad ; ; ; dfrac{1}{5} = dfrac{1}{5}; checkmark end{split}]

When we work with numbers and the exponent is less than or equal to 3, we will apply the exponent. When the exponent is greater than 3, we leave the answer in exponential form.

Example (PageIndex{1}):

Simplify: (a) (dfrac{x^{10}}{x^{8}}) (b) (dfrac{2^{9}}{2^{2}})

**Solution**

To simplify an expression with a quotient, we need to first compare the exponents in the numerator and denominator.

(a)

Since 10 > 8, there are more factors of x in the numerator. | $$dfrac{x^{10}}{x^{8}}$$ |

Use the quotient property with m > n, (dfrac{a^{m}}{a^{n}} = a^{m − n}). | $$x^{ extcolor{red}{10-8}}$$ |

Simplify. | $$x^{2}$$ |

(b)

Since 9 > 2, there are more factors of 2 in the numerator. | $$dfrac{2^{9}}{2^{2}}$$ |

Use the quotient property with m > n, (dfrac{a^{m}}{a^{n}} = a^{m − n}). | $$2^{ extcolor{red}{9-2}}$$ |

Simplify. | $$2^{7}$$ |

Notice that when the larger exponent is in the numerator, we are left with factors in the numerator.

Exercise (PageIndex{1}):

Simplify: (a) (dfrac{x^{12}}{x^{9}}) (b) (dfrac{7^{14}}{7^{5}})

- Answer a
(x^3)

- Answer b
(7^9)

Exercise (PageIndex{2}):

Simplify: (a) (dfrac{y^{23}}{y^{17}}) (b) (dfrac{8^{15}}{8^{7}})

- Answer a
(y^6)

- Answer b
(8^8)

Example (PageIndex{2}):

Simplify: (a) (dfrac{b^{10}}{b^{15}}) (b) (dfrac{3^{3}}{3^{5}})

**Solution**

To simplify an expression with a quotient, we need to first compare the exponents in the numerator and denominator.

(a)

Since 15 > 10, there are more factors of b in the denominator. | $$dfrac{b^{10}}{b^{15}}$$ |

Use the quotient property with n > m, (dfrac{a^{m}}{a^{n}} = dfrac{1}{a^{n − m}}). | $$dfrac{ extcolor{red}{1}}{b^{ extcolor{red}{15-10}}}$$ |

Simplify. | $$dfrac{1}{b^{5}}$$ |

(b)

Since 5 > 3, there are more factors of 3 in the denominator. | $$dfrac{3^{3}}{3^{5}}$$ |

Use the quotient property with n > m, (dfrac{a^{m}}{a^{n}} = dfrac{1}{a^{n − m}}). | $$dfrac{ extcolor{red}{1}}{3^{ extcolor{red}{5-3}}}$$ |

Simplify. | $$dfrac{1}{3^{2}}$$ |

Apply the exponent. | $$dfrac{1}{9}$$ |

Notice that when the larger exponent is in the denominator, we are left with factors in the denominator and 1 in the numerator.

Exercise (PageIndex{3}):

Simplify: (a) (dfrac{x^{8}}{x^{15}}) (b) (dfrac{12^{11}}{12^{21}})

- Answer a
(frac{1}{x^7})

- Answer b
(frac{1}{12^10})

Exercise (PageIndex{4}):

Simplify: (a) (dfrac{m^{17}}{m^{26}}) (b) (dfrac{7^{8}}{7^{14}})

- Answer a
(frac{1}{m^9})

- Answer b
(frac{1}{7^6})

Example (PageIndex{3}):

Simplify: (a) (dfrac{a^{5}}{a^{9}}) (b) (dfrac{x^{11}}{x^{7}})

**Solution**

(a)

Since 9 > 5, there are more a's in the denominator and so we will end up with factors in the denominator. | $$dfrac{a^{5}}{a^{9}}$$ |

Use the quotient property with n > m, (dfrac{a^{m}}{a^{n}} = dfrac{1}{a^{n − m}}). | $$dfrac{ extcolor{red}{1}}{a^{ extcolor{red}{9-5}}}$$ |

Simplify. | $$dfrac{1}{a^{4}}$$ |

(b)

Notice there are more factors of x in the numerator, since 11 > 7. So we will end up with factors in the numerator. | $$dfrac{x^{11}}{x^{97}}$$ |

Use the quotient property with m > n, (dfrac{a^{m}}{a^{n}} = a^{m − n}). | $$a^{ extcolor{red}{11-7}}$$ |

Simplify. | $$x^{4}$$ |

Exercise (PageIndex{5}):

Simplify: (a) (dfrac{b^{19}}{b^{11}}) (b) (dfrac{z^{5}}{z^{11}})

- Answer a
(b^8)

- Answer b
(frac{1}{z^6})

Exercise (PageIndex{6}):

Simplify: (a) (dfrac{p^{9}}{p^{17}}) (b) (dfrac{w^{13}}{w^{9}})

- Answer a
(frac{1}{p^8})

- Answer b
(w^4)

## Simplify Expressions with Zero Exponents

A special case of the Quotient Property is when the exponents of the numerator and denominator are equal, such as an expression like (dfrac{a^{m}}{a^{m}}). From earlier work with fractions, we know that

[dfrac{2}{2} = 1 qquad dfrac{17}{17} = 1 qquad dfrac{-43}{-43} = 1]

In words, a number divided by itself is 1. So (dfrac{x}{x}) = 1, for any x (x ≠ 0), since any number divided by itself is 1.

The Quotient Property of Exponents shows us how to simplify (dfrac{a^{m}}{a^{n}}) when m > n and when n < m by subtracting exponents. What if m = n?

Now we will simplify (dfrac{a^{m}}{a^{m}}) in two ways to lead us to the definition of the **zero exponent**. Consider first (dfrac{8}{8}), which we know is 1.

$$dfrac{8}{8} = 1$$ | |

Write 8 as 2^{3}. | $$dfrac{2^{3}}{2^{3}} = 1$$ |

Subtract exponents. | $$2^{3-3} = 1$$ |

Simplify. | $$2^{0} = 1$$ |

We see (dfrac{a^{m}}{a^{n}}) simplifies to a^{0} and to 1. So a^{0} = 1.

Definition: Zero Exponent

If a is a non-zero number, then a^{0} = 1. Any nonzero number raised to the zero power is 1.

In this text, we assume any variable that we raise to the zero power is not zero.

Example (PageIndex{4}):

Simplify: (a) 12^{0} (b) y^{0}

**Solution**

The definition says any non-zero number raised to the zero power is 1.

(a) 12^{0}

Use the definition of the zero exponent. | 1 |

(b) y^{0}

Use the definition of the zero exponent. | 1 |

Exercise (PageIndex{7}):

Simplify: (a) 17^{0} (b) m^{0}

- Answer a
1

- Answer b
1

Exercise (PageIndex{8}):

Simplify: (a) k^{0} (b) 29^{0}

- Answer a
1

- Answer b
1

Now that we have defined the zero exponent, we can expand all the Properties of Exponents to include whole number exponents.

What about raising an expression to the zero power? Let's look at (2x)^{0}. We can use the product to a power rule to rewrite this expression.

(2x)^{0} | |

Use the Product to a Power Rule. | 2^{0}x^{0} |

Use the Zero Exponent Property. | 1 • 1 |

Simplify. | 1 |

This tells us that any non-zero expression raised to the zero power is one.

Example (PageIndex{5}):

Simplify: (7z)^{0}.

**Solution**

Use the definition of the zero exponent. | 1 |

Exercise (PageIndex{9}):

Simplify: (−4y)^{0}.

- Answer
1

Exercise (PageIndex{10}):

Simplify: (left(dfrac{2}{3} x ight)^{0}).

- Answer
1

Example (PageIndex{6}):

Simplify: (a) (−3x^{2}y)^{0} (b) −3x^{2}y^{0}

**Solution**

(a) (−3x^{2}y)^{0}

The product is raised to the zero power. | (−3x^{2}y)^{0} |

Use the definition of the zero exponent. | 1 |

(b) −3x^{2}y^{0}

Notice that only the variable y is being raised to the zero power. | −3x^{2}y^{0} |

Use the definition of the zero exponent. | −3x^{2 }• 1 |

Simplify. | −3x^{2} |

Exercise (PageIndex{11}):

Simplify: (a) (7x^{2}y)^{0} (b) 7x^{2}y^{0}

- Answer a
1

- Answer b
(7x^2)

Exercise (PageIndex{12}):

Simplify: (a) −23x^{2}y^{0} (b) (−23x^{2}y)^{0}

- Answer a
(-23x^2)

- Answer b
1

## Simplify Expressions Using the Quotient to a Power Property

Now we will look at an example that will lead us to the Quotient to a Power Property.

$$left(dfrac{x}{y} ight)^{3}$$ | |

This means | $$dfrac{x}{y} cdot dfrac{x}{y} cdot dfrac{x}{y}$$ |

Multiply the fractions. | $$dfrac{x cdot x cdot x}{y cdot y cdot y}$$ |

Write with exponents. | $$dfrac{x^{3}}{y^{3}}$$ |

Notice that the exponent applies to both the numerator and the denominator. We see that (left(dfrac{x}{y} ight)^{3}) is (dfrac{x^{3}}{y^{3}}). We write:

[left(dfrac{x}{y} ight)^{3} = dfrac{x^{3}}{y^{3}}]

This leads to the Quotient to a Power Property for Exponents.

Definition: Quotient to a Power Property of Exponents

If a and b are real numbers, b ≠ 0, and m is a counting number, then

[left(dfrac{a}{b} ight)^{m} = dfrac{a^{m}}{b^{m}}]

To raise a fraction to a power, raise the numerator and denominator to that power.

An example with numbers may help you understand this property:

[egin{split} left(dfrac{2}{3} ight)^{3} &stackrel{?}{=} dfrac{2^{3}}{3^{3}} dfrac{2}{3} cdot dfrac{2}{3} cdot dfrac{2}{3} &stackrel{?}{=} dfrac{8}{27} dfrac{8}{27} &= dfrac{8}{27}; checkmark end{split}]

Example (PageIndex{7}):

Simplify: (a) (left(dfrac{5}{8} ight)^{2}) (b) (left(dfrac{x}{3} ight)^{4}) (c) (left(dfrac{y}{m} ight)^{3})

**Solution**

(a) (left(dfrac{5}{8} ight)^{2})

Use the Quotient to a Power Property, (left(dfrac{a}{b} ight)^{m} = dfrac{a^{m}}{b^{m}}). | $$dfrac{5^{ extcolor{red}{2}}}{8^{ extcolor{red}{2}}}$$ |

Simplify. | $$dfrac{25}{64}$$ |

(b) (left(dfrac{x}{3} ight)^{4})

Use the Quotient to a Power Property, (left(dfrac{a}{b} ight)^{m} = dfrac{a^{m}}{b^{m}}). | $$dfrac{x^{ extcolor{red}{4}}}{3^{ extcolor{red}{4}}}$$ |

Simplify. | $$dfrac{x^{4}}{81}$$ |

(c) (left(dfrac{y}{m} ight)^{3})

Raise the numerator and denominator to the third power. | $$dfrac{y^{ extcolor{red}{3}}}{m^{ extcolor{red}{3}}}$$ |

Exercise (PageIndex{13}):

Simplify: (a) (left(dfrac{7}{9} ight)^{2}) (b) (left(dfrac{y}{8} ight)^{3}) (c) (left(dfrac{p}{q} ight)^{6})

- Answer a
(dfrac{49}{81})

- Answer b
(dfrac{y^3}{512})

- Answer c
(dfrac{p^6}{q^6})

Exercise (PageIndex{14}):

Simplify: (a) (left(dfrac{1}{8} ight)^{2}) (b) (left(dfrac{-5}{m} ight)^{3}) (c) (left(dfrac{r}{s} ight)^{4})

- Answer a
(dfrac{1}{64})

- Answer b
(-dfrac{125}{m^3})

- Answer c
(dfrac{r^4}{s^4})

## 10.4 Divide Monomials

Earlier in this chapter, we developed the properties of exponents for multiplication. We summarize these properties here.

### Summary of Exponent Properties for Multiplication

Now we will look at the exponent properties for division. A quick memory refresher may help before we get started. In Fractions you learned that fractions may be simplified by dividing out common factors from the numerator and denominator using the Equivalent Fractions Property . This property will also help us work with algebraic fractions—which are also quotients.

### Equivalent Fractions Property

As before, we'll try to discover a property by looking at some examples.

Consider x 5 x 2 and x 2 x 3 What do they mean? x ⋅ x ⋅ x ⋅ x ⋅ x x ⋅ x x ⋅ x x ⋅ x ⋅ x Use the Equivalent Fractions Property. x ⋅ x ⋅ x ⋅ x ⋅ x x ⋅ x ⋅ 1 x ⋅ x ⋅ 1 x ⋅ x ⋅ x Simplify. x 3 1 x Consider x 5 x 2 and x 2 x 3 What do they mean? x ⋅ x ⋅ x ⋅ x ⋅ x x ⋅ x x ⋅ x x ⋅ x ⋅ x Use the Equivalent Fractions Property. x ⋅ x ⋅ x ⋅ x ⋅ x x ⋅ x ⋅ 1 x ⋅ x ⋅ 1 x ⋅ x ⋅ x Simplify. x 3 1 x

Notice that in each case the bases were the same and we subtracted the exponents.

### Quotient Property of Exponents

A couple of examples with numbers may help to verify this property.

### Example 10.45

#### Solution

To simplify an expression with a quotient, we need to first compare the exponents in the numerator and denominator.

Notice that when the larger exponent is in the numerator, we are left with factors in the numerator.

### Example 10.46

#### Solution

Notice that when the larger exponent is in the denominator, we are left with factors in the denominator and 1 1 in the numerator.

### Example 10.47

#### Solution

### Simplify Expressions with Zero Exponents

A special case of the Quotient Property is when the exponents of the numerator and denominator are equal, such as an expression like a m a m . a m a m . From earlier work with fractions, we know that

### Zero Exponent

Any nonzero number raised to the zero power is 1 . 1 .

In this text, we assume any variable that we raise to the zero power is not zero.

### Example 10.48

#### Solution

The definition says any non-zero number raised to the zero power is 1 . 1 .

Now that we have defined the zero exponent, we can expand all the Properties of Exponents to include whole number exponents.

What about raising an expression to the zero power? Let's look at ( 2 x ) 0 . ( 2 x ) 0 . We can use the product to a power rule to rewrite this expression.

This tells us that any non-zero expression raised to the zero power is one.

### Example 10.49

#### Solution

### Example 10.50

#### Solution

### Simplify Expressions Using the Quotient to a Power Property

Now we will look at an example that will lead us to the Quotient to a Power Property.

Notice that the exponent applies to both the numerator and the denominator.

This leads to the Quotient to a Power Property for Exponents.

### Quotient to a Power Property of Exponents

To raise a fraction to a power, raise the numerator and denominator to that power.

An example with numbers may help you understand this property:

### Example 10.51

#### Solution

### Simplify Expressions by Applying Several Properties

We'll now summarize all the properties of exponents so they are all together to refer to as we simplify expressions using several properties. Notice that they are now defined for whole number exponents.

### Summary of Exponent Properties

### Example 10.52

#### Solution

### Example 10.53

#### Solution

### Example 10.54

#### Solution

### Example 10.55

#### Solution

Here we cannot simplify inside the parentheses first, since the bases are not the same.

### Example 10.56

#### Solution

### Example 10.57

Simplify: ( y 2 ) 3 ( y 2 ) 4 ( y 5 ) 4 . ( y 2 ) 3 ( y 2 ) 4 ( y 5 ) 4 .

#### Solution

Simplify: ( y 4 ) 4 ( y 3 ) 5 ( y 7 ) 6 . ( y 4 ) 4 ( y 3 ) 5 ( y 7 ) 6 .

Simplify: ( 3 x 4 ) 2 ( x 3 ) 4 ( x 5 ) 3 . ( 3 x 4 ) 2 ( x 3 ) 4 ( x 5 ) 3 .

### Divide Monomials

We have now seen all the properties of exponents. We'll use them to divide monomials. Later, you'll use them to divide polynomials.

### Example 10.58

Find the quotient: 56 x 5 ÷ 7 x 2 . 56 x 5 ÷ 7 x 2 .

#### Solution

Find the quotient: 63 x 8 ÷ 9 x 4 . 63 x 8 ÷ 9 x 4 .

Find the quotient: 96 y 11 ÷ 6 y 8 . 96 y 11 ÷ 6 y 8 .

When we divide monomials with more than one variable, we write one fraction for each variable.

### Example 10.59

Find the quotient: 42 x 2 y 3 −7 x y 5 . 42 x 2 y 3 −7 x y 5 .

#### Solution

Find the quotient: −84 x 8 y 3 7 x 10 y 2 . −84 x 8 y 3 7 x 10 y 2 .

Find the quotient: −72 a 4 b 5 −8 a 9 b 5 . −72 a 4 b 5 −8 a 9 b 5 .

### Example 10.60

Find the quotient: 24 a 5 b 3 48 a b 4 . 24 a 5 b 3 48 a b 4 .

#### Solution

Find the quotient: 16 a 7 b 6 24 a b 8 . 16 a 7 b 6 24 a b 8 .

Find the quotient: 27 p 4 q 7 −45 p 12 q . 27 p 4 q 7 −45 p 12 q .

Once you become familiar with the process and have practiced it step by step several times, you may be able to simplify a fraction in one step.

### Example 10.61

Find the quotient: 14 x 7 y 12 21 x 11 y 6 . 14 x 7 y 12 21 x 11 y 6 .

#### Solution

Find the quotient: 28 x 5 y 14 49 x 9 y 12 . 28 x 5 y 14 49 x 9 y 12 .

Find the quotient: 30 m 5 n 11 48 m 10 n 14 . 30 m 5 n 11 48 m 10 n 14 .

In all examples so far, there was no work to do in the numerator or denominator before simplifying the fraction. In the next example, we'll first find the product of two monomials in the numerator before we simplify the fraction.

### Example 10.62

Find the quotient: ( 3 x 3 y 2 ) ( 10 x 2 y 3 ) 6 x 4 y 5 . ( 3 x 3 y 2 ) ( 10 x 2 y 3 ) 6 x 4 y 5 .

#### Solution

Remember, the fraction bar is a grouping symbol. We will simplify the numerator first.

Find the quotient: ( 3 x 4 y 5 ) ( 8 x 2 y 5 ) 12 x 5 y 8 . ( 3 x 4 y 5 ) ( 8 x 2 y 5 ) 12 x 5 y 8 .

Find the quotient: ( −6 a 6 b 9 ) ( −8 a 5 b 8 ) −12 a 10 b 12 . ( −6 a 6 b 9 ) ( −8 a 5 b 8 ) −12 a 10 b 12 .

### Media

#### ACCESS ADDITIONAL ONLINE RESOURCES

### Section 10.4 Exercises

#### Practice Makes Perfect

**Simplify Expressions Using the Quotient Property of Exponents**

In the following exercises, simplify.

**Simplify Expressions with Zero Exponents**

In the following exercises, simplify.

**Simplify Expressions Using the Quotient to a Power Property**

In the following exercises, simplify.

**Simplify Expressions by Applying Several Properties**

In the following exercises, simplify.

( 3 x 4 ) 3 ( 2 x 3 ) 2 ( 6 x 5 ) 2 ( 3 x 4 ) 3 ( 2 x 3 ) 2 ( 6 x 5 ) 2

( −2 y 3 ) 4 ( 3 y 4 ) 2 ( −6 y 3 ) 2 ( −2 y 3 ) 4 ( 3 y 4 ) 2 ( −6 y 3 ) 2

**Divide Monomials**

In the following exercises, divide the monomials.

48 x 11 y 9 z 3 36 x 6 y 8 z 5 48 x 11 y 9 z 3 36 x 6 y 8 z 5

64 x 5 y 9 z 7 48 x 7 y 12 z 6 64 x 5 y 9 z 7 48 x 7 y 12 z 6

( 10 u 2 v ) ( 4 u 3 v 6 ) 5 u 9 v 2 ( 10 u 2 v ) ( 4 u 3 v 6 ) 5 u 9 v 2

( 6 m 2 n ) ( 5 m 4 n 3 ) 3 m 10 n 2 ( 6 m 2 n ) ( 5 m 4 n 3 ) 3 m 10 n 2

( 6 a 4 b 3 ) ( 4 a b 5 ) ( 12 a 8 b ) ( a 3 b ) ( 6 a 4 b 3 ) ( 4 a b 5 ) ( 12 a 8 b ) ( a 3 b )

( 4 u 5 v 4 ) ( 15 u 8 v ) ( 12 u 3 v ) ( u 6 v ) ( 4 u 5 v 4 ) ( 15 u 8 v ) ( 12 u 3 v ) ( u 6 v )

#### Mixed Practice

27 a 7 3 a 3 + 54 a 9 9 a 5 27 a 7 3 a 3 + 54 a 9 9 a 5

32 c 11 4 c 5 + 42 c 9 6 c 3 32 c 11 4 c 5 + 42 c 9 6 c 3

32 y 5 8 y 2 − 60 y 10 5 y 7 32 y 5 8 y 2 − 60 y 10 5 y 7

48 x 6 6 x 4 − 35 x 9 7 x 7 48 x 6 6 x 4 − 35 x 9 7 x 7

63 r 6 s 3 9 r 4 s 2 − 72 r 2 s 2 6 s 63 r 6 s 3 9 r 4 s 2 − 72 r 2 s 2 6 s

56 y 4 z 5 7 y 3 z 3 − 45 y 2 z 2 5 y 56 y 4 z 5 7 y 3 z 3 − 45 y 2 z 2 5 y

#### Everyday Math

#### Writing Exercises

#### Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ On a scale of 1–10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

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## 10.6 Introduction to Factoring Polynomials

Earlier we multiplied factors together to get a product. Now, we will be reversing this process we will start with a product and then break it down into its factors. Splitting a product into factors is called factoring.

In The Language of Algebra we factored numbers to find the least common multiple (LCM) of two or more numbers. Now we will factor expressions and find the

*greatest common factor*of two or more expressions. The method we use is similar to what we used to find the LCM.### Greatest Common Factor

The greatest common factor (GCF) of two or more expressions is the largest expression that is a factor of all the expressions.

First we will find the greatest common factor of two numbers.

### Example 10.80

Find the greatest common factor of 24 24 and 36 . 36 .

#### Solution

**Step 1:**Factor each coefficient into primes. Write all variables with exponents in expanded form.Factor 24 and 36. **Step 2:**List all factors--matching common factors in a column.In each column, circle the common factors. Circle the 2, 2, and 3 that are shared by both numbers. **Step 3:**Bring down the common factors that all expressions share.Bring down the 2, 2, 3 and then multiply. **Step 4:**Multiply the factors.The GCF of 24 and 36 is 12. Notice that since the GCF is a factor of both numbers, 24 24 and 36 36 can be written as multiples of 12 . 12 .

Find the greatest common factor: 54 , 36 . 54 , 36 .

Find the greatest common factor: 48 , 80 . 48 , 80 .

In the previous example, we found the greatest common factor of constants. The greatest common factor of an algebraic expression can contain variables raised to powers along with coefficients. We summarize the steps we use to find the greatest common factor .

### How To

#### Find the greatest common factor.

- Step 1. Factor each coefficient into primes. Write all variables with exponents in expanded form.
- Step 2. List all factors—matching common factors in a column. In each column, circle the common factors.
- Step 3. Bring down the common factors that all expressions share.
- Step 4. Multiply the factors.

### Example 10.81

Find the greatest common factor of 5 x and 15 . 5 x and 15 .

#### Solution

Factor each number into primes.

Circle the common factors in each column.

Bring down the common factors.The GCF of 5x and 15 is 5. Find the greatest common factor: 7 y , 14 . 7 y , 14 .

Find the greatest common factor: 22 , 11 m . 22 , 11 m .

In the examples so far, the greatest common factor was a constant. In the next two examples we will get variables in the greatest common factor.

### Example 10.82

Find the greatest common factor of 12 x 2 12 x 2 and 18 x 3 . 18 x 3 .

#### Solution

Factor each coefficient into primes and write

the variables with exponents in expanded form.

Circle the common factors in each column.

Bring down the common factors.

Multiply the factors.The GCF of 12 x 2 and 18 x 3 is 6 x 2 The GCF of 12 x 2 and 18 x 3 is 6 x 2 Find the greatest common factor: 16 x 2 , 24 x 3 . 16 x 2 , 24 x 3 .

Find the greatest common factor: 27 y 3 , 18 y 4 . 27 y 3 , 18 y 4 .

### Example 10.83

Find the greatest common factor of 14 x 3 , 8 x 2 , 10 x . 14 x 3 , 8 x 2 , 10 x .

#### Solution

Factor each coefficient into primes and write

the variables with exponents in expanded form.

Circle the common factors in each column.

Bring down the common factors.

Multiply the factors.The GCF of 14 x 3 and 8 x 2 , and 10 x is 2 x The GCF of 14 x 3 and 8 x 2 , and 10 x is 2 x Find the greatest common factor: 21 x 3 , 9 x 2 , 15 x . 21 x 3 , 9 x 2 , 15 x .

Find the greatest common factor: 25 m 4 , 35 m 3 , 20 m 2 . 25 m 4 , 35 m 3 , 20 m 2 .

### Factor the Greatest Common Factor from a Polynomial

### Distributive Property

The form on the left is used to multiply. The form on the right is used to factor.

So how do we use the Distributive Property to factor a polynomial? We find the GCF of all the terms and write the polynomial as a product!

### Example 10.84

#### Solution

**Step 1:**Find the GCF of all the terms of the polynomial.Find the GCF of 2x and 14. **Step 2:**Rewrite each term as a product using the GCF.Rewrite 2x and 14 as products of their GCF, 2.

2 x = 2 ⋅ x 2 x = 2 ⋅ x

14 = 2 ⋅ 7 14 = 2 ⋅ 7**Step 3:**Use the Distributive Property 'in reverse' to factor the expression.2 ( x + 7 ) 2 ( x + 7 ) **Step 4:**Check by multiplying the factors.Check: Notice that in Example 10.84, we used the word

*factor*as both a noun and a verb:### How To

#### Factor the greatest common factor from a polynomial.

- Step 1. Find the GCF of all the terms of the polynomial.
- Step 2. Rewrite each term as a product using the GCF.
- Step 3. Use the Distributive Property ‘in reverse’ to factor the expression.
- Step 4. Check by multiplying the factors.

### Example 10.85

#### Solution

Rewrite each term as a product using the GCF. Use the Distributive Property 'in reverse' to factor the GCF. Check by multiplying the factors to get the original polynomial. The expressions in the next example have several factors in common. Remember to write the GCF as the product of all the common factors.

### Example 10.86

#### Solution

Now we’ll factor the greatest common factor from a trinomial . We start by finding the GCF of all three terms.

### Example 10.87

#### Solution

In the next example, we factor a variable from a binomial .

### Example 10.88

#### Solution

When there are several common factors, as we’ll see in the next two examples, good organization and neat work helps!

### Example 10.89

#### Solution

### Example 10.90

#### Solution

### Example 10.91

Factor: 14 x 3 + 8 x 2 − 10 x . 14 x 3 + 8 x 2 − 10 x .

#### Solution

Previously, we found the GCF of 14 x 3 , 8 x 2 , and 10 x 14 x 3 , 8 x 2 , and 10 x to be 2 x . 2 x .

Factor: 18 y 3 − 6 y 2 − 24 y . 18 y 3 − 6 y 2 − 24 y .

Factor: 16 x 3 + 8 x 2 − 12 x . 16 x 3 + 8 x 2 − 12 x .

When the leading coefficient , the coefficient of the first term, is negative, we factor the negative out as part of the GCF.

### Example 10.92

#### Solution

When the leading coefficient is negative, the GCF will be negative. Ignoring the signs of the terms, we first find the GCF of 9 *y*and 27 is 9.Since the expression −9y−27 has a negative leading coefficient, we use −9 as the GCF. − 9 y − 27 − 9 y − 27 Rewrite each term using the GCF. Factor the GCF. − 9 ( y + 3 ) − 9 ( y + 3 ) Pay close attention to the signs of the terms in the next example.

### Example 10.93

#### Solution

The leading coefficient is negative, so the GCF will be negative. Since the leading coefficient is negative, the GCF is negative, −4 *a*.−4 a 2 + 16 a −4 a 2 + 16 a Rewrite each term. Factor the GCF. − 4 a ( a − 4 ) − 4 a ( a − 4 ) Check on your own by multiplying. ### Media

#### ACCESS ADDITIONAL ONLINE RESOURCES

### Section 10.6 Exercises

#### Practice Makes Perfect

**Find the Greatest Common Factor of Two or More Expressions**In the following exercises, find the greatest common factor.

**Factor the Greatest Common Factor from a Polynomial**In the following exercises, factor the greatest common factor from each polynomial.

#### Everyday Math

**Revenue**A manufacturer of microwave ovens has found that the revenue received from selling microwaves a cost of p p dollars each is given by the polynomial −5 p 2 + 150 p . −5 p 2 + 150 p . Factor the greatest common factor from this polynomial.#### Writing Exercises

#### Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?

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## Matemaths

It was the end of February when we were told that schools would close and that we should invent a new way to teach overnight.

After two months I’m still struggling. I love technology and I think it has great potential, but most of us made a mistake. We thought that we could continue teaching in the old way, only behind a screen instead of in class, but kids get bored. Sometimes I get this feeling that all my students are sleeping, and there is nothing I can do about it…almost.

A few days ago, I ran into boom cards www.boomlearning.com.

It allows you to create interactive lessons and do everything you want: drag and drop, fill in the blanks, multiple choice.

There are thousands of decks already created by other teachers, some are free others you have to pay for, but they are beautiful, engaging, and fun and you can create your own decks, so you can decide the level perfect for your class.

Another bright side of it is that you can sell the decks that you create.

I teach middle and high school and from what I understand boom cards are not so common to be used for older students while they are really popular for elementary school.

Boom Cards are the perfect combination of engagement and efficiency for elementary teachers. The platform provides a unique digital experience for the students as it presents content in perfect bite-sized pieces. Bright visuals guide the students through essential learning tasks one question at a time and the students feel like they’re playing the latest app. The entire process is streamlined with the ability for students to input their responses instantaneously.

The very fact that kids aren’t overwhelmed with a list of questions or associated information is one of the reasons educators are flocking towards the cards.

The control that Boom provides teachers is unprecedented as the cards guide your students through your learning objectives and records all responses creating feedback opportunities and usable data for assessment.

For students learning in a digital environment it is an absolutely vital tool and as they re-enter the classroom it will remain a staple of the modern elementary classroom.

## Mrs. Miller's Math Mavens - 8th grade

**Power of a Power Property**

**Words**

**To fi****nd a power of a power, multiply the exponents.**

**Numbers**

**[(4)^6]^3****=****(4)^(18)**

**Algebra**

**[(a)^m]^n****=****(a)^(mn)**

**Power of a Product Property**

**Words**

**To find a power of a product, find the power of each factor****and multiply.**

**Numbers**

**(3 ⋅ 2)^5 = (3^5)****⋅****(2^5)**

**Algebra**

**(ab)^m = (a^m)****⋅****(b^m)****How can you divide two powers that have****the same base?****10****.4 Zero and Negative Exponents***I can**evaluate expressions**involving numbers with**zero as an exponent.**I can evaluate expressions**involving negative**integer exponents.***How can you evaluate a nonzero number with****an exponent of zero?****How can you evaluate a nonzero number with a negative****integer exponent?****Words****For any nonzero number a, (a)^0 = 1. The power (0)^0****is undefined.****Numbers****(4)^0 = 1****Algebra****(a)^0 = 1, where a ≠ 0****Negative Exponents****Words****For any integer n and any nonzero number a, (a)^(−n) is the****reciprocal of (a)^n.****How can you write a number in****scientific notation?****Writing Numbers in Scientific Notation****Step 1: Move the decimal point so it is located to the right of the****leading nonzero digit.****Step 2: Count the number of places you moved the decimal point.****This indicates the exponent of the power of 10.****Number Greater Than or Equal to 10**

**Use a positive exponent when****you move the decimal point****to the left.**

**8600 = 8.6 × 10^3****Number Between 0 and 1**

**Use a negative exponent when****you move the decimal point to****the right.**

**0.0024 = 2.4 × 10^()****I can***add, subtract, multiply,**and divide numbers**written in scientific**notation.*

Sullivan, Struve & Mazzarella

With training in mathematics, statistics, and economics, Michael Sullivan, III has a varied teaching background that includes 27 years of instruction in both high school and college-level mathematics. He is currently a full-time professor of mathematics at Joliet Junior College. Michael has numerous textbooks in publication, including an Introductory Statistics series and a Precalculus series, which he writes with his father, Michael Sullivan.

Michael believes that his experiences writing texts for college-level math and statistics courses give him a unique perspective as to where students are headed once they leave the developmental mathematics tract. This experience is reflected in the philosophy and presentation of his developmental text series. When not in the classroom or writing, Michael enjoys spending time with his three children, Michael, Kevin, and Marissa, and playing golf. Now that his two sons are getting older, he has the opportunity to do both at the same time!

Kathy Struve has been a classroom teacher for nearly 35 years, first at the high school level and, for the past 27 years, at Columbus State Community College. Kathy embraces classroom diversity: diversity of students’ age, learning styles, and previous learning success. She is aware of the challenges of teaching mathematics at a large, urban community college, where students have varied mathematics backgrounds and may enter college with a high level of mathematics anxiety.

Kathy served as Lead Instructor of the Developmental Algebra sequence at Columbus State, where she developed curriculum, conducted workshops, and provided leadership to adjunct faculty in the mathematics department. She embraces the use of technology in instruction, and has taught web and hybrid classes in addition to traditional face-to-face and emporium-style classes. She is always looking for ways to more fully involve students in the learning process. In her spare time Kathy enjoys spending time with her two adult daughters, her four granddaughters, and biking, hiking, and traveling with her husband.

Born and raised in San Diego county, Janet Mazzarella spent her career teaching in culturally and economically diverse high schools before taking a position at Southwestern College 25 years ago. Janet has taught a wide range of mathematics courses, from arithmetic through calculus for math/science/engineering majors and has training in mathematics, education, engineering, and accounting.

Janet has worked to incorporate technology into the curriculum by participating in the development of Interactive Math and Math Pro. At Southwestern College, she helped develop the self-paced developmental mathematics program. In addition, Janet was the Dean of the School of Mathematics, Science, and Engineering, the Chair of the Mathematics Department, the faculty union president, and the faculty coordinator for Intermediate Algebra. In the past, free time consisted of racing motorcycles off-road in the Baja 500 and rock climbing, but recently she has given up the adrenaline rush of these activities for the thrill of traveling in Europe.

**Jessica Bernards, Contributor**Jessica Bernards has been teaching mathematics since 2005. She began her career at the high school level and then transitioned to teaching at Portland Community College in 2010. She has taught a wide range of mathematics courses from Developmental Math up to Calculus and has created curriculum for all of these levels. Additionally, Jessica is a member of AMATYC's Project ACCCESS Cohort 9 where she developed a Math Study Skills Program which is now used across the nation. In 2017, she was the honored recipient of the Leila and Simon Peskoff AMATYC Award for her work with Project ACCCESS.

When not working, Jessica loves spending time with her husband and two boys in the Pacific Northwest. She enjoys running races, cooking, and hiking and is also an active member of her community coordinating a neighborhood group that brings local moms together.

**Wendy Fresh, Contributor**Wendy Fresh has been full-time instructor at Portland Community College since 1997 and has taught a wide range of classes from Developmental Math through Calculus, both on campus and online. Before teaching at PCC, Wendy began her teaching career in 1992, teaching high school at both rural and urban schools. Her love of creating curriculum to make the classroom come alive, has led her to working with technologies that can be incorporated into her many courses.

She earned her Bachelor’s Degree in Mathematics Education from the University of Oregon and her Master’s Degree in the Teaching of Mathematics from Portland State University. When not teaching, Wendy loves hanging out at home with her husband and two college age “kids”. In addition, she enjoys running, gardening, watching soccer and reading.

Monomials A monomial is an algebraic expression that consists of only one term. (A term is a numerical or literal expression with its own sign.) For instance, 9x , -4a², and 3mpx³ are all monomials. The number in front of the variable is called the numerical coefficient. In -9xy, -9 is the coefficient. Adding and subtracting monomials To add or subtract monomials, followContinue reading “Monomials part 1”

Multiplying Positive and Negatives numbers We can only do arithmetic in the usual way. To calculate 5(−2), we have to do 5· 2 = 10 — and then decide on the sign. Is it +10 or −10? For the answer, we have the following Rule of Signs. Rules of signs Like signs produce a positiveContinue reading “Relative numbers (part 2)”

## Factor and Coefficient

### What is Factor?

Each combination of the constants and variables, which form a term, is called a

**Factor**.**For examples**(i) 7, x and 7x are factors of 7x, in which

7 is constant (numerical) factor and x is variable (literal) factor.(ii) In

**&ndash5x****2y**, the numerical factor is &ndash5 and literal factors are : x, y, xy, x2 and x2y.²

**Coefficient :**Any factor of a term is called the coefficient of the remaining term.

**For example :**(i) In

**7x**7 is coefficient of x(ii) In

**&ndash5x****2y**5 is coefficient of &ndashx2y &ndash5 is coefficient of x2y.**Ex. 1**Write the coefficient of :coefficient of x0 is 7.

**Ø**The greatest power (exponent) of the terms of a polynomial is called degree of the polynomial.

**For example :**(a) In polynomial 5×2 &ndash 8×7 + 3x :

(i) The power of term 5×2 = 2

(ii) The power of term &ndash8×7 = 7

Since, the greatest power is 7, therefore degree of the polynomial 5×2 &ndash 8×7 + 3x is 7

(b) The degree of polynomial :

(iii) 2m &ndash 7m8 + m13 is 13 and so on.

**v****EXAMPLES****v****Ex.2**Find which of the following algebraic expression is a polynomial.Since, the power of the first term () is , which is not a whole number.

Since, the exponent of the second term is

1/3, which in not a whole number. Therefore, the given expression is not a polynomial.**Ex.3**Find the degree of the polynomial :**Sol.**(i) Since the term with highest exponent (power) is 8×7 and its power is 7.The degree of given polynomial is 7.

(ii) The highest power of the variable is 15

**(A)**__Based on degree :__If degree of polynomial is

x 3 + 3x 2 &ndash7x+8, 2x 2 +5x 3 +7,

x 4 + y 4 + 2x 2 y 2 , x 4 + 3,&hellip

**(B)**__Based on Terms :__If number of terms in polynomial is

2 + 7y 6 , y 3 + x 14 , 7 + 5x 9 ,&hellip

x 3 &ndash2x + y, x 31 +y 32 + z 33 ,&hellip..

**Note**: (1) Degree of constant polynomials(

**Ex.**5, 7, &ndash3, 8/5, &hellip) is zero.(2) Degree of zero polynomial (zero = 0

= zero polynomial) is not defined.If a polynomial has only one variable then it is called polynomial in one variable.

**Ex.**P(x) = 2x 3 + 5x &ndash 3 Cubic trinomialQ(x) = 7x 7 &ndash 5x 5 &ndash 3x 3 + x + 3 polynomial of

S(t) = t 2 + 3 Quadratic Binomial

for quadratic ax 2 + bx + c, a ¹ 0

for cubic ax 3 + bx 2 + cx + d, a ¹ 0

(i) Remainder obtained on dividing polynomial p(x) by x &ndash a is equal to p(a) .

(ii) If a polynomial p(x) is divided by (x + a) the remainder is the value of p(x) at x = &ndasha.

(iii) (x &ndash a) is a factor of polynomial p(x) if p(a) = 0

(iv) (x + a) is a factor of polynomial p(x) if p(&ndasha) = 0

(v) (x &ndash a)(x &ndash b) is a factor of polynomial p(x),

**v****EXAMPLES****v****Ex.4**Find the remainder when 4×3 &ndash 3×2 + 2x &ndash 4 is divided by**Sol.**Let p(x) = 4×3 &ndash 3×2 + 2x &ndash 4(a) When p(x) is divided by (x &ndash 1), then by remainder theorem, the required remainder will be p(1)

(b) When p(x) is divided by (x + 2), then by remainder theorem, the required remainder will be p (&ndash2).

(c) When p(x) is divided by, then by remainder theorem, the required remainder will be

**Ø**For a polynomial f(x) = 3×2 &ndash 4x + 2.To find its value at x = 3

replace x by 3 everywhere.

So, the value of f(x) = 3×2 &ndash 4x + 2 at x = 3 is

Similarly, the value of polynomial

(ii) at x = 0 is f(0) = 3(0)2 &ndash 4(0) + 2

**Ex.5**Find the value of the polynomial 5x &ndash 4×2 + 3 at:**Sol.**Let p(x) = 5x &ndash 4×2 + 3.(i) At x = 0, p(0) = 5 × 0 &ndash 4 × (0)2 + 3

**Ø**If for x = a, the value of the polynomial p(x) is 0 i.e., p(a) = 0 then x = a is a zero of the polynomial p(x).**For example :**(i) For polynomial p(x) = x &ndash 2 p(2) = 2 &ndash 2 = 0

x = 2 or simply 2 is a zero of the polynomial

(ii) For the polynomial g(u) = u2 &ndash 5u + 6

g(3) = (3)2 &ndash 5 × 3 + 6 = 9 &ndash 15 + 6 = 0

3 is a zero of the polynomial g(u)

Also, g(2) = (2)2 &ndash 5 × 2 + 6 = 4 &ndash 10 + 6 = 0

2 is also a zero of the polynomial

(a) Every linear polynomial has one and only one zero.

(b) A given polynomial may have more than one zeroes.

(c) If the degree of a polynomial is n the largest number of zeroes it can have is also n.

**For example**:If the degree of a polynomial is 5, the polynomial can have at the most 5 zeroes if the degree of a polynomial is 8 largest number of zeroes it can have is 8.

(d) A zero of a polynomial need not be 0.

**For example :**If f(x) = x2 &ndash 4,Here, zero of the polynomial f(x) = x2 &ndash 4 is 2 which itself is not 0.

(e) 0 may be a zero of a polynomial.

**For example :**If f(x) = x2 &ndash x,Here 0 is the zero of polynomial

**v****EXAMPLES****v****Ex.6**Verify whether the indicated numbers are zeroes of the polynomial corresponding to them in the following cases :**Sol.**(i) p(x) = 3x + 1x = &ndash is a zero of p(x) = 3x + 1.

and, p(2) = (2 + 1) (2 &ndash 2) = 3 × 0 = 0

x = &ndash1 and x = 2 are zeroes of the given polynomial.

x = 0 is a zero of the given polynomial

x = &ndash is a zero of the given polynomial.

x = is not a zero of the given polynomial.

**Ex.7**Find the zero of the polynomial in each of the following cases :**Sol.**To find the zero of a polynomial p(x) means to solve the polynomial equation p(x) = 0.(i) For the zero of polynomial p(x) = x + 5

x = &ndash5 is a zero of the polynomial

p(x) = x + 5.x = is a zero of p(x) = 2x + 5.

Let us consider linear polynomial ax + b. The graph of y = ax + b is a straight line.

**For example :**The graph of y = 3x + 4 is a straight line passing through (0, 4) and (2, 10).(i) Let us consider the graph of y = 2x &ndash 4 intersects the x-axis at x = 2. The zero 2x &ndash 4 is 2. Thus, the zero of the polynomial 2x &ndash 4 is the x-coordinate of the point where the graph y = 2x &ndash 4 intersects the x-axis.

(ii) A general equation of a linear polynomial is

ax + b. The graph of y = ax + b is a straight line which intersects the x-axis at .Zero of the polynomial ax + b is the x-coordinate of the point of intersection of the graph with x-axis.

(iii) Let us consider the quadratic polynomial

x2 &ndash 4x + 3. The graph of x2 &ndash 4x + 3 intersects the x-axis at the point (1, 0) and (3, 0). Zeroes of the polynomial x2 &ndash 4x + 3 are the

x-coordinates of the points of intersection of the graph with x-axis.The shape of the graph of the quadratic polynomials is È and the curve is known as parabola.

(iv) Now let us consider one more polynomial

&ndashx2 + 2x + 8. Graph of this polynomial intersects the x-axis at the points

(4, 0), (&ndash2, 0). Zeroes of the polynomial &ndashx2 + 2x + 8 are the x-coordinates of the points at which the graph intersects the x-axis. The shape of the graph of the given quadratic polynomial is Ç and the curve is known as parabola.The zeroes of a quadratic polynomial

ax 2 + bx + c he x-coordinates of the points where the graph of y = ax 2 + bx + c intersects the x-axis.**Cubic polynomial :**Let us find out geometrically how many zeroes a cubic has.Let consider cubic polynomial

The graph of the cubic equation intersects the

x-axis at three points (1, 0), (2, 0) and (3, 0). Zeroes of the given polynomial are the

x-coordinates of the points of intersection with the x-axis.The cubic equation x3 &ndash x2 intersects the x-axis at the point (0, 0) and (1, 0). Zero of a polynomial x3 &ndash x2 are the x-coordinates of the point where the graph cuts the x-axis.

Zeroes of the cubic polynomial are 0 and 1.

Cubic polynomial has only one zero.

**In brief :**A cubic equation can have 1 or 2 or 3 zeroes or any polynomial of degree three can have at most three zeroes.**Remarks :**In general, polynomial of degree n, the graph of y = p(x) passes x-axis at most at n points. Therefore, a polynomial p(x) of degree n has at most n zeroes.**v****EXAMPLES****v****Ex.8**Which of the following correspond to the graph to a linear or a quadratic polynomial and find the number of zeroes of polynomial.**Sol.**(i) The graph is a straight line so the graph is of a linear polynomial. The number of zeroes is one as the graph intersects the x-axis at one point only.(ii) The graph is a parabola. So, this is the graph of quadratic polynomial. The number of zeroes is zero as the graph does not intersect the x-axis.

(iii) Here the polynomial is quadratic as the graph is a parabola. The number of zeroes is one as the graph intersects the x-axis at one point only (two coincident points).

(iv) Here, the polynomial is quadratic as the graph is a parabola. The number of zeroes is two as the graph intersects the x-axis at two points.

(v) The polynomial is linear as the graph is straight line. The number of zeroes is zero as the graph does not intersect the x-axis.

(vi) The polynomial is quadratic as the graph is a parabola. The number of zeroes is 1 as the graph intersects the x-axis at one point (two coincident points) only.

(vii)The polynomial is quadratic as the graph is a parabola. The number of zeroes is zero, as the graph does not intersect the x-axis.

(viii) Polynomial is neither linear nor quadratic as the graph is neither a straight line nor a parabola is one as the graph intersects the x-axis at one point only.

(ix) Here, the polynomial is quadratic as the graph is a parabola. The number of zeroes is one as the graph intersects the x-axis at one point only (two coincident points).

(x) The polynomial is linear as the graph is a straight line. The number of zeroes is one as the graph intersects the x-axis at only one point.

## 10.6: Divide Monomials (Part 1) - Mathematics

Review and practice for Final Exam by using the Final Exam study guide.

Review and practice for Final Exam by using the Final Exam study guide.

Assess students' level of understanding of important concepts learned during the school year.

Final Exam for Periods 1-2

Assess students' level of understanding of important concepts learned during the school year.

Communicate to students the results of their Final Exam..

13.3 The fundamental counting principle

By the end of this lesson, students will be able to count the number of choices that can be made from sets

Quiz on Sections 13.1-13.3 for Period 3

Assess students' knowledge of the probability concepts learned in Sections 13.1-13.3

Quiz on Sections 13.1-13.3 for Periods 1/2 Review for Final Exam

Review for the Keystone Exam

Review problems on the Predictive Keystone Exam given on Friday 05/16/ 2015

Complete problems #7-20 on a new Review packet

Keystone Exam in Algebra 1

Answer students' questions on the Keystone Exam based on their needs.

Keystone Exam in Algebra 1

13.2 Counting the elements of sets (pp. 654-660)

By the end of this lesson, students will be able to 1) find the union and intersection of sets 2) count the elements of sets 3) apply the addition of probability principle

4.3 Introduction to probability

Find the experimental probability that an event will occur.

Practice for the Keystone Exam

Use "usatestprep" to prepare students for the Keystone Exam

Practice for the Keystone Exam

Use "usatestprep" to prepare students for the Keystone Exam

13.1 Theoretical probability (pp. 648-653)

1) list or describe the sample space of an experiment 2) find the theoretical probability of a favorable outcome

13.2 Counting the elements of sets (pp. 654-660)

1) find the union and intersection of sets 2) count the elements of sets 3) apply the addition of probability principle

Review and practice on Sections 12.6-12.7

Use the tangent, sine and cosine ratios to solve problems

Review for Test on Chapter 12

Complete problems # 1-14 of Pretest during class

For Period 1-2: p.643: # 2-30 even For Period 3: Complete problems 15-25 of Pretest.

Practice for the Keystone Exam.

Review diverse concepts learned during the school year.

Assess students' knowledge of the concepts learned from Chapter 12.

Keystone Exam preparation

Use "usatestprep" to prepare students for the Keystone Exam.

1) Define and use the equations of circles 2) Use the coordinate plane to investigate the diagonals of a rectangle and the midsegment of a triangle

Review and practice on the distance, midpoint and circle formulas.

Use the distance, midpoint, and circle formulas to solve problems.

Identify and use the tangent ratio in a right triangle.

12.6 The tangent function (Continued)

Find unknown side and angle measures in right triangles.

12.7 The sine and cosine functions

1) Define the sine and cosine ratios in a right triangle 2) Find unknown side and angle measures in right triangles.

12.3 The Pythagorean theorem (pp. 591-597)

Find the side length of a right triangle given the length of its other two sides.

12.3 The Pythagorean theorem (Continued)

Apply the Pythagorean theorem to real-world problems.

Second administration of the SLO Quiz

SLO Quiz + Review for Quiz on S. 12.1-12.3

Assess students' knowledge

12.1 Operations with radicals (pp. 576-582)

1) Identify and estimate square root 2) Define and write square root in simplest radical forms.

p.581: # 24-45 multiples of 3

12.1 Operations with radicals (Continued)

1) Write square-root expressions in simplest radical form 2) Perform mathematical operations with radicals

pp.581-582: 48-78 multiples of 3

Review for the Benchmark Test

Review the following concepts: slope, linear equations, linear functions.

Review Chapter 5 Test on page 271

12.2 Square root functions & radical equations (Continued)

By the end of this lesson, students will be able to solve equations by using radicals.

p. 589: # 30-36 even, 40, 42, 50, 52, 54-57

11.2 Rational expressions and functions

Define and illustrate the use of rational expressions and functions.

11.2 Rational expressions and functions (Continued)

Graph non-trivial rational functions

11.3 Simplifying rational expressions.

Factor the numerator and the denominator to simplify rational expressions

11.3 Simplifying rational expressions (Continued)

1) State the restrictions on the variable of the simplified rational expression 2) Extend simplification techniques to other algebraic fractions

Assess students' knowledge of rational functions and simplification of rational functions.

Students will be able to define and use two different forms of inverse variation to study real-world situations

11.2 Rational expressions and functions

Students will be able to define and illustrate the use of rational expressions and functions.

10.5 The quadratic formula (Continued)

Solve quadratic equations by using the quadratic formula

10.6 Graphing quadratic inequalities (pp. 511-515)

By the end of this lesson, students will be able to solve and graph quadratic inequalities and test solution regions.

For Periods 1-2: p.514: # 12-33 multiples of 3 For Period 3: Complete the first twelve problems on Review packet.

Review and practice for test on Ch 10

Review all concepts/formulas learned in Ch 10

Assess students' knowledge of quadratic functions.

Students will be able to define and use two different forms of inverse variation to study real-world situations

Review and practice for Test on Chapter 9

Prepare students for Test on Chapter 9

Assess students' knowledge of polynomials and factoring techniques

10.4 Solving equations of the form x 2 +bx+c=0 (pp.498-503)

By the end of this lesson, students will be able to solve quadratic equations by completing the square or by factoring.

10.4 Solving equations of the form x 2 +bx+c=0 (pp.498-503)

By the end of this lesson, students will be able to solve quadratic equations by completing the square or by factoring.

10.5 The quadratic formula (pp.506-510)

By the end of this lesson, students will be able to evaluate the discriminant to determine how many roots a quadratic equation has and whether it can be factored..

10.1 Graphing parabolas (pp. 480-485)

Discover how adding a constant to the parent function y = x 2 affects the graph of the function.

10.1 Graphing parabolas (Continued)

Use the zeros of a quadratic function to find the vertex of the graph of the function.

pp. 484-485: # 32-40 even, 42-45

10.2 Solving equations by using the square root (pp. 486-491)

By the end of this lesson, students will be able to solve equation of the form ax 2 = k..

10.2 Solving equations by using the square root (continued)

Solve equation of the form ax 2 = k., where x is replaced by an algebraic expression.

10.3 Completing the square (pp. 492-497)

1) Form a perfect-square trinomial from a given quadratic binomial 2) write a given quadratic function in Vertex form.

For periods 1-2: p.496: 18-42 even for Period 3: complete the first 15 problems on Pretest on loose leaf with your work!

9.7 Factoring quadratic trinomials (pp. 458-463)

1) Factor quadratic trinomials by using Guess-and-Check 2) Factor quadratic trinomials by using the grouping process.

pp.462-463: 10-24 even, 27-36 multiples of 3

9.7 Factoring quadratic trinomials (Continued)

Students will be able to factor trinomials by using the methods learned in Section 9.7

9.8 Solving equations by factoring (pp. 464-469) + Quiz on Sections 9.3-9.6

1) Find the zeros of a function 2) Solve equations by factoring.

Assess students' knowledge of the concepts learned in chapter 7

Find products of binomials by using the FOIL method and mentally simplify special products of binomials.

9.4 Polynomial Functions (pp. 443-447)

1) Define polynomial functions 2) solve problems involving polynomial functions

pp. 446-447: # 9-18 multiples of 3, 19-25

9.5 Common Factors (pp.448-451)

By the end of this lesson, students will be able to factor a polynomial by using the greatest common factor

9.5 Factoring special polynomials

1) Factor perfect-square trinomials 2) Factor the difference of two squares.

7.6 Classic puzzles in two variables

Solve traditional math puzzles in two variables

Review and Practice for Test on Chapter 7

Students practice class work as a preparation for Chapter 7 Test.

Test on Chapter 7 postponed until Monday due to extreme weather.

Students complete work on systems of linear equations from the Keystone preparation workbook.

7.4 Consistent & inconsistent Systems

Identify consistent, inconsistent, dependent & independent systems of linear equations

p. 342: # 12-27 multiples of 3, 30, 31, 38

* 7.5 Systems of inequalities (pp. 345-352)

1) Graph the solution to a linear inequality 2) graph the solution to a system of linear inequalities

7.5 Systems of inequalities (Continued)

By the end of this lesson, students will be able to graph the solution to a system of linear inequalities

7.5 Systems of inequalities (Continued)

By the end of this lesson, students will be able to graph the solution to a system of linear inequalities

Complete Practice worksheet to be distributed in class: Problems 1-9

7.6 Classic puzzles in two variables

Solve traditional math puzzles in two variables

For Periods 1-2: pp. 358-359: # 11, 13-16 for Period 3: Complete the first 15 problems on Pretest.

7.1 Graphing systems of equations

Graph systems of equations and solve graphically systems of equations

7.2 The substitution method (pp.326-330)

Find the exact solution to a system of linear equations by using the substitution method

7.3 The elimination method (pp 331-337)

Use the elimination method to solve a system of equations

Philadelphia School District Test

Practice on Solving systems of linear equations

Use the graphing, substitution or elimination methods to solve systems of linear equations

Mid-Year Exam Review packet: # 66-79

Mid-Year Exam for Period 1/2

Exam for Periods 1/2 For Period 3: Practice for Benchmark 2 Test

Mid-Year Exam for Period 3

Exam for Period 3 For 1/2: Practice for Benchmark 2 Test

7.1 Graphing systems of equations

By the end of this lesson, students will be able to graph systems of equations and solve graphically systems of equations

6.5 Absolute-value equations and inequalities.

Use technology to check the solutions to Absolute-value equations and inequalities

Complete the first 20 problems of The Review for Mid-Year Exam packet

Complete problems # 25-40 of The Review for Mid-Year Exam packet

Complete problems # 45-60 of The Review for Mid-Year Exam packet

6.3 Compound inequalities (Continued)

By the end of this lesson, students will be able to use compound inequalities to solve problems

Quiz, and practice on solving compound inequalities

6.4 Absolute value functions

By the end of this lesson, students will be able to learn by exploring the features of absolute value functions, and the basic transformations of the absolute value functions

pp.298: # 12-45 multiples of 3

6.5 Absolute-value equations and inequalities

By the end of this lesson, students will be able to solve absolute-value equations and inequalities, and express the solution as a range of values on the number line

## Example problems for broad differentiation

**Broad differentiation example problem 1:**

Find the third derivative value of the given function f(x) = 5x 4 - 7x 2 + 61x - 9**Solution:**

Given function is f(x) = 5x 4 - 7x 2 + 61x - 9

Differentiate the given function with respect to x, we get

f'(x) = 20x 3 - 14x + 61

Again differentiate the given function for finding the second derivative, we get

f''(x) = 60x 2 - 14

For finding the third derivative, again differentiate the given function

f'''(x) = 120x**Answer:**

The final answer is 120x**Broad differentiation example problem 2:**

Find the third derivative value of the given function f(x) = 52x 4 - 17x 2 + 33x**Solution:**

Given function is f(x) = 52x 4 - 17x 2 + 33x

Differentiate the given function with respect to x, we get

f'(x) = 208x 3 - 34x + 33

Again differentiate the given function for finding the second derivative, we get

f''(x) = 624x 2 - 34

For finding the third derivative, again differentiate the given function

f'''(x) = 1248x**Answer:**

The final answer is 1248xBroad differentiation example problem 3:

Find the fourth derivative value of the given function f(x) = 10x 4 + 4x 3 + 14x 2 - x**Solution:**

Given function is f(x) = 10x 4 + 4x 3 + 14x 2 - x

Differentiate the given function with respect to x, we get

f'(x) = 40x 3 + 12x 2 + 28x - 1

Again differentiate the given function for finding the second derivative, we get

f''(x) = 120x 2 + 24x + 28

For finding the third derivative, again differentiate the given function

f'''(x) = 240x + 24

For fourth derivative, we get

f''''(x) = 240**Answer:**

The final answer is 240

## Formula for how to rewrite radicals:

**Example for how to rewrite radicals:****Rewrite radical****( `sqrt (1225)` )****Given:****( `sqrt (1225)` )****Solution:**Given question says, radical (1225),

When, we take**radical for 1225**, we obtain**25*49.**

Because,**( `^nsqrt (ab)` ) = ( ` ^nsqrt (a)` )****( ` ^nsqrt (b)` )**

`sqrt (1225)`**= `sqrt (25)` `sqrt (49)`**

**=****`sqrt(5)`***** `sqrt(7)`**

`sqrt (1225)`**= `35`**

Thus, we can do**how to rewrite radicals**in the prefered way.**Example for how to rewrite radicals:****Rewrite radical****( `^3sqrt (512)` )****Given:****( `^3sqrt (512)` )****Solution:**Given question says, cubic root of (512),

When, we take**cubic root for 512,**we obtain**8.**

Because, `8*8*8 = 512`

`^3sqrt(512)` ` =` `8^3`

Therefore,**cubic root for (512**) = 8^3

Thus,we can do**how to rewrite radicals**in the prefered way**Example for how to rewrite radicals:****Rewrite radical****`^3sqrt (729)`****Given:****( `^3sqrt (729)` )****Solution:**Given question says,**cubic root of (729)**,

When, we take cubic root for 729, we obtain 9.

Because, `9*9*9 = 729`

`^3sqrt(729) = 9^3`

Therefore,**cubic root of (729) = 9^3****Example for how to rewrite radicals:****Rewrite radical****`^4sqrt (1296)`****Given:****( `^4sqrt (1296)` )****Solution:**Given question says,**fourth root of (1296)**,

When, we take fourth radical for 1296, we obtain 6.

Because,**`6*6*6*6 = 1296`**

**`^4sqrt(1296 )= 6^4`**

Therefore,**fourth root of (1296) = 6^4****Example for how to rewrite radicals: Rewrite radical****`^5sqrt (3125)`****Given:****( `^5sqrt (3125)` )****Solution:**Given question says,**fifth root of (3125)**,

When, we take Fifth radical for 3125, we obtain 8.

Because, `5*5*5*5*5 = 3125`

`^5sqrt(3125) = 5^5`

Therefore,**fifth root of (3125) = 5**

## Watch the video: division of monomials part 1 (January 2022).