## 7.3: Factoring trinomials of the form ax2 + bx + c - Mathematics

· Factor trinomials with a leading coefficient of 1.

· Factor trinomials with a common factor.

· Factor trinomials with a leading coefficient other than 1.

A **polynomial** with three terms is called a **trinomial**. Trinomials often (but not always!) have the form *x* 2 + *bx* + *c*. At first glance, it may seem difficult to factor trinomials, but you can take advantage of some interesting mathematical patterns to factor even the most difficult-looking trinomials.

So, how do you get from 6*x* 2 + 2*x* – 20 to (2*x* + 4)(3*x* −5)? Let’s take a look.

**Factoring Trinomials: x 2 + bx + c**

Trinomials in the form *x* 2 + *bx* + *c* can often be factored as the product of two **binomials**. Remember that a binomial is simply a two-term polynomial. Let’s start by reviewing what happens when two binomials, such as (*x* + 2) and (*x* + 5), are multiplied.

**Multiply ( x + 2)(x + 5).**

Use the FOIL method to multiply binomials.

*x* 2 + 5*x* + 2*x* +10

Then combine like terms 2*x* and 5*x*.

Factoring is the reverse of multiplying. So let’s go in reverse and factor the trinomial *x* 2 + 7*x* + 10. The individual terms *x* 2 , 7*x*, and 10 share no common factors. So look at rewriting *x* 2 + 7*x* + 10 as *x* 2 + 5*x* + 2*x* + 10.

And, you can group pairs of factors: (*x* 2 + 5*x*) + (2*x* + 10)

Factor each pair: *x*(*x* + 5) + 2(*x* + 5)

Then factor out the common factor *x* + 5: (*x* + 5)(*x* + 2)

Here is the same problem done in the form of an example:

**Factor x 2 + 7x +10.**

*x* 2 + 5*x* + 2*x* +10

Rewrite the middle term 7*x* as 5*x* + 2*x*.

*x* ( *x* + 5) + 2( *x* + 5 )

Group the pairs and factor out the common factor *x* from the first pair and 2 from the second pair.

Factor out the common factor

How do you know how to rewrite the middle term? Unfortunately, you can’t rewrite it just any way. If you rewrite 7*x* as 6*x* + *x*, this method won’t work. Fortunately, there's a rule for that.

**Factoring Trinomials in the form x 2 + bx + c**

To factor a trinomial in the form *x* 2 + *bx* + *c*, find two integers, *r* and *s*, whose product is *c* and whose sum is *b*.

Rewrite the trinomial as *x* 2 + *rx* + *sx* + *c* and then use grouping and the distributive property to factor the polynomial. The resulting factors will be (*x* + *r*) and (*x* + *s*).

For example, to factor *x* 2 + 7*x* +10, you are looking for two numbers whose sum is 7 (the coefficient of the middle term) and whose product is 10 (the last term).

Look at factor pairs of 10: 1 and 10, 2 and 5. Do either of these pairs have a sum of 7? Yes, 2 and 5. So you can rewrite 7*x* as 2*x* + 5*x*, and continue factoring as in the example above. Note that you can also rewrite 7*x* as 5*x* + 2*x*. Both will work.

Let’s factor the trinomial *x* 2 + 5*x* + 6. In this polynomial, the *b* part of the middle term is 5 and the *c* term is 6. A chart will help us organize possibilities. On the left, list all possible factors of the *c* term, 6 on the right you'll find the sums.

**Factors whose product is 6**

** Sum of the factors**

There are only two possible factor combinations, 1 and 6, and 2 and 3. You can see that 2 + 3 = 5. So 2*x* + 3*x* = 5*x*, giving us the correct middle term.

**Factor x 2 + 5x + 6.**

*x* 2 + 2*x* + 3*x* + 6

Use values from the chart above. Replace 5*x* with 2*x* + 3*x*.

(*x* 2 + 2*x*) + (3*x* + 6)

*x* (*x* + 2) + (3*x* + 6)

Factor *x* out of the first pair of terms.

*x* (*x* + 2) + 3(*x* + 2)

Factor 3 out of the second pair of terms.

Note that if you wrote *x* 2 + 5*x* + 6 as *x* 2 + 3*x* + 2*x* + 6 and grouped the pairs as (*x* 2 + 3*x*) + (2*x* + 6) then factored, *x*(*x* + 3) + 2(*x* + 3), and factored out *x* + 3, the answer would be (*x* + 3)(*x* + 2). Since multiplication is commutative, the order of the factors does not matter. So this answer is correct as well they are equivalent answers.

Finally, let’s take a look at the trinomial *x* 2 + *x* – 12. In this trinomial, the *c* term is −12. So look at all of the combinations of factors whose product is −12. Then see which of these combinations will give you the correct middle term, where *b* is 1.

**Factors whose product is** **−** **12**

** Sum of the factors**

**4** • **−** **3 =** **−** **12**

There is only one combination where the product is −12 and the sum is 1, and that is when *r* = 4, and *s* = −3. Let’s use these to factor our original trinomial.

**Factor x 2 + x – 12**

*x* 2 + 4*x* + − 3*x* – 12

Rewrite the trinomial using the values from the chart above. Use values *r* = 4 and *s* = − 3.

(*x* 2 + 4*x*) + ( − 3*x* – 12)

*x* (*x* + 4) + ( − 3*x* – 12)

Factor x out of the first group.

*x* (*x* + 4) – 3(*x* + 4)

Factor − 3 out of the second group.

In the above example, you could also rewrite *x* 2 + *x* – 12 as *x* 2 – 3*x* + 4*x* – 12 first. Then factor *x*(*x* – 3) + 4(*x* – 3), and factor out (*x* – 3) getting (*x* – 3 )(*x* + 4). Since multiplication is commutative, this is the same answer.

Factoring trinomials is a matter of practice and patience. Sometimes, the appropriate number combinations will just pop out and seem so obvious! Other times, despite trying many possibilities, the correct combinations are hard to find. And, there are times when the trinomial cannot be factored.

While there is no foolproof way to find the right combination on the first guess, there are some tips that can ease the way.

**Tips for Finding Values that Work**

When factoring a trinomial in the form *x* 2 + *bx* + *c*, consider the following tips.

Look at the *c* term first.

o If the *c* term is a positive number, then the factors of *c* will both be positive or both be negative. In other words, *r* and *s* will have the same sign.

o If the *c* term is a negative number, then one factor of *c* will be positive, and one factor of *c* will be negative. Either *r* or *s* will be negative, but not both.

Look at the *b* term second.

o If the *c* term is positive and the *b* term is positive, then both *r* and *s* are positive.

o If the *c* term is positive and the *b* term is negative, then both *r* and *s* are negative.

o If the *c* term is negative and the *b* term is positive, then the factor that is positive will have the greater absolute value. That is, if *|r| > |s|,* then *r* is positive and *s* is negative.

o If the *c* term is negative and the *b* term is negative, then the factor that is negative will have the greater absolute value. That is, if *|r| > |s|,* then *r* is negative and *s* is positive.

After you have factored a number of trinomials in the form *x* 2 + *bx* + *c*, you may notice that the numbers you identify for *r* and *s* end up being included in the factored form of the trinomial. Have a look at the following chart, which reviews the three problems you have seen so far.

** x**

**2**

**+ 7**

*x*+ 10** x**

**2**

**+ 5**

*x*+ 6** x**

**2**

**+**

*x*- 12** r**

**and**

*s*valuesNotice that in each of these examples, the *r* and *s* values are repeated in the factored form of the trinomial.

So what does this mean? It means that in trinomials of the form *x* 2 + *bx* + *c* (where the coefficient in front of *x* 2 is 1), if you can identify the correct *r* and *s* values, you can effectively skip the grouping steps and go right to the factored form. You may want to stick with the grouping method until you are comfortable factoring, but this is a neat shortcut to know about!

Jess is trying to use the grouping method to factor the trinomial *v* 2 – 10*v* + 21. How should she rewrite the central *b* term, − 10*v*?

Incorrect. Because the *c* term is positive and the *b* term is negative, both terms should be negative. (Notice that using the integers 7 and 3, 7 + 3 = +10, so this would provide the term 10*v* instead of − 10*v*.) The correct answer is − 7*v* – 3*v*.

Correct. Because the *c* term is positive and the *b* term is negative, both terms should be negative. Check: using the integers − 7 and − 3, − 7 + − 3 = − 10 and − 7 • − 3 = 21, so this provides both terms − 10*v* and 21 correctly.

Incorrect. Because the *c* term is positive and the *b* term is negative, both terms should be negative. (Notice that using the integers − 7 and 3, − 7 + 3 = − 4 and − 7 • 3 = − 21, so this would provide − 4v instead of − 10*v* and − 21 instead of 21.) The correct answer is

Incorrect. Because the *c* term is positive and the *b* term is negative, both terms should be negative. (Notice that using the integers 7 and − 3, 7 + − 3 = 4 and 7 • − 3 = − 21, so this would provide 4*v* instead of − 10*v* and − 21 instead of 21.) The correct answer is

**Identifying Common Factors**

Not all trinomials look like *x* 2 + 5*x* + 6, where the coefficient in front of the *x* 2 term is 1. In these cases, your first step should be to look for common factors for the three terms.

**Factor out Common Factor**

2*x* 2 + 10*x* + 12

− 5*a* 2 − 15*a* − 10

*c* 3 – 8*c* 2 + 15c

*c* (*c* 2 – 8*c* + 15)

*c* (*c* – 5)(*c* – 3)

*y* 4 – 9*y* 3 – 10*y* 2

y 2 (*y* – 10)(*y* + 1)

Notice that once you have identified and pulled out the common factor, you can factor the remaining trinomial as usual. This process is shown below.

**Factor 3 x 3 – 3x 2 – 90x.**

3(*x* 3 – *x* 2 – 30*x*)

Since 3 is a common factor for the three terms, factor out the 3.

3*x*(*x* 2 – *x* – 30)

*x* is also a common factor, so factor out *x*.

3*x*(*x* 2 – 6*x* + 5*x* – 30)

Now you can factor the trinomial

*x* 2 – *x* – 30. To find *r* and *s*, identify two numbers whose product is − 30 and whose sum is − 1.

The pair of factors is − 6 and 5. So replace – *x* with − 6*x* + 5*x*.

3*x*[(*x* 2 – 6*x*) + (5x – 30)]

Use grouping to consider the terms in pairs.

3*x*[(*x*(*x* – 6) + 5(*x* – 6)]

Factor *x* out of the first group and factor 5 out of the second group.

3*x*(*x* – 6)(*x* + 5)

Then factor out *x* – 6.

3*x*(*x* – 6)(*x* + 5)

**Factoring Trinomials: ax 2 + bx + c**

The general form of trinomials with a leading coefficient of *a* is *ax* 2 + *bx* + *c*. Sometimes the factor of *a* can be factored as you saw above this happens when *a* can be factored out of all three terms. The remaining trinomial that still needs factoring will then be simpler, with the leading term only being an *x* 2 term, instead of an *ax* 2 term.

However, if the coefficients of all three terms of a trinomial don’t have a common factor, then you will need to factor the trinomial with a coefficient of something other than 1.

**Factoring Trinomials in the form ax 2 + bx + c**

To factor a trinomial in the form *ax* 2 + *bx* + *c*, find two integers, *r* and *s*, whose sum is *b* and whose product is *ac.* Rewrite the trinomial as *ax 2* + *rx* + *sx* + *c* and then use grouping and the distributive property to factor the polynomial.

This is almost the same as factoring trinomials in the form *x* 2 + *bx* + *c*, as in this form *a* = 1. Now you are looking for two factors whose product is *a* • *c*, and whose sum is *b*.

Let’s see how this strategy works by factoring 6*z* 2 + 11*z* + 4.

In this trinomial, *a* = 6, *b* = 11, and *c* = 4. According to the strategy, you need to find two factors, *r* and *s*, whose sum is *b* (11) and whose product is *ac* (or 6 • 4 = 24). You can make a chart to organize the possible factor combinations. (Notice that this chart only has positive numbers. Since *ac* is positive and *b* is positive, you can be certain that the two factors you're looking for are also positive numbers.)

## Key Concepts

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## Method 1: Factoring x2 + bx + c

1. Learn FOIL multiplication. You might have already learned the FOIL method, or "First, Outside, Inside, Last," to multiply expressions like (x+2)(x+4). It's useful to know how this works before we get to factoring:

- Multiply the First terms: (x+2)(x+4) = x2 + __
- Multiply the Outside terms: (x+2)(x+4) = x2+4x + __
- Multiply the Inside terms: (x+2)(x+4) = x2+4x+2x + __
- Multiply the Last terms: (x+2)(x+4) = x2+4x+2x+8
- Simplify: x2+4x+2x+8 = x2+6x+8

2. Understand factoring. When you multiply two binomials together in the FOIL method, you end up with a trinomial (an expression with three terms) in the form ax2+bx+c, where a, b, and c are ordinary numbers. If you start with an equation in the same form, you can factor it back into two binomials.

- If the equation isn't written in this order, move the terms around so they are. For example, rewrite 3x - 10 + x2 as x2 + 3x - 10.
- Because the highest exponent is 2 (x2, this type of expression is "quadratic."

3. Write a space for the answer in FOIL form. For now, just write (__ __)(__ __) in the space where you'll write the answer. We'll fill this out as we go.

4. Fill out the First terms. For simple problems, where the first term of your trinomial is just x2, the terms in the First position will always be x and x. These are the factors of the term x2, since x times x = x2.

- We'll cover more complicated problems in the next section, including trinomials that begin with a term like 6x2 or -x2. For now, follow the example problem.

5. Use factoring to guess at the Last terms. If you go back and reread the FOIL method step, you'll see that multiplying the Last terms together gives you the final term in the polynomial (the one with no x). So to factor, we need to find two numbers that multiply to form the last term.

- In our example x2 + 3x - 10, the last term is -10.
- What are the factors of -10? What two numbers multiplied together equal -10?
- There are a few possibilities: -1 times 10, 1 times -10, -2 times 5, or 2 times -5. Write these pairs down somewhere to remember them.
- Don't change our answer yet. It still looks like this: (x __)(x __).

6. Test which possibilities work with Outside and Inside multiplication. We've narrowed the Last terms down to a few possibilities. Use trial and error to test each possibility, multiplying the Outside and Inside terms, and comparing the result to our trinomial. For example:

## Factoring Trinomials with Common Factors

It is a good practice to first factor out the GCF, if there is one. Doing this produces a trinomial factor with smaller coefficients. As we have seen, trinomials with smaller coefficients require much less effort to factor. This commonly overlooked step is worth identifying early.

**Example 6:**Factor: 12 x 2 − 27 x + 6 .**Solution:**Begin by factoring out the GCF.After factoring out 3, the coefficients of the resulting trinomial are smaller and have fewer factors.

After some thought, we can see that the combination that gives the coefficient of the middle term is 4 ( − 2 ) + 1 ( − 1 ) = − 8 − 1 = − 9 .

The factor 3 is part of the factored form of the original expression be sure to include it in the answer.

It is a good practice to consistently work with trinomials where the leading coefficient is positive.

**Example 7:**Factor: − x 2 + 2 x + 15 .**Solution:**In this example, the leading coefficient is −1. Before beginning the factoring process, factor out the −1:At this point, factor the remaining trinomial as usual, remembering to write the −1 as a factor in your final answer. Because 3 + (−5) = −2, use 3 and 5 as the factors of 15.

Answer: − 1 ( x + 3 ) ( x − 5 ) . The check is left to the reader.

**Example 8:**Factor: − 60 a 2 − 5 a + 30 .**Solution:**The GCF of all the terms is 5. However, in this case factor out −5 because this produces a trinomial factor where the leading coefficient is positive.Focus on the factors of 12 and 6 that combine to give the middle coefficient, 1.

After much thought, we find that 3 ⋅ 3 − 4 ⋅ 2 = 9 − 8 = 1 . Factor the remaining trinomial.

Answer: − 5 ( 4 a + 3 ) ( 3 a − 2 ) . The check is left to the reader.

**Try this!**Factor: 24 + 2 x − x 2 .### Video Solution

## Get Factoring Ax2 Bx C Worksheet Answer Key

Algebra and Applications P. Pathak Worksheet 3 Section 5. 3 Factoring Trinomials of the form ax2 bx c Factor completely. 1. 2x2 5x 3 27. 22x2 29x 6 2. 2x2 5x 2 28. 20z 2 7z 6 3. 2y 2 13y 20 29. 2x2 1xy 10y 2 4. 2y 2 11y 15 30. 2x2 11xy 12y 2 5. 2t2 7t 15 31. 3x2 28xy 32y 2 6. 2t2 9t 35 7. 2x2 3x 20 33. 5x2 27xy 10y 2 8. 2x2 11x 21 34. 5x2 6xy 8y 2 9. 3y 2 13y 10 35. 7x2 10xy 3y 2 10. 3x2 17x 20 36. 6x2 7xy 3y 2 11. 3y 2 17y 28 37. 2x3 5x2 12x 12. 3y 2 13y 14 38. 3x3 19x2 20x 13. 5y 2 23y 24 39. 36x3 12x2 15x 14. 5x2 12x 32 40. 6x3 10x2 4x 15. 5y 2 17y 14 41. 18x3 21x2 9x 16. 5y 2 11y 12 42. 12t3 10t2 12t 17. 4x2 25x 25 43. 12t3 22t2 6t 18. 4y 2 5y 12 44. 15t3 18t2 24t 19. 4y 2 4y 15 45. 5x3 y 10x2 y 2 15xy 3 20. 4x2 4x 35 46. 6x5 y 25x4 y 2 4x3 y 3 21. 6x2 7x 20 47. 12x4 y 3 11x3 y 4 2x2 y 5 22. 6y 2 5y 21 48. 12x3 y 3 28x2 y 4 8xy 5 23. 8y 2 14y 15 49. x3 5x2 6x 24. 8x2 6x 5 50. y 3 3y 2 2y 25. 12y 2 y 6 51. 1. 2x2 5x 3 27. 22x2 29x 6 2. 2x2 5x 2 28. 20z 2 7z 6 3. 2y 2 13y 20 29. 2x2 1xy 10y 2 4. 2y 2 11y 15 30. 2x2 11xy 12y 2 5. 2t2 7t 15 31. 3x2 28xy 32y 2 6. 2t2 9t 35 7. 2x2 3x 20 33. 5x2 27xy 10y 2 8. 2x2 11xy 12y 2 5. 2t2 7t 15 31. 3x2 28xy 32y 2 6. 2t2 9t 35 7. 2x2 3x 20 33. 5x2 27xy 10y 2 8. 2x2 11x 21 34. 5x2 6xy 8y 2 9. 3y 2 13y 10 35. 7x2 10xy 3y 2 10. 3x2 17x 20 36. 6x2 7xy 3y 2 11. 2x2 11x 21 34. 5x2 6xy 8y 2 9. 3y 2 13y 10 35. 7x2 10xy 3y 2 10. 3x2 17x 20 36. 6x2 7xy 3y 2 11. 3y 2 17y 28 37. 2x3 5x2 12x 12. 3y 2 13y 14 38. 3x3 19x2 20x 13. 5y 2 23y 24 39. 36x3 12x2 15x 14. 3y 2 17y 28 37. 2x3 5x2 12x 12. 3y 2 13y 14 38. 3x3 19x2 20x 13. 5y 2 23y 24 39. 36x3 12x2 15x 14. 5x2 12x 32 40. 6x3 10x2 4x 15. 5y 2 17y 14 41. 18x3 21x2 9x 16. 5y 2 11y 12 42. 12t3 10t2 12t 17. 5x2 12x 32 40. 6x3 10x2 4x 15. 5y 2 17y 14 41. 18x3 21x2 9x 16. 5y 2 11y 12 42. 12t3 10t2 12t 17. 4x2 25x 25 43. 12t3 22t2 6t 18. 4y 2 5y 12 44. 15t3 18t2 24t 19. 4y 2 4y 15 45. 5x3 y 10x2 y 2 15xy 3 20. 4x2 25x 25 43. 12t3 22t2 6t 18. 4y 2 5y 12 44. 15t3 18t2 24t 19. 4y 2 4y 15 45. 5x3 y 10x2 y 2 15xy 3 20. 4x2 4x 35 46. 6x5 y 25x4 y 2 4x3 y 3 21. 6x2 7x 20 47. 12x4 y 3 11x3 y 4 2x2 y 5 22. 6y 2 5y 21 48. 4x2 4x 35 46. 6x5 y 25x4 y 2 4x3 y 3 21. 6x2 7x 20 47. 12x4 y 3 11x3 y 4 2x2 y 5 22. 6y 2 5y 21 48. 12x3 y 3 28x2 y 4 8xy 5 23. 8y 2 14y 15 49. x3 5x2 6x 24. 8x2 6x 5 50. y 3 3y 2 2y 25. 12y 2 y 6 51. 1. 2x2 5x 3 27. 22x2 29x 6 2. 2x2 5x 2 28. 20z 2 7z 6 3. 2y 2 13y 20 29. 2x2 1xy 10y 2 4. 2y 2 11y 15 30. 2x2 11xy 12y 2 5. 2t2 7t 15 31. 3x2 28xy 32y 2 6. 2t2 9t 35 7. 2x2 3x 20 33. 5x2 27xy 10y 2 8. 2x2 11x 21 34. 5x2 6xy 8y 2 9. 3y 2 13y 10 35. 7x2 10xy 3y 2 10. 3x2 17x 20 36. 6x2 7xy 3y 2 11. 2x2 11xy 12y 2 5. 2t2 7t 15 31. 3x2 28xy 32y 2 6. 2t2 9t 35 7. 2x2 3x 20 33. 5x2 27xy 10y 2 8. 2x2 11x 21 34. 5x2 6xy 8y 2 9. 3y 2 13y 10 35. 7x2 10xy 3y 2 10. 3x2 17x 20 36. 6x2 7xy 3y 2 11. 3y 2 17y 28 37. 2x3 5x2 12x 12. 3y 2 13y 14 38. 3x3 19x2 20x 13. 5y 2 23y 24 39. 36x3 12x2 15x 14..

## Steps to Factorize the Trinomial of Form ax 2 + bx + c?

1. Note down the given expression and compare it with the basic expression ax 2 + bx + c.

2. Note down the product and sum terms and find the two numbers.

3. Depends on the values of two numbers, expand the given expression.

4. Factor out the common terms.

5. Finally, we will get the product of two terms which is equal to the trinomial expression.### Solved Examples on Factoring Trinomials of the Form ax 2 + bx + c

Solution:

The Given expression is 2s 2 + 9s + 10.

By comparing the given expression 2s 2 + 9s + 10 with the basic expression ax 2 + bx + c.

Here, a = 2, b = 9, and c = 10.

The sum of two numbers is p + q = b = 9 = 5 + 4.

The product of two number is p * q = a * c = 2 * 10 = 20 = 5 * 4.

From the above two instructions, we can write the values of two numbers p and q as 5 and 4.

Then, 2s 2 + 9s + 10 = 2s 2 + 5s + 4s + 20.

= 2s (s + 5) + 4 (s + 5).

Factor out the common terms.Then, 2s 2 + 9s + 10 = (2s + 4) (s + 5).

Solution:

The Given expression is 6s 2 + 7s – 3.

By comparing the given expression 6s 2 + 7s – 3 with the basic expression ax 2 + bx + c.

Here, a = 6, b = 7, and c = 3.

The sum of two numbers is p + q = b = 7 = 9 – 2.

The product of two number is p * q = a * c = 6 * 3 = 18 = 9 * 2.

From the above two instructions, we can write the values of two numbers p and q as 9 and 2.

Then, 6s 2 + 7s -3 = 6s 2 + 9s – 2s – 3.

= 6s 2 – 2s + 9s – 3.

= 2s (3s – 1) + 3(3s – 1).

Factor out the common terms.Then, 6s 2 + 7s – 3 = (3s – 1) (2s + 3).

2. Factorize the trinomial.

Solution:

The Given expression is 2x 2 + 7x + 3.

By comparing the given expression 2x 2 + 7x + 3 with the basic expression ax 2 + bx + c.

Here, a = 2, b = 7, and c = 3.

The sum of two numbers is p + q = b = 7 = 6 + 1.

The product of two number is p * q = a * c = 2 * 3 = 6 = 6 * 1.

From the above two instructions, we can write the values of two numbers p and q as 6 and 1.

Then,2x 2 + 7x + 3 = 2x 2 + 6x + x + 3.

= 2x (x + 3) + (x + 3).

Factor out the common terms.Then, 2x 2 + 7x + 3 = (x + 3) (2x + 1).

Solution:

The Given expression is 3s 2 – 4s – 4.

By comparing the given expression 3s 2 – 4s – 4 with the basic expression ax 2 + bx + c.

Here, a = 3, b = – 4, and c = – 4.

The sum of two numbers is p + q = b = – 4 = – 6 + 2.

The product of two number is p * q = a * c = 3 * (- 4) = – 12 = (- 6) * 2.

From the above two instructions, we can write the values of two numbers p and q as – 6 and 2.

Then, 3s 2 – 4s – 4 = 3s 2 – 6s + 2s – 4.

= 3s (s – 2) + 2(s – 2).

Factor out the common terms.

## Factoring Polynomials Lesson 3 Trinomials ax2 + bx + c Practice & Game PREP FREE

ax2 + bx + c. Links to my other factoring lessons can be found below. This lesson is a fun way to teach factoring polynomials guaranteed to keep kids engaged and they LOVE the competition of the game at the end.

This lesson is 100% Prep Free.

-A full Interactive PowerPoint Lesson including examples with step by step explanations. (For a sample click preview)

-20 Practice Questions worked into the lesson

-Checkpoints to determine whether students are ready to move on or need more practice (additional questions included if more practice is needed)

-Fun interactive game that includes 20 questions (For a look at the game click preview)

- Student notes pages for lesson and game

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## Introduction

Quadratic expressions may be used to model physical properties of a large bridge, the trajectory of a baseball or rocket, and revenue and profit of a business. By factoring these expressions, specific characteristics of the model can be identified. In this chapter, you will explore the process of factoring expressions and see how factoring is used to solve certain types of equations.

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- Authors: Lynn Marecek, MaryAnne Anthony-Smith, Andrea Honeycutt Mathis
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- Book title: Elementary Algebra 2e
- Publication date: Apr 22, 2020
- Location: Houston, Texas
- Book URL: https://openstax.org/books/elementary-algebra-2e/pages/1-introduction
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## If you are using algebra tiles to factor a trinomial of the form ax2 + bx + c, when would you need to bring in zero pairs? Why?

If the value of c is negative, you would need zero pairs to model the factorization of the polynomial. The x-tiles on the board determine what the constants are in the factors. The product of these constants is equal to the value of c, so you would need positive tiles on one side of the x-squared tile and negative x-tiles on the other side to have opposite signs on the constants. Opposite signs on the constants will result in a negative value for c when multiplying the factors.

The trinomial form ax2 +bx + c is a quadratic form. We need to bring in zero pairs because we need to get the roots of that quadratic equation. The roots is referred to as the solutions or the possible value of x to make the equation true

## Watch the video: Factoring Trinomials with Lead Term of 1: Examples 5-7 (January 2022).