Factoring - Polynomials - Idiot's Guides - Algebra I

Idiot's Guides: Algebra I (2015)

Part IV. Polynomials

Chapter 13. Factoring

In This Chapter

·  Factoring out a greatest common factor

·  Factoring trinomials by reversing the FOIL rule

·  Factoring polynomials produced by special multiplication patterns

Throughout algebra, we talk a great deal about opposite operations. That makes sense, because inverse operations are the key to solving equations. We undo addition by subtraction and multiplication by dividing. That strategy is successful as long as we know what to subtract, or what the divisor should be.

Sometimes, however, we can see that an expression was probably created by multiplying, but we don’t know either of the numbers that were multiplied. Without some number to use as the divisor, we can’t really divide, so we’re left to try to figure out what the numbers might have been.

If you’re looking at the number 210 and you know that it was formed by multiplying 6 times some number, it’s a simple matter to divide 210 by 6 and find that the other number is 35. On the other hand, if you’re looking at 210 and you know it was created by multiplying two numbers but you don’t know what either of the numbers was, you can only look for all the pairs of numbers that multiply to 210. You’ll often find that there are several pairs of numbers. The number 210 could be 1·210, 2·105, 3·70, 5·42, 6·35, 7·30, 10·21, or 14·15.

In this chapter, we’ll investigate a similar search, but we’ll work with polynomials rather than numbers. We’ll look at a polynomial, think of it as a product and try to find two polynomials that will produce that product. There will be a lot of trial and error involved, so patience is required, but we’ll also call on what we know about polynomial multiplication and division and on rules for exponents.

Greatest Common Factor

Your goal throughout this chapter (and in several that follow) will be to rewrite a polynomial as the product of two polynomials of lower degree. This is the process called factoring. The first type of polynomial multiplication you learned about was a monomial multiplied by a polynomial, and you did that by using the distributive property. The first type of factoring you’ll look at is rewriting a polynomial as a monomial multiplied by a polynomial. You’ll want to find out whether the polynomial you’re trying to factor could have been created by multiplying using the distributive property, and if so, what the monomial was. Once you have that, you can divide to find the other polynomial.


Factoring is the process of rewriting an integer as the product of two integers, or a polynomial as the product of two polynomials of lesser degree.

Let’s start with something simple, or at least, short: 12x + 8. Imagine a multiplication problem that uses the distributive law and that has 12x + 8 as its answer.

You need to fill in the blanks, and like looking for the factors of 210, there could be more than one possibility. The largest possible polynomial that will fit into the monomial slot in this problem is the greatest common factor of the two terms 12x and 8. That means it’s a monomial that divides both 12x and 8.

Like many other problems, this one can be broken down into smaller tasks. Look first at the coefficients and find the largest number that divides both, the largest factor they have in common. The largest number that divides both 12 and 8 is 4, because 12 = 4·3 and 8 = 4·2. Next consider the variable part of the terms. 12x contains x to the first power, but 8 is a constant, so we can either say it has no x or we can say it has x to the zero power. The greatest common factor will contain the smaller of the two powers of x, so it will have x0. The greatest common factor of 12x and 8 is the 4 that is the greatest common factor of 12 and 8 multiplied by the x0, the largest power of x that could divide both terms without creating negative exponents.

Now you know that the greatest common factor, or GCF, of 12x + 8 is 4x0, or simply 4, so the problem looks like this.

You can get the rest by dividing .

That means the multiplication problem that creates 12x + 8 is

You can check our factoring by multiplying and verifying that it equals 12x + 8.


You should find that with practice, you won’t need to formally divide by the common factor. You actually thought about it already when you chose the common factor, and should just need to check your arithmetic.

Let’s try one a little more challenging. Let’s factor 15x3 − 25x2 + 10x and take it step by step.

First, what’s the largest number that divides 15, -25, and 10? Because 15 = 5·3, −25 = 5(−5), and 10 = 5·2, the GCF is 5.

Second, what’s the largest power of x that will divide x3, x2, and x, without creating negative exponents? That would be the lowest power of the group, so x.

Third, identify the GCF of the polynomial. In this case, 5x.

Fourth, divide the original polynomial by the GCF.

Finally, write the polynomial in factored form, as if it were a multiplication problem.

15x3 − 25x2 + 10x = 5x (3x2 − 5x + 2)


If the common factor is positive, the signs of each term of the polynomial will stay the same. If the common factor is negative, all the signs of the polynomial will reverse. It’s all or nothing.

Some people find it easier to factor by looking at a version of the polynomial written in an expanded form, like this.

15x3 − 25x2 + 10x = 5·3 ·x·x·x − 5·5 ·x·x + 5·2·x

In this expanded version, you can identify what appears in every term. Every term has a 5 and an x.

This identifies the common factor, which you can remove from the individual terms and place in front, once.

Don’t be surprised to see more than one variable in problems like these. When you first see a polynomial like 21ky5 − 35ky3 + 49ky2, the presence of two variables makes it look difficult. Once you identify the greatest common factor, however, the expression looks more manageable. All three coefficients are divisible by 7, all three terms have a single k, and each term contains at least y2. The greatest common factor for 21ky5 − 35ky3 + 49ky2 is 7ky2. Divide 21ky5 − 35ky3 + 49ky2 by 7ky2 to find the other factor.

The polynomial 21ky5 − 35ky3 + 49ky2 can be factored as 7ky2 (3y3 − 5y + 7). There are still two variables in the expression, but now we can see it as a monomial with two variables times a trinomial in y.


Always look for a greatest common factor first. It may be the only type of factoring possible, or there may be other ways to factor the polynomial, but removing the GCF will make any other factoring easier to find.

If there is no way to factor a polynomial except to say that it equals itself times 1, the polynomial is prime, just as a number is prime when its only factors are itself and 1. The number 11 is prime because it can only be written as 11·1, and x2 − 7xy2 + 5y2 is prime because there is no way to factor it except 1(x2 − 7xy2 + 5y2).


Factor each expression by factoring out the greatest common factor. If no factoring is possible, state that the polynomial is prime.

1. 12x2 − 18x

2. 21t2 − 35

3. 16y3 − 56 y2 + 72y

4. 5ax2 − 25ax + 15a

5. 32x5 − 48x4 + 16x3

6. 11 − 77t + 22t2

7. πr2 + πrh

8. 15x2 − 3x − 5

9. y2 + 2xx

10. 4a3 + 8a2 − 24a

The first type of polynomial multiplication we learned was the distributive property and the first type of factoring we explored was greatest common factor. The relationship between the two is clear. Factoring out the GCF is based on the assumption that the polynomial was created by doing a distributive multiplication.

The second type of multiplication we encountered was the FOIL rule, so it makes sense to look next at the factoring pattern that connects to FOIL. We’re going to take this one in steps, starting with simple trinomials that fit closely to a pattern, and then branching out into others that require more trial and error.

Factoring x2 + bx + c

When you multiply two simple binomials, like (x + 5)(x − 3), the FOIL rule tells you to do four multiplications, but usually there are like terms to combine, and you end up with a trinomial.

When you see a trinomial with an x2-term, an x-term, and a constant term, it’s reasonable to think that it might have come from that sort of multiplication.

The simpler the trinomial is, the easier it is to determine what its factors are. If the x2 term is just x2, you can be confident that each of the factors began with just a simple x. Then you can look at the constant term of the trinomial and make a list of factor pairs. The numbers that can multiply to this constant are possibilities for the constant terms of the binomials.

If you start out with x2 + 8x + 15, the x2 tells you the factors start with x’s.

Then you can look at the 15, and realize that to get 15, you need to multiply either 1·15 or 3·5. That means you have two possibilities.

x2 + 8x + 15 = (x + 1)(x + 15) or x2 + 8x + 15 = (x + 3)(x + 5)

How do we decide which is correct? The obvious answer is multiply. Multiply (x + 1)(x + 15) and see if you get x2 + 8x + 15. No? Try multiplying (x + 3)(x + 5). Did that one give you x2 + 8x + 15? It should. But having to multiply to find out which one works could be very time-consuming, especially if there are a lot of possible factor pairs. Noticing the pattern of what usually happens will lead to a shortcut. When you multiply two simple binomials of the form (x + p)(x + q), the resulting trinomial starts with an x2, ends with the product of the numbers in the p and q positions, and has a middle term, an x-term, whose coefficient is the sum of p and q.

(x + p)(x + q) = x2 + px + qx + pq = x2 + (p + q)x + pq

In the example, you know that both 1·15 and 3·5 multiply to 15, but only 3 and 5 will add to give you 8x.


When you start to look for factors, focus on the absolute values of the numbers. If the constant term is positive, the factors of the constant will add to the middle term. If the constant is negative, the middle term is the difference between them. Once you’ve found them, you can place the signs. If the constant is positive, both factors get the sign of the middle term. If the constant is negative, one gets a plus and one a minus. Test to see which arrangement gives the correct middle term.

Here’s another example. Let’s try to factor x2x − 6, assuming it came from the multiplication of two binomials of the form (x + p)(x + q). Multiplying x times x for the First will give you x2, so you just have to find the constants. To get a 6, you could use 1·6 or 2·3. You want the result to be −6, so you need to multiply a positive number by a negative number, but which number is positive and which is negative? Trying to get −6 rather than 6 actually increases our list of possibilities. You can use −1·6, 1·(−6), −2·3 or 2·(−3).

You need a pair of numbers that will multiply to −6 and also add to −1, the coefficient of the x-term. Only one pair of numbers, 2·(−3), meet both requirements. The correct factoring is x2x − 6 = (x + 2)(x − 3).

Step by step, here’s how to tackle this sort of factoring problem. To factor a polynomial of the form x2 + bx + c:

1. Set up two parentheses for the two binomial factors. x2 + bx + c = (x )(x ) Start each with an x, so that x·x = x2.

2. List factor pairs for the absolute value of the constant term, |c|.

3. If c is a positive number, choose the factor pair that adds to |b|.
x2 − 7x + 12 = (x 3)(x 4) because 3·4 = 12 and 3 + 4 = 7.

4. If c is a negative number, choose the pair that subtracts to |b|.
x2 + 2x − 15 = (x 5)(x 3) because 5·3 = 15 and 5 − 3 = 2.

5. Put one of the factors in each parenthesis. x2 + bx + c = (x p)(x q)

6. If c is positive, both of the factors get the sign of b. x2 − 7x + 12 = (x − 3)(x − 4) because the x-term is negative.

7. If c is negative, one factor will be positive and one negative. The larger factor gets the sign of b. x2 + 2x − 15 = (x + 5)(x − 3). The bigger factor of 5 gets the positive sign of 2x and the smaller factor gets the negative sign.

8. Multiply using the FOIL rule to check.


Factor each trinomial to the product of two binomials. If the trinomial cannot be factored, state that it is prime.

11. x2 + 7x + 12

12. y2 − 6x + 8

13. t2 + 9t + 20

14. y2 − 2y + 1

15. x2 + 12x + 27

16. t2t − 12

17. x2 + 2x − 15

18. y2 − 8y − 20

19. t2 + 10t − 11

20. x2 − 12x − 28

When the coefficient of the squared term is 1, the simple square term, x2, clearly is equal to x·x. When that coefficient is anything other than 1, however, you have to think more carefully about how to produce the square term, and also about the ramifications of that change for the rest of the factoring process.

Factoring ax2 + bx + c

Let’s look at this first from a multiplication point of view. If you multiply (x − 9)(x + 1), the First multiplication gives you x·x = x2, Outer is x·1 = x, Inner is −9·x = −9x, and Last is −9·1 = −9. The like terms combine to give you (x − 9)(x + 1) = x2 − 8x − 9.

Now let’s look at the change that occurs when you change one coefficient. If you change (x − 9)(x + 1) to (2x− 9)(x + 1), it causes a change in both the First multiplication and the Outer.

(2x − 9) (x + 1) = 2x2 + 2x − 9x − 9 = 2x2 − 7x − 9

Adding a coefficient in front of one x has changed both the x-squared term and the x-term. Keep that in mind when you’re trying to factor a trinomial with a leading coefficient that is not 1. Notice that where you put the new coefficient also affects the x-term. (2x − 9)(x + 1) is not the same product as (x − 9)(2x + 1). As you’ve seen, (2x− 9)(x + 1) = 2x2 − 7x− 9, but (x − 9)(2x + 1) = 2x2 + x − 18x − 9 = 2x2 − 17x− 9.


If your middle term has a small coefficient, look for factors that are not very different. For a large middle term, you’re more likely to need one large and one small factor. If your constant is 35, for example, try 1 and 35 if you need a large middle term, but 5 and 7 for a small one.

Let’s try factoring 3x2 + 8x − 35. Our earlier plan of looking for a pair of factors of 35 that will add or subtract to 8 is not going to work when the coefficient of x2 is not 1. We’re going to have to depend much more on trial and error. We’ll start by setting up the first term of each binomial. To produce a leading coefficient of 3, we’ll need 3·1.

3x2 + 8x − 35 = (3x )(x )

Next we look at the possible factor pairs for 35. We could use 1·35 or 5·7. Neither of these pairs add or subtract to 8, but that’s okay. The 3 will affect the Outer product before the Inner and Outer combine to make 8. The constant term of 35 is actually -35, which means one factor will be positive and one negative, and that will make it look like the Inner and Outer products are subtracting to 8.

Now it’s trial and error time, but you can use a little number sense as well. The first factor pair, 1 and 35, doesn’t subtract to 8 on its own, and multiplying the 35 by 3 will only make things worse. (3x 1)(x 35) has no chance of producing 3x2 + 8x − 35, and you can see that even before you put the signs in.

There are four possibilities.

(3x 1)(x 35)

(3x 35)(x 1)

(3x 5)(x 7)

(3x 7)(x 5)

You don’t need to check the First or Last products, because you set these up based on those. You just need to look at the Outer and Inner and see if they’ll give you the correct middle term.

Now that you’ve determined the numbers to need use, you have to position the signs. You need a plus and a minus to produce the −35 in 3x2 + 8x − 35, but here again our old rule about the larger number getting the sign of the middle term may not hold up. The leading coefficient can confuse the issue. Instead, look at the Outer and Inner products. The larger of those is going to get the sign of the middle term, in this case, the plus.

In the examples we’ve looked at so far, the coefficient of the squared term was a prime number. There was only one possibility of factors for that coefficient. When that coefficient is not prime, and there are several possibilities for factors of the lead term, there’s more trial and error involved and it’s important that you do it in an organized manner.

Let’s factor 6x2 + 13x + 5. The lead term, 6x2, could be factored as 1x·6x or as 2x·3x, but fortunately the constant term, 5, is prime, so you’ll only have 1·5 to think about there. Work in pencil and have a good eraser handy, because this is a trial and error process. You’re going to need to change things.

Choose a pair of factors for the lead term. Set up your parentheses.

6x2 + 13x + 5 = (x )(6x )

Place the factors of the constant term.

6x2 + 13x + 5 = (x 1)(6x 5)

Find the Inner and Outer products.

Can the Inner and Outer add or subtract to the middle term? In this case, 6x and 5x, won’t combine to 13x.

Switch the positions of the factors of the constants.

6x2 + 13x + 5 = (x 1)(6x 5)

Try the Inner and Outer again.

These still won’t combine to our middle term of 13x. If you had other possibilities for the factors of the constant term, you’d repeat this trial process for each factor pair. Don’t forget to do the switch of position before you give up on a factor pair.

If you’ve tried all the possibilities for the constant term, it’s tempting to think the trinomial is not factorable, but before you declare it a prime polynomial, you need to try that other pair of factors for the lead term. Start over, this time with 2x·3x for the lead.

6x2 + 13x + 5 = (2x )(3x )

Place the factors of the constant and check the Inner and the Outer.

This will give you 13x, so you just need to place the signs, but if this didn’t work, you’d switch the 1 and the 5 and try again. In this case, the factoring is 6x2 + 13x + 5 = (2x + 1)(3x + 5).


Make lists of the possibilities for factors of the lead term and factors of the constant term. Choose a pair of factors for the lead term and stay with it until you have tried every pair of factors for the constant, in both orders. Only then should you cross off that pair of lead factors and try the next one. Repeat all the pairs of factors for the constant. Check off each pair as you try it so you don’t miss anything.

To factor ax2 + bx + c:

1. List factors of the lead term and factors of the constant.

2. Place the first set of factors for the lead term and the first set of factors for the constant.

3. Check the Inner and Outer product and see if they can combine to form the middle term.

4. If not, switch the positions of the factors of the constant term and check the middle term again.

5. Repeat for each possible factor pair for the constant term.

6. If all these fail, move to the next pair of factors for the lead term, and retry the factor pairs for the constant.

7. Repeat until you find Inner and Outer products that will create the middle term.

8. Place the signs.


Factor each trinomial to the product of two binomials. If the trinomial cannot be factored, state that the trinomial is prime.

21. 8x2 + 2x − 1

22. 3a2 + 2a − 8

23. 7y2 + 20y − 3

24. 6x2 + 5x − 6

25. 3t2 − 11t − 15

26. 2x2 − 15x + 7

27. 3y2 − 14y − 5

28. 10t2 − 23t + 12

29. 4x2 − 17x − 15

30. 10x2 + 7x − 12

Factoring Special Forms

When we talked about multiplication of polynomials, we mentioned two special cases. Knowing that multiplying the sum and difference of the same two terms gives you a difference of squares, or that there is a pattern to the perfect square trinomial, can save you time in multiplying. That knowledge can also help you to factor.

Perfect Square Trinomial

If you can recognize a perfect square trinomial when you see one, you’ll save time factoring. Remember (x + b)2 = x2 + 2bx + b2 and (ax + b)2 = a2x2 + 2abx + b2. When you look at x2 + 10x + 25 and notice that both x2 and 25 are perfect squares, look at the middle term. When you see that 10x = 2·5x, you can identify x2 + 10x + 25 as a perfect square trinomial and know that it factors as x2 + 10x + 25 = (x + 5)2.

If you’re asked to factor 36x2 − 84x + 49, you might be reluctant to start trial and error, because of how many factor pairs exist for 36. Do you notice, however that 36 is 62? And that 49 is 72? Make a quick check of the middle term. Because 2·6·7 = 84 , this looks like a perfect square. 36x2 − 84x + 49 = (6x − 7)2.

Difference of Squares

When you’re confronted with a polynomial like x2 − 81 or 100x2 − 121 to factor, you may feel helpless at first. How do you check the middle term if there isn’t one? If you know the special cases from multiplication, however, you will recognize x2 − 81 as x2 − 92 and 100x2 − 121 as 102x2 − 112, and realize that each of them will factor to the sum and difference of the same two terms.

x2 − 81 = (x + 9)(x − 9) and 100x2 − 121 = (10x + 11)(10x− 11).


When factoring a second degree expression with only two terms, it will be a difference of squares, or there will be a common factor, (or both), or it will be prime. For the difference of squares, both terms must be perfect squares. Non-squares will not work. It must be the difference of squares. The sum of squares does not factor.


Use your knowledge of special forms to factor each expression.

31. x2 − 64

32. y2 + 12y + 36

33. 4x2 − 121

34. a2 − 16a + 64

35. 9y2 − 49

36. 25t2 + 30t + 9

37. 81t2 − 16

38. 16x2 − 56x + 49

39. 100x2 − 1

40. 36z2 + 60z + 25

The Least You Need to Know

·  Always check for a common factor for all the terms of a polynomial. Remove it by writing the polynomial as a product for the distributive property.

·  Factor x2 + bx + c = (x + p)(x + q), where p· q = c and p + q = b.

·  You can factor ax2 + bx + c by making a list of the factors of a and a list of the factors of c. Try factor pairs in an organized fashion until you find the combination that produces your middle term.

·  Remember special forms: a2 x2 + 2abx + b2 = (ax + b)2 and a2 x2b2 = (ax + b)(axb).