## Factor Quadratics: x^2 + bx + c

## Introduction

The most basic kind of factoring involves taking an expression and finding the greatest common factor of each of the terms in the expression. In the example to the right, each expression has a common factor of 3x, and this term can be “factored” out of the original expression.

The three terms in the expression y^{2} + 8y + 12 do not have any common factors. This expression can be factored using other methods and the result is (y + 6)(y + 2).

The first step in understanding factoring is understanding how to multiply the factors (y + 6) and (y + 2) to get the original expression above.

The lesson shows strategies of how to factor trinomials like y^{2} + 8y + 12. Before moving on, take a minute and look carefully at the multiplication of the (y + 6) and (y + 2) above. Can you guess how the 8y and the 12 (in y^{2} + 8y + 12) relate to the (y + 6) and (y + 2)?

## Lesson

In the introduction, (y + 6) and (y + 2) were multiplied to get an answer of y

^{2}+ 8y + 12. You may notice that:

- 6 + 2 = 8 (our 8y)
- 6 ∙ 2 = 12 (our 12)

These two facts can be used to factor many trinomials.

**Example 1****:** Factor the trinomials

**Scroll over the problem above to reveal the answer.*

Each trinomial in example 1 is in the form ax^{2} + bx + c, where a = 1 (in other words, x^{2} + bx + c). When factoring trinomials of this type, it is just a matter of finding two numbers that add up to b and multiply to equal c.

In example 1, each number was a positive number. Now that you have had a little practice factoring trinomials, see if you can do a problem that involves negatives:

Working with negatives simply means more possibilities when finding factors. In example 2, some of the numbers are negative. See if you can figure out the answer, then scroll over it to see if you were correct.

**Example 2****:** Factor the trinomials

**Scroll over the problem above to reveal the answer.*

**Special Cases in Factoring**

Consider the expressions x^{2} + 1 and x^{2} – 1. Only one of these is factorable. Can you guess which one it is?

You may think that x^{2} + 1 would factor to (x + 1)(x + 1), but the math just doesn’t work out. The expression x^{2} + 1 is actually the same as x^{2} + 0x + 1. To factor it, there would need to be two numbers that multiplied to 1 and added to zero. Since there are no such numbers, it is nonfactorable.

The expression x^{2 }– 1 can also be written as x^{2} + 0x – 1. This can be factored and the answer must contain two numbers that multiply to -1 and add to zero. The only two integers that multiply to -1 are 1 and -1. These two numbers add up to 0, so they work for factoring x^{2 }– 1. This is called the difference of two squares.

The chart above shows how the factors (x + 1) and (x – 1) can be combined to make these special cases. The same works for (x + a) and (x – a), where a is any positive whole number.

**Example 3****:** Factor each 2^{nd} degree polynomial. Write “not factorable” if it cannot be factored.

**Scroll over the problem above to reveal the answer.*

## Try It

Factor each polynomial. Write “not factorable” if it cannot be factored.

1) x^{2} + 10x + 16

2) x^{2} – 25

3) y^{2} – y – 20

4) y^{2} – 100

5) z^{2} + 11z + 10

6) z^{2} + 36

7) a^{2} – 18x + 80

8) a^{2} + 11a – 60

*Hint: It is easy to peek at the answers below, so it is recommended that you try all problems on a separate piece of paper, then check answers all at once.

Scroll down for answers:

Answers:

1) x^{2} + 10x + 16 = (x + 8)(x + 2)

2) x^{2} – 25 = (x + 5)(x – 5)

3) y^{2} – y – 20 = (y + 4)(y – 5)

4) y^{2} – 100 = (y + 10)(y – 10)

5) z^{2} + 11z + 10 = (z + 10)(z + 1)

6) z^{2} + 36 » not factorable

7) a^{2} – 18x + 80 = (a – 10)(z – 8)

8) a^{2} + 11a – 60 = (a + 15)(a – 4)

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