# Factoring a Quadratic Polynomial, Part 1

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Throughout the rest of this section, we're going to look at techniques that you can use to factor specific kinds of polynomials. In this lesson, we're going to look at one of the simplest cases: a quadratic polynomial where the first coefficient is zero. Remember that a quadratic polynomial is a second degree polynomial with one variable. For example:

x2 + 3x + 4x2 + 2x - 63x2 - 42x + 2

The first two are examples of what we'll be looking at in this lesson. We'll take up the third kind in the next one.

In keeping with the nature of WCE's short courses, I'm not going to go into detail about why this kind of polynomial factors the way it does. Instead, I'm going to focus solely on how to do the factoring. This kind of polynomial is always going to factor into something that looks like:

(x + d)(x + e)

The trick is finding the right values of d and e so taht when you multiply the two binomials together you get the original expression. To do that we need to find values for d and e so that when you multiply them together you get the number at the end of the original expression and when you add them together you get the middle term. The procedure you should follow goes like this:

 Factor: x2 + 6x + 8 1. Write down all the numbers that multiply together to give the constant term in the original polynomial. In my example, that would be 8. Notice how I included negative numbers as well as positive ones. 1 x 8-1 x -82 x 4-2 x -4 2. Look for a pair that adds up to the number in front of the x term in the original expression. In my example that would be 6. 2 + 4 = 6 3. Substitute those numbers into (x + d)(x + e). for d and e and you'll have the factored version of the original polynomial. (x + 2)(x + 4)

# Example 1

Factor x2 + 14x + 24.

First we need to write down all of the numbers that multiply together to give 24:

 1 x 24 2 x 12 3 x 8 4 x 6 -1 x -24 -2 x -12 -3 x -8 -4 x -6

One of those pairs is going to be the numbers in our factorization. To figure out which we need to add all of them together and find the pair that adds up to 14, the number in front of the x in the original expression.

 1 + 24 = 25 2 + 12 = 14 3 + 8 = 11 4 + 6 = 24 -1 + -24 = -25 -2 + -12 = -14 -3 + -8 = -11 -4 + -6 = -24

The only pair of numbers that multiplies to 24 and adds up to 14 is 2 and 12 so our factorization must be

(x + 12)(x + 2)

# Example 3

Factor x2 - 6x - 16.

Just like in the previous examples, we need all of the numbers that multiply to -16.

 -1 x 16 -2 x 8 -4 x 4 1 x -16 2 x -8

Now we'll find the sums of each pair.

 -1 + 16 = 15 -2 + 8 = 6 -4 + 4 = 0 1 + -16 = -15 2 + -8 = -6

The only pair of numbers that multiplies to -16 and adds up to -6 is 2 and -8 so our factorization must be

(x - 2)(x + 8)

# Example 2

Factor x2 - 7x + 12.

We'll start this one the same way we did the previous example - by writing out all the integers that multiply to give us 12.

 1 x 12 2 x 6 3 x 4 -1 x -12 -2 x -6 -3 x -4

Now we'll find the sums of each pair.

 1 + 12 = 13 2 + 6 = 8 3 + 4 = 7 -1 + -12 = -13 -2 + -6 = -8 -3 + -4 = -7

The only pair of numbers that multiplies to 12 and adds up to -7 is -3 and -4 so our factorization must be

(x + -3)(x + -4)

Putting both a + and a - sign together like that is an awkward way to write it. As a final step, you would want to simplify it to:

(x - 3)(x - 4)

# Example 4

Factor x2 + x + 18.

We'll start this one the same way we did the previous example - by writing out all the integers that multiply to give us 12.

 1 x 18 2 x 9 3 x 6 -1 x -18 -2 x -9 -3 x -6

Now we'll find the sums of each pair.

 1 + 18 = 19 2 + 9 = 11 3 + 6 = 9 -1 + -18 = -19 -2 + -9 = -11 -3 + -6 = -9

We're looking for a pair of numbers that add up to 1 but a quick look at our list shows that there aren't any. That can happen. When you can't find a pair of numbers that multiplies to the constant number and adds up to the number in front of the x then your answer would be that the polynomial is "prime" or that "it can't be factored".

# Video Lectures

The next step up in our polynomial factoring algorithm is a three term polyonmial, specifically a quadratic polynomial whose leading coefficient is 1: x2 + ax + c.
There are several different scenarios that can come up when you're trying to factor a quadratic equation so we're going to spend some time looking at specific examples.

## Dynamic Tutorial - Factoring a Quadratic Polynomial

Directions: This solution has 4 steps. To see a description of each step click on the boxes on the left side below. To see the calculations, click on the corresponding box on the right side. Try working out the solution yourself and use the descriptions if you need a hint and the calculations to check your solution.