# Factor trinomials where the coefficient of x – quadratic is one ( a = 1)

Factoring with three terms, or trinomials, is the most important type of factoring to be able to master. As factoring is multiplication backwards we will start with a multiplication problem and look at how we can reverse the process.

Example 1:

Solve (x + 6)(x – 4).

Solution:

Distribute (x +6) through second parenthesis

x(x + 6) -4(x + 6)

Distribute each monomial through parenthesis

$x^2+6x-4x-24$

Our Solution is $x^2+2x-24$.

You may notice that if you reverse the last three steps the process looks like grouping. This is because it is grouping! The GCF of the left two terms is x and the GCF of the second two terms is – 4. The way we will factor trinomials is to make them into a polynomial with four terms and then factor by grouping. This is shown in the following example, the same problem worked backwards.

Also Read : Factor polynomials with four terms using grouping

Example 2:

Factorize $x^2+2x-24$.

Solution:

Split middle term into +6x – 4x

$x^2+6x-4x-24$

Grouping: GCF on left is x, on right is -4

x(x + 6) -4(x + 6)

(x + 6) is the same, factor out this GCF

Our solution is (x + 6)(x – 4).

The trick to make these problems work is how we split the middle term. Why did we pick + 6x – 4x and not + 5x – 3x? The reason is because 6x – 4x is the only combination that works! So how do we know what is the one combination that works? To find the correct way to split the middle term we will use what is called the ac method. In the next lesson we will discuss why it is called the ac method. The way the ac method works is we find a pair of numbers that multiply to a certain number and add to another number. Here we will try to multiply to get the last term and add to get the coefficient of the middle term. In the previous example that would mean we wanted to multiply to – 24 and add to + 2. The only numbers that can do this are 6 and – 4 (6 · – 4 = – 24 and 6 + ( – 4) = 2). This process is shown in the next few examples

Example 3:

Factorize $x^2+9x+18$.

Solution:

Want to multiply to 18, add to 9

$x^2+6x+3x+18$.

6 and 3, split the middle term

x(x + 6) +3(x + 6)

the solution is (x + 6)(x + 3).

Example 4:

Factorize $x^2-4x+3$

Solution:

Want to multiply to 3, add to -4

$x^2-3x-x+3$

-3 and -1, split the middle term

Factor by grouping

x(x – 3) – 1(x – 3)

the solution is (x – 3)(x – 1).

Example 5:

Factorize $x^2-8x-20$.

Solution:

Want to multiply to -20, add to -8

-10 and 2, split the middle term

$x^2-10x+20-20$.

Factor by grouping

a(a – 7b) – 2b(a – 7b)

the solution is (a – 7b)(a – 2b).

As the past few examples illustrate, it is very important to be aware of negatives as we find the pair of numbers we will use to split the middle term. Consider the following example, done incorrectly, ignoring negative signs

Example 6:

Factorize $x^2+5x-6$.

Solution:

Want to multiply to 6, add 5

2 and 3, split the middle term

$x^2+2x+3x-6$

Factor by grouping

x(x + 2) + 3(x – 2)

binomials do not match!

Because we did not use the negative sign with the six to find our pair of numbers, the binomials did not match and grouping was not able to work at the end. Now the problem will be done correctly.

$x^2+5x-6$.

Want to multiply to -6, add to 5

6 and -1, split the middle term

$x^2+6x-x-6$.

Factor by grouping

x(x + 6) – 1(x + 6)

Our Solution is ( x + 6 ) ( x – 1).