# Factoring Trinomials – Examples and Practice Problems

Trinomials are three-term polynomials. Generally, when we mention trinomials, we mean quadratic trinomials. Quadratic trinomials can be factored by finding numbers, which when multiplied or added match the original trinomial.

Here, we will review the process used to factor trinomials. We will also look at several examples with answers of factoring trinomials to understand the use of the aforementioned process.

##### ALGEBRA

Relevant for

Exploring examples of factoring trinomials.

See examples

##### ALGEBRA

Relevant for

Exploring examples of factoring trinomials.

See examples

## Summary of factoring trinomials

The general form of a quadratic trinomial is written as $latex a{{x}^2}+bx+c$, where a, b, and c are constants. In the following exercises, we will consider the case when the value of a is 1, that is, when we have $latex a=1$ or $latex a=-1$. Therefore, the general form of this case is reduced to:

$latex {{x}^2}+bx+c$

The basic strategy used to factor these types of trinomials is to find two numbers, which when multiplied, result in the constant number c. Also, when we add these two numbers, we must obtain the constant b, the coefficient of the x term.

## Factoring trinomials – Examples with answers

The strategy mentioned above is used to solve the following examples of factoring trinomials. It is recommended that you try to solve the exercises yourself before looking at the solution.

### EXAMPLE 1

Factor the trinomial $latex {{x}^2}+5x+6$ as a product of two binomials.

We see that the coefficient of the quadratic term is 1, so this is easy to factor. We need to identify the other relevant constants. Notice that $latex b=5$ and $latex c =5$.

Then, we have to find two numbers which when multiplied are equal to the value of the constant term $latex c=6$ and which when added are equal to the value of $latex b = 5$.

Given that the product of the two numbers must be positive, the numbers must be either both positive or negative.

We can use several tries and errors to find the correct combination. The following are some possible combinations:

Since the correct combination of numbers is 2 and 3, our final answer is:

$latex (x+2)(x+3)$

We can verify this by multiplying this expression:

$latex (x+2)(x+3)$

$latex ={{x}^2}+2x+3x+6$

$latex ={{x}^2}+5x+6$

We obtained the original trinomial, so the factorization is correct.

### EXAMPLE 2

Factor the trinomial $latex {{x}^2}+6x+8$.

In this trinomial we have $latex b=6$ and $latex c=8$. Therefore, we have to find two numbers that have a product equal to the constant term $latex c=8$ and a sum equal to the value of $latex b=6$.

The product of both numbers must be positive, so the numbers must be either both positive or negative.

Let’s look at the following possible combinations:

Therefore, the correct combination of numbers is 2 and 4, and the factorization is:

$latex (x+2)(x+4)$

Verifying, we have:

$latex (x+2)(x+4)$

$latex ={{x}^2}+2x+4x+8$

$latex ={{x}^2}+6x+8$

The factorization obtained is correct.

### EXAMPLE 3

Obtain the factorization of the trinomial $latex {{x}^2}-2x-8$.

We have the coefficients $latex b=-2$ and $latex c=-8$. Therefore, we have to find two numbers that have a product equal to the value of the constant term $latex c=-8$ and that have a sum equal to the value of $latex b=-2$.

Here, the product of the two numbers must be negative, so one number must be positive and the other negative.

We see the following possible combinations:

The correct combination of numbers is 2 and -4, and the factorization is:

$latex (x+2)(x-4)$

Checking this, we have:

$latex (x+2)(x-4)$

$latex ={{x}^2}+2x-4x-8$

$latex ={{x}^2}-2x-8$

We obtained the original trinomial, so the factorization is correct.

### EXAMPLE 4

Factor the trinomial $latex {{x}^2}+4x-12$.

We identify the coefficients $latex b=4$ and $latex c=-12$.

Then, we have to find two numbers which when multiplied are equal to $latex c=-12$ and which when added are equal to $latex b=4$. The product of these numbers is negative, so one number must be positive and the other negative.

We can use the following table to try different possible combinations:

The correct combination of numbers is 6 and -2, so we have:

$latex (x+6)(x-2)$

Checking this, we have:

$latex (x+6)(x-2)$

$latex ={{x}^2}+6x-2x-12$

$latex ={{x}^2}+4x-12$

The factorization is correct since we obtained the original trinomial.

### EXAMPLE 5

Factor the trinomial $latex {{x}^2}-8x+15$.

We have the coefficients $latex b=-8$ and $latex c=15$. Therefore, we have to find two numbers whose product is equal to $latex c=15$ and whose sum is equal to $latex b=-8$.

The product of the two numbers must be positive, so the numbers could be either both positive or negative. However, we see that the sum is negative, so the numbers must both be negative

Testing some combinations we have:

Since the correct combination of numbers is -5 and -3, we have:

$latex (x-5)(x-3)$

Verifying, we have:

$latex (x-5)(x-3)$

$latex ={{x}^2}-5x-3x+15$

$latex ={{x}^2}-8x+15$

We obtained the original trinomial, so the factorization is correct.

### EXAMPLE 6

Obtain the factorization of the trinomial $latex {{x}^2}+17x+16$.

We identify the relevant constants $latex b=17$ and $latex c=16$.

Then, we must find two numbers with a product equal to the value of the constant term $latex c=16$ and with a sum equal to the value of $latex b=17$. The product must be positive, so the numbers must be either both positive or negative.

However, we see that the value of b (the result of the sum) is greater than the product, so we can only get this if both numbers are positive.

Since the correct combination of numbers is 1 and 16, our final answer is:

$latex (x+1)(x+16)$

We can verify this by multiplying this expression:

$latex (x+1)(x+16)$

$latex ={{x}^2}+1x+16x+16$

$latex ={{x}^2}+17x+16$

We obtained the original trinomial, so the factorization is correct.

## Factoring trinomials – Practice problems

Practice the above strategy by factoring the following quadratic trinomials. If you need help, you can look at the solved examples above carefully.