
## Section3.3Factoring Trinomials with Leading Coefficients of 1

Factoring a trinomial of form $x^2+bx+c\text{,}$ where $b$ and $c$ are integers, is essentially the reversal of a FOIL process. For this reason, we can develop a strategy by investigating a FOIL expansion. Let's expand $(x+4)(x+6)\text{.}$

\begin{align*} (x+4)(x+6)\amp=x^2+6x+4x+24\\ \amp=x^2+10x+24 \end{align*}

Let's observe that the linear coefficient, $10\text{,}$ is the sum of $4$ and $6$ whereas the constant term, $24\text{,}$ is the product of $4$ and $6\text{.}$ Since we expanded $(x+4)(x+6)\text{,}$ we can infer that finding the constants in a factorization of trinomials of form $x^2+bx+c$ is dependent upon determining two numbers that sum to $b$ and multiply to $c\text{.}$

From the factor pairs and sums shown in Table 3.3.1, we can infer the following factorizations.

\begin{align*} x^2+7x+6\amp=(x+1)(x+6)\\ x^2+5x+6\amp=(x+2)(x+3)\\ x^2-7x+6\amp=(x-1)(x-6)\\ x^2-5x+6\amp=(x-2)(x-3) \end{align*}
 factors sum $1,6$ $7$ $2,3$ $5$ $-1,-6$ $-7$ $-2,-3$ $-5$

Just as noteworthy, since the factor pairs shown in Table 3.3.1 form an exhaustive list, we cannot factor any trinomial of form $x^2+bx+6$ where b is not one of the four numbers shown in the sum column. Such trinomials are said to be prime. For example, $x^2+3x+6$ is prime.

###### Example3.3.2

Factor $x^2-8x+15$

Solution

Our first task is to determine a factor pair that multiplies to $15$ and adds to $-8\text{.}$ Since the product is positive and the sum is negative, we are searching for two negative numbers. The pair that works is $-3$ and $-5\text{.}$ This gives us:

\begin{equation*} x^2-8x+15=(x-3)(x-5)\text{.} \end{equation*}
###### Example3.3.3

Factor $w^2-4w-12$

Solution

Since our factor pair needs to multiply to a negative value $(-12)\text{,}$ one the numbers in the pair must be positive and the other negative. Let's note that $6$ and $2$ multiply to 12 and have a difference of $4\text{.}$ This can be helpful in determining that the factor pair that multiplies to $-12$ and adds to $-4$ is $-6$ and $2\text{.}$ This gives us:

\begin{equation*} w^2-4w-12=(w-6)(w+2)\text{.} \end{equation*}

### Subsection3.3.1Exercises

Factor each trinomial. Check your answer by expanding the factorization. If the trinomial cannot be factored, state that it is prime.

###### 1

$x^2+6x+5$

Solution

$x^2+6x+5=(x+2)(x+3)$

###### 2

$t^2-6t+8$

Solution

$t^2-6t+8=(t-2)(t-4)$

###### 3

$x^2+8x+4$

Solution

$x^2+8x+4$ is prime.

###### 4

$y^2-14y-32$

Solution

$y^2-14y-32=(y-16)(y+2)$

###### 5

$x^2+6x-20$

Solution

$x^2+6x-20$ is prime.

###### 6

$x^2+15x-100$

Solution

$x^2+15x-100=(x+20)(x-5)$

###### 7

$w^2-18w+45$

Solution

$w^2-18w+45=(w-15)(w-3)$

###### 8

$x^2+21x+80$

Solution

$x^2+21x+80=(x+16)(x+5)$

###### 9

$b^2-9b-400$

Solution

$b^2-9b-400=(b-25)(b+16)$