Skip to main content

Section 3.3 Factoring 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
\(1,6\) \(7\)
\(2,3\) \(5\)
\(-1,-6\) \(-7\)
\(-2,-3\) \(-5\)
Table 3.3.1. Factor pairs of \(6\)

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.

Example 3.3.2.

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*}
Example 3.3.3.

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*}
Figure 3.3.4. Practice Factoring Trinomials with a Leading Coefficient of One

Exercises Exercises

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

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