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Section 4.3 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 Figure 4.3.1, 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
\(1,6\) \(7\)
\(2,3\) \(5\)
\(-1,-6\) \(-7\)
\(-2,-3\) \(-5\)
Figure 4.3.1. Factor pairs of \(6\)

Just as noteworthy, since the factor pairs shown in Figure 4.3.1 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.

A common strategy for organizing the work associated with factoring trinomials with a leading coefficient of \(1\) is to write the constant term at the top of a large X and the linear coefficient at the bottom of the X. Go ahead and press "Show Goal" in Figure 4.3.2 to see what I mean. Now we can focus on determining the two numerical factors. I find it useful to first establish the signs on the two factors. If the number at the top of the X is negative, then we need one positive numerical factor and one negative numerical factor. If the number at the top of the X is positive, then both numerical factors have the same sign, and that sign agrees with the number at the bottom of the X. Go ahead and press "Show Numerical Factors" and "Show Factorization" to see the remainder of the process. Go through a few examples and then try doing it on your own before you see the steps on the computer.

Figure 4.3.2. Practice Factoring Trinomials with a Leading Coefficient of One
Example 4.3.3.

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 4.3.4.

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 4.3.5 gives you another opportunity to generate random practice exercises.

Figure 4.3.5. 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

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