Chapter 5.4: Factoring Trinomials where a ≠ 1

Mike LePine

Chapter 5.4: Factoring Trinomials where a ≠ 1

Factoring trinomials where the leading term is not 1 is only slightly more difficult than when the leading coefficient is 1. The method used to factor the trinomial is unchanged.

Example 1

Factor the trinomial $3x^2 + 11x + 6$ .

Start by multiplying the coefficients from the first and the last terms. This is $3\cdot 6$ , which yields 18.

The next task is to find all possible integers that multiply to 18 and their sums.

$\begin{array}{cc} \text{multiply to }18\hspace{0.25in}&\text{sum of these integers} \\ 1\cdot 18&19 \\ 2\cdot 9&11 \\ 3\cdot 6&9 \\ 6\cdot 3&9 \\ 9\cdot 2&11 \\ 18\cdot 1&19 \end{array}$

Look for the pair of integers that multiplies to 18 and adds to 11, so that it matches the equation that you started with.

For this example, the pair is $2\cdot 9$ , which adds to 11.

Now take the original trinomial $3x^2 + 11x + 6$ and break the $11x$ into $2x$ and $9x$ .

Rewrite the original trinomial as $3x^2 + 2x + 9x + 6$ .

Now, split this into two binomials as done in the previous section and factor.

$3x^2 + 2x$ yields $x(3x + 2)$ and $9x + 6$ yields $3(3x + 2)$ .

$3x^2 + 2x + 9x + 6$ becomes $x(3x + 2) + 3(3x + 2)$ .

$x(3x + 2) + 3(3x + 2)$ factors to $(3x + 2)(x + 3)$ .

$3x^2 + 11x + 6 = (3x + 2)(x + 3)$

The master product method works for any integer breakup of the polynomial. Slightly more complicated are questions that involve two different variables in the original polynomial.

Example 2

Factor the trinomial $4x^2 - xy - 5y^2$ .

Start by multiplying the coefficients from the first and the last terms. This is $4\cdot -5$ , which yields −20.

The next task is to find all possible integers that multiply to −20 and their sums.

$\begin{array}{cc} \text{multiply to }-20\hspace{0.25in}&\text{sum of these integers} \\ -1\cdot 20&\phantom{-}19 \\ -2\cdot 10&\phantom{-}8 \\ -4\cdot 5&\phantom{-}1 \\ -5\cdot 4&-1 \\ -10\cdot 2&-8 \\ -20\cdot 1&-19 \end{array}$

Look for the pair of integers that multiplies to −20 and adds to −11, so that it matches the equation that you started with.

For this example, the pair is $-5\cdot 4$ , which adds to −1.

Now take the original trinomial $4x^2 - xy - 5y^2$ and break the $-xy$ into $-5xy$ and $4xy$ .

Rewrite the original trinomial as $4x^2 - 5xy + 4xy - 5y^2$ .

Now, split this into two binomials as done in the previous section and factor.

$4x^2 - 5xy$ yields $x(4x - 5y)$ and $4xy - 5y^2$ yields $y(4x - 5y)$ .

$4x^2 - xy - 5y^2$ becomes $x(4x - 5y) + y(4x - 5y)$ .

$x(4x - 5y) + y(4x - 5y)$ factors to $(x + y) (4x - 5y)$ .

$4x^2 - xy - 5y^2 = (x + y) (4x - 5y)$

There are a number of variations potentially encountered when factoring trinomials. For instance, the original terms might be mixed up. There could be something like $-10x + 3x^2 + 8$ that is not in descending powers of $x$ . This requires reordering in descending powers before beginning to factor.

$-10x + 3x^2 + 8 \longrightarrow 3x^2 -10x + 8 \text{ (factorable form)}$

It might also be necessary to factor out a common factor before starting. The polynomial above can be written as $30x^2 -100x + 80$ , in which a common factor of 10 should be factored out prior to factoring.

$\text{This turns }30x^2 -100x + 80 \text{ into }10(3x^2 -10x + 8)$ .

There are also slight variations on the common factored binomial that can be illustrated by factoring the trinomial $3x^2 -10x + 8$ .

Example 3

Factor the trinomial $3x^2 - 10x + 8$ .

Start by multiplying the coefficients from the first and the last terms. This is $3\cdot 8$ , which yields 24.

The next task is to find all possible integers that multiply to 24 and their sums (knowing that the middle coefficient must be negative).

$\begin{array}{cc} \text{multiply to }24 \hspace{0.25in}&\text{sum of these integers} \\ -1\cdot -24&-25 \\ -2\cdot -12&-14 \\ -3\cdot -8&-11 \\ -4\cdot -6&-10 \\ -6\cdot -4 & -10 \\ -8\cdot -3&-11 \\ -12\cdot -2 & -14 \\ -24\cdot -1&-25 \end{array}$

Look for the pair of integers that multiplies to 24 and adds to −10, so that it matches the equation that you started with.

For this example, the pair is $-4\cdot -6$ , which adds to −10.

Now take the original trinomial $3x^2 - 10x + 8$ and break the $-10x$ into $-4x$ and $-6x$ .

Rewrite the original trinomial as $3x^2 - 4x - 6x + 8$ .

Now, split this into two binomials as done in the previous section and factor.

$3x^2 - 4x$ yields $x(3x - 4)$ , but $-6x + 8$ yields $2(-3x + 4)$ .

$x(3x - 4)$ and $2(-3x + 4)$ are a close match, but their signs are different.

The way to deal with this is to factor out a negative, specifically, −2 instead of 2.

$-6x + 8$ can be factored two ways: $2(-3x + 4)$ and $-2(3x - 4)$ .

Choose the second factoring, so the common factor matches.

$3x^2 - 10x + 8$ becomes $x(3x - 4) + -2(3x - 4)$ .

$x(3x - 4) + -2(3x - 4)$ factors to $(3x - 4)(x - 2)$ .

$3x^2 - 10x + 8 = (3x - 4)(x - 2)$

Example 4

Factor the following trinomials, which are both variations of the trinomial seen before in 7.4.3:

$3x^2 - 14x + 8$
The pair of numbers that can be used to break it up is −2 and −12.
$\begin{array}{lll} 3x^2-14x+8\text{ breaks into}&3x^2-2x-12x+8& \\ &x(3x-2)-4(3x-2)&\text{Common factor is }(3x-2) \\ &(3x-2)(x-4)&\text{Left over when factored} \end{array}$
$3x^2 - 11x + 8$
The pair of numbers that can be used to break it up is −3 and −8.
$\begin{array}{lll} 3x^2-11x+8\text{ breaks into}&3x^2-3x-8x+8& \\ &3x(x-1)-8(x-1)&\text{Common factor is }(x-1) \\ &(x-1)(3x-8)&\text{Left over when factored} \end{array}$