Factoring Trinomial whose leading coefficient is NOT 1

The trinomial in the form

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

where $a \ne 1$, can be factored by using the grouping method.

First, we need to find two numbers $m$ and $n$ such that:

$$\eqalign{
m + n &= b \cr
mn &= ac \cr} $$

Next, substitute the middle term using the newly found two numbers that satisfy the given conditions:

$$a{x^2} + mx + nx + c$$

Then, we separate the terms into two groups of binomials.

$$\left( {a{x^2} + mx} \right) + \left( {nx + c} \right)$$

Factor out the GCF of each binomial. If you have done it correctly, you should end up with two binomials that are the same. The final step is to factor out this common binomial to get the factors of the original trinomial.

I know the steps may seem overwhelming at first, but when you see them in practice, they’re actually pretty straightforward.

Example 1: Factor $6{x^2} – 7x – 3$

Note that $a=6$, $b=-7$, and $c=-3$. Let’s find the two numbers $m$ and $n$ such that

$$\eqalign{
m + n &= b = – 7 \cr
mn &= ac = – 18 \cr} $$

These are the factor pairs of $18$.

$$\eqalign{
& 1 \cdot 18 \cr
& 2 \cdot 9 \cr
& 3 \cdot 6 \cr} $$

The second factor pair seems to be the one we need, but we need to adjust the signs. Because the product is negative, either $2$ or $9$ must be negative. In this case, we want the larger number to be negative since the desired sum is negative.

The two numbers that we want are $2$ and $-9$ because

$$\eqalign{
& 2 + \left( { – 9} \right) = – 7 \cr
& \left( 2 \right)\left( { – 9} \right) = – 18 \cr} $$

Replace the middle term using the two numbers we identified. Notice that ( -7x = 2x – 9x ).

$$\eqalign{
& 6{x^2}{\color{blue} – 7x} – 3 \cr
& 6x{}^2 {\color{blue}+ 2x – 9x} – 3 \cr} $$

Separate the binomials using grouping symbols.

$$\left( {6{x^2} + 2x} \right) – \left( {9x + 3} \right)$$

If possible, factor the greatest common factor (GCF) of each binomial.

$$2x\left( {3x + 1} \right) – 3\left( {3x + 1} \right)$$

Finally, factor out the common binomial that arises.

$$\left( {3x + 1} \right)\left( {2x – 3} \right)$$

Therefore,

$$6{x^2} – 7x – 3 = \left( {3x + 1} \right)\left( {2x – 3} \right)$$

Example 2: Factor $12{x^2} – 34x + 10$.

Before we begin factoring the trinomial itself, it’s important to address the Greatest Common Factor (GCF). In this case, the trinomial has a GCF of $2$, so we should factor that out first.

$$12{x^2} – 34x + 10 = 2\left( {6{x^2} – 17x + 5} \right)$$

For now, we’ll set aside the factor of $2$ and concentrate on the main trinomial. We’ll reattach that factor $2$ at the end.

So for the trinomial ${6{x^2} – 17x + 5}$, we want to find the two numbers $m$ and $n$ that satisfy the given conditions.

$$\eqalign{
m + n &= – 17 \cr
mn &= 30 \cr} $$

To achieve a positive product and a negative sum with two numbers, both numbers must be negative. Through trial and error, we can determine that the numbers $-2$ and $-15$ satisfy these conditions because

$$\eqalign{
\left( { – 2} \right) + \left( { – 15} \right) &= – 17 \cr
\left( { – 2} \right)\left( { – 15} \right) &= 30 \cr} $$

Replace the middle term by the two numbers we found.

$$\eqalign{
& 6{x^2} {\color{blue}- 17x} + 5 \cr
& 6{x^2}{\color{blue} – 2x – 15x} + 5 \cr} $$

Separate them into two binomials using parentheses.

$$\left( {6{x^2} – 2x} \right) – \left( {15x – 5} \right)$$

We can factor out $2x$ from the first binomial and $5$ from the second binomial.

$$2x\left( {3x – 1} \right) – 5\left( {3x – 1} \right)$$

Finally, factor out the common binomial $3x-1$.

$$\left( {3x – 1} \right)\left( {2x – 5} \right)$$

To complete our final answer, be sure to reattach the number $2$ that we factored out at the start. Therefore, our final answer is

$$12{x^2} – 34x + 10 = 2\left( {3x – 1} \right)\left( {2x – 5} \right)$$