The trinomial of the form

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

where \(a \ne 1\) can be factored using the grouping method.

First, we need to find two numbers \(m\) and \(n\) such that

$$\eqalign{

m + n &= b \cr

mn &= ac \cr} $$

Then, substitute the middle term using the newly found two numbers that satisfy the given conditions.

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

Next, 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 given trinomial.

I understand that the steps can be intimidating. But when we see it in action, it's actually not too bad.

**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} $$

It looks like the second factor pair is what we want. We just need to tweak the signs. Since the product is negative, either \(2\) or \(9\) must be negative. In this scenario, we want the larger number to be negative because the intended 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 found. Note that \( - 7x = 2x - 9x\).

$$\eqalign{

& 6{x^2}{\color{blue} - 7x} - 3 \cr

& 6x{}^2 {\color{blue}+ 2x - 9x} - 3 \cr} $$

Separe 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 \(12{x^2} - 34x + 10\).

By the way, before we even start factoring the trinomial itself we need to make sure that we deal with the GCF first. This trinomial has the greatest common factor of \(2\). Let's factor it out first.

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

For now, we will disregard \(2\) while we focus on the main trinomial. We will just reattach that factor \(2\) in 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} $$

For two numbers to have a positive product and a negative sum, they must be both negative numbers. By trial and error, we should come up with \(-2\) and \(-15\) since

$$\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 write our final answer, don't forget to attach the number \(2\) that we factored out from the very beginning. Hence, our final answer is

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