Suppose we have a trinomial of the form

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

where the leading coefficient \(a\) is equal to \(1\).

Since \(a=1\), we can rewrite the trinomial as

$$\large{{x^2} + bx + c}$$

These are the simple steps on how to factor out a trinomial of this form.

- Find two numbers such that the product is \(c\) while the sum equals \(b\).
- Use the two numbers found in the previous step to break the trinomial as a product of two binomials.

**Example 1:** Factor \({x^2} + 8x + 12\)

Let's find two numbers such that the product is \(12\) while the sum equals \(8\).

Because the product is positive and the sum is positive, we expect the two numbers to be both positive.

These are the factor pairs of \(12\).

$$\eqalign{

& 1 \cdot 12 \cr

& 2 \cdot 6 \cr

& 3 \cdot 4 \cr} $$

The factor pair that adds up to \(8\) is the second from the top which are \(2\) and \(6\).

$$\eqalign{

& 2 \cdot 6 = 12\,\, \checkmark \cr

& 2 + 6 = 8\,\,\checkmark \cr} $$

Therefore, the factors are

$${x^2} + 8x + 12 = \left( {x + 2} \right)\left( {x + 6} \right)$$

**Example 2:** Factor \({x^2} - 3x - 54\)

We need to find two numbers with a product of \(-54\) and a sum of \(-3\).

If you think about it, the two numbers must have opposite signs and the negative number must have the larger absolute value.

For now, we disregard the sign of \(-54\). We want to list all the factor pairs of just \(54\).

$$\eqalign{

& 1 \cdot 54 \cr

& 2 \cdot 27 \cr

& 3 \cdot 18 \cr

& 6 \cdot 9 \cr} $$

Which of the factor pairs have a difference of \(3\)? Yes, that's the bottom factor pair which are the numbers \(6\) and \(9\). Since we want to have a sum of \(-3\), we will need to make \(9\) negative. Therefore, the two numbers would be \(6\), and \(-9\).

$$\eqalign{

& 6\left( { - 9} \right) = - 54 \cr

& 6 + \left( { - 9} \right) = - 3 \cr} $$

So the factorization becomes

$${x^2} - 3x - 54 = \left( {x + 6} \right)\left( {x - 9} \right)$$

**Example 3:** Factor \({x^2} - 11x + 24\)

For two numbers to have a positive product and a negative sum, they must be both negative numbers.

These are the factor pairs of \(24\).

$$\eqalign{

& 1 \cdot 24 \cr

& 2 \cdot 12 \cr

& 3 \cdot 8 \cr

& 4 \cdot 6 \cr} $$

Notice that the third factor pair from the top almost gives us what we are looking for. We just need to make them both negative.

$$\eqalign{

& \left( { - 3} \right)\left( { - 8} \right) = 24 \cr

& \left( { - 3} \right) + \left( { - 8} \right) = - 11 \cr} $$

Thus

$${x^2} - 11x + 24 = \left( {x - 3} \right)\left( {x - 8} \right)$$