Factoring Trinomial whose leading coefficient is 1

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

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

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

Let’s find two numbers such that their product is \(12\) and their 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 should have opposite signs, with the negative number having the larger absolute value.

For now, we can set aside the negative sign of \(-54\) and concentrate on listing all the factor pairs of \(54\).

$$\eqalign{
& 1 \cdot 54 \cr
& 2 \cdot 27 \cr
& 3 \cdot 18 \cr
& 6 \cdot 9 \cr} $$

Which factor pairs have a difference of \(3\)? That would be the pair \(6\) and \(9\). Since we need a sum of \(-3\), we should make \(9\) negative. Therefore, the two numbers are \(6\) 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, both numbers must be negative.

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

$$\eqalign{
& 1 \cdot 24 \cr
& 2 \cdot 12 \cr
& 3 \cdot 8 \cr
& 4 \cdot 6 \cr} $$

Observe that the third factor pair from the top is nearly what we need. We just have to make both numbers negative.

$$\eqalign{
& \left( { – 3} \right)\left( { – 8} \right) = 24 \cr
& \left( { – 3} \right) + \left( { – 8} \right) = – 11 \cr} $$

Thus

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