A parabola that cuts the $x$ -axis at integer values of $x$ will have a quadratic expression that factors easily. The factored form is useful for both graphing and for solving quadratic equations.

If a parabola does not cut the $x$ axis at all, or if the $x$ -intercepts are irrational numbers, factoring the quadratic expression is unlikely to be a helpful process.

Review

The general form for a quadratic expression is

$ax^2+bx+c$

When the coefficient of $x^2$ , $a$ is 1 we have

$x^2+bx+c$

. These expressions are the most straightforward to factor.

Example 1:

$x^2+7x+10 =(x+p)(x+q)$

We need to find numbers $p$ and $q$ such that

$p\times q = 10 \quad \text{and}\quad p+q=7$

A little thought or writing on paper leads us to the values $2$ and $5$ because $2\times 5 = 10$ and $2 + 5 = 7$ .

Therefore we have

$x^2+7x+10= (x+2)(x+5)$

Check this answer by expanding the brackets with FOIL or otherwise to confirm that the expression on the left hand side is equivalent to the expression on the right hand side.

Example 2

$x^2+3x-10 =(x+p)(x+q)$

We need to find numbers $p$ and $q$ such that

$p\times q = -10 \quad \text{and}\quad p+q=3$

The values $-2$ and $5$ work here because $-2\times 5 = -10$ and $-2 + 5 = 3$ .

Therefore we have

$x^2+3x-10= (x-2)(x+5)$

Example 3: Difference of two squares

Check that this works by multiplying out the brackets:

$x^2-16 =(x-4)(x+4)$

In general

$x^2-p^2=(x-p)(x+p)$

Or, to be more general

$A^2-B^2=(A-B)(A+B)$

Difference of two squares questions

$25x^2-16$
$4y^2-9x^2$
$x^4-16$

Show Answers

Practice when a = 1

applet link

When the coefficient a is not 1: easy

These can also be quite straightforward if there turns out to be a common factor in all three terms, or if it can be understood as the difference between two squares, or if there is no constant term and $x$ is a common factor.

Example 4: No constant term

$4x^2-16x$

The highest common factor of these two terms is $4x$ .

$4x^2-16x=4x(x-4)$

Example 5: There is a common factor between all three terms

$2x^2-10x+8$

First we notice that 2 divides all three terms. Let’s factor it out:

$2x^2-10x+8=2(x^2-5x+4)$

Now, within the brackets, we have a trinomial with $a=1$ and so we simply find a factor pair that multiplies to 4 and adds to -5. I’m thinking -4 and -1 will do it!

$2x^2-10x+8=2(x^2-5x+4)=2(x-4)(x-1)$

When the coefficient a is not 1: not so easy

If none of these apply, the factoring can be quite tricky.

Here are four methods:

factoring by decomposition
factoring using the grid method
factoring using an auxiliary equation (in progress)
factoring be limited trial on error, described below.

Factoring by Limited Trial on Error

First, remember that factoring is about efficiency. If some time has been spent and no factors have been found, we may use an alternative technique (graphing, quadratic formula, completing the square…).

Here is how to do it by limited trial on error, that is, by testing the limited number of possibilities until a solution is found:

Example 6:

$3x^2-11x-20$

We need to examine the factor pairs of 3 and of 20: (1,3) is the only factor pair of 3. Factor pairs of 20 include (1,20); (2,10); (4,5)

We can begin

$3x^2-11x-20 = (3x \dots)(x \dots)$

as (3,1) is the only factor pair of 3. Considering FOIL, the ‘F’ has now been dealt with. As we are only going to insert a factor pair of 20 the ‘L’ will also be dealt with.

We only need to consider the ‘O’ and the ‘I’. These must add to -11. We have three factor pairs to consider, knowing that one of the pairs must be negative to yield $-20$ as the last term.

Let’s try the factor pair (1,20) in all configurations:

$\begin{align*} &(3x +1)(x-20)=3x^2-59x-20 \dots \text{no!}\\[10 pt] &(3x -1)(x+20)=3x^2+59x-20 \dots \text{no!}\\[10 pt] &(3x -20)(x+1)=3x^2-17x-20 \dots \text{also no!}\\[10 pt] &(3x +20)(x -1)=3x^2+17x-20 \dots \text{alas, no!}\end{align}$

Let’s try the factor pair (2, 10) in all configurations but this time, just consider the O and the I terms, and we need not swap the signs as both give $\pm$ the same value:

$\begin{align*} &(3x +2)(x-10)= \dots -30x+2x \dots \text{no!}\\[10 pt] &(3x +10)(x -2)= \dots -6x+10x \dots \text{also no!}\end{align}$

Now the factor pair (4,5):

$\begin{align*} &(3x +4)(x-5)= \dots -15x+4x \dots \text{yes!}\\[10 pt] &(3x +5)(x -4)= \dots -12x+5x \dots \text{(didn't need to do this)}\end{align}$

Therefore $3x^2-11x-20=(3x+4)(x-5)$ .

Although this might seem tedious to go through the factor pairs, with a little practice one gets quite quick at checking the possibilities. If you have good multiplication table recall – or if you want to have better recall (use a printed multiplication table to help not a calculator), this is a good method for you.

Practice Factoring

Kuta software worksheet a=1; Kuta software worksheet a not 1

Applet link

Practice to find zeros (roots)

As we saw on the page Factored form to graph, the factored form leads us quickly to the $x$ -intercepts of a parabola – if the $x$ -intercepts are integers or reasonably simple fractions. Use this applet to practice finding the roots of the parabola by factoring the quadratic expression.