Factoring Tricks#

This page is designed to walk you through all of the factoring tricks we’ll need. Each tab discusses a new factoring trick, and you can find an exercise generator at the bottom of each tab. Practice with each type of problem until you’re comfortable with the relevant trick, and then move on to the next tab!

Note that if we FOIL the expression \((x+m)(x+n)\), then we get:

\[(x+m)(x+n) = x^2 + nx + mx + mn = x^2 + \underbrace{(m+n)}_{b} x + \underbrace{(mn)}_{c}.\]

When factoring, we want to reverse this pattern. That is, given an \(x^2 + bx + c\), we’d like to find the numbers \(m\) and \(n\) so that \(x^2 + bx + c = (x+m)(x+n)\). But to go from the \(ax^2 + bx + c\) form back to the factored form \((x + m)(x + n)\), we need a few things to work out:

  • The leading coefficient \(a\) needs to be 1. In other words, this trick only works on things like \(x^2 + bx + c\), not on \(7x^2 + bx + c\) or \(-2x^2 + bx + c\).

  • We’re looking for two numbers, \(m\) and \(n\), but they have to add up/multiply to the right things:

    • The sum of \(m\) and \(n\) needs to be \(b\)

    • The product of \(m\) and \(n\) needs to be \(c\).

In other words, we’re looking for factors of \(c\) that add up to \(b\).

If we can find numbers \(m, n\) with \(m\cdot n = c\) and \(m + n = b\), then the original function will factor like \(x^2 + bx + c = (x + m)(x + n)\).

To make things more concrete, here’s an example.

Example 1

Let’s try to factor \(x^2 + 5x + 6\). This is a quadratic function in general form with \(a = 1\), \(b = 5\), and \(c = 6\). Since \(a = 1\), we can try out the factoring trick above. We just need to find factors of \(c = 6\) that add up to \(b = 5\).

The factors of \(6\) that add up to \(5\) are \(2, 3\) (when you multiply them, you get 6, and when you add them, you get 5).

Therefore, \(x^2 + 5x + 6 = (x + 2)(x + 3)\). We can FOIL the right-hand side to check our answer, if we want.

Exercise (Random Generator)

Click the Randomize button to start!

Reroll

Hint

In this section, we’ll discuss an extension of the trick for factoring monic quadratics. What happens if \(a \neq 1\)?

flowchart LR A["`Find factors $m$, *n* of *ac* that add up to *b*.`"] --> B[Split up middle term *bx* into *mx + nx*.] B --> C[Factor by grouping.]

Fixes and more coming soon!

Exercise (Random Generator)

Click the Randomize button to start!

Reroll

Hint

Here we’ll discuss factoring tricks involving a difference of squares or a sum/difference of cubes.

Differences of squares can be factored using:

(1)#\[A^2 - B^2 = (A - B)(A + B).\]

Here’s an example.

Example 2

Let’s try to factor \(f(x) = 9x^2 - 16\).

The first thing to notice is that both \(9\) and \(16\) are squares: \(3^2\) and \(4^2\). In fact, we can rewrite the original \(f(x)\) as a difference of squares.

\[f(x) = 9x^2 - 16 = (3x)^2 - (4)^2.\]

Now, applying the factoring trick (1) above, we get \(f(x) = (3x)^2 - (4)^2 = (3x - 4)(3x + 4)\).

Exercise (Random Generator)

Click the Randomize button to start!

Reroll

Hint

There are also factoring tricks for sums/differences of cubes.

(2)#\[\begin{split}A^3 + B^3 = (A + B)(A^2 - AB + B^2) \\\\ A^3 - B^3 = (A - B)(A^2 + AB + B^2)\end{split}\]

Here’s an example.

Example 3

Let’s try to factor \(f(x) = 27x^3 + 64\).

The first thing to notice is that \(27\) and \(64\) are cubes: \(27 = 3^3\) and \(64 = 4^3\). In fact, we can rewrite the original \(f(x)\) as a sum of cubes.

\[f(x) = 27x^3 + 64 = (3x)^3 + (4)^3.\]

Now, applying the factoring trick (2) above (specifically, the top formula, since this is a sum of cubes), we get

\[\begin{split} \begin{align*} f(x) = (3x)^3 + (4)^3 &= (3x + 4)((3x)^2 - (3x)(4) + (4)^2) && \text{(factoring trick)}\\ &= (3x + 4)(9x^2 - 12x + 16). && \text{(clean up)} \end{align*} \end{split}\]

Try some for yourself.

Exercise (Random Generator)

Click the Randomize button to start!

Reroll

Hint

This exercise randomizer combines all of the randomizers from the previous tabs. Are you ready to test yourself on all of the factoring tricks?

Exercise (Random Generator)

Click the Randomize button to start!

Reroll

Hint

Warning

The randomized solutions aren’t smart enough (yet?) to spot common factors. For example, if you were asked to factor \(9x^2 - 9\), it’s best to start by factoring out the common factor of \(9\) to get \(9(x^2 - 1)\), and then you can finish factoring the \(x^2 - 1\) part using the difference of squares trick.

The solutions for the randomizer will ignore any common factors like this - they’re only going to use the factoring tricks provided on this page.