Putting It All Together

In one sentence

Factorization is not luck. Given an expression, ask yourself a few questions in a fixed order. After each step, ask them all again, from the top, for every factor.

First take out a common factor, then try a formula, then the cross method, then grouping. If none of them works, think of splitting or adding terms, or a substitution. Finally, check all the way down.

The flowchart above is this rule of thumb. Only two things matter: the order and asking all the way down.

Predict first

To factor $3 x^{2} - 12$ , what should the first step be?

Use the difference of squaresTake out the common factor

3

Use the cross method

Choose an answer first; then you can see the result.

One expression, three rounds of questions

Step 1 / 4

First question: is there a common factor?
Both terms of $x^{5} - 16 x$ contain $x$ . Whatever formula it might look like later, take that out first.

$x^{5} - 16 x$
Done? Ask again from the top
Use the formula, then ask again
All the way down

Stop 1 · Treat a block as one letter

When a block keeps appearing in an expression, give it a name and the whole expression shrinks.

Example: factor (x² + x)² − 8(x² + x) + 12

Step 1 / 4

Find the repeated block
$x^{2} + x$ appears twice. Do not rush to expand: that gives a quartic, which is harder.

$(x^{2} + x)^{2} - 8 (x^{2} + x) + 12$
Give it a name
Substitute back
Check all the way down

This is substitution: treat a part as a whole, and the problem drops from degree four to degree two.

Stop 2 · More than one road

One problem, several solutions

Factor $x^{6} - 1$ .

Start with a difference of squares

$x^{6} - 1 = (x^{3})^{2} - 1 = (x^{3} - 1) (x^{3} + 1)$

Then use the difference and sum of cubes:

$(x - 1) (x^{2} + x + 1) (x + 1) (x^{2} - x + 1)$

CostBoth steps are straight formula work: the least effort.

Start with a difference of cubes

$x^{6} - 1 = (x^{2})^{3} - 1 = (x^{2} - 1) (x^{4} + x^{2} + 1)$

$x^{2} - 1 = (x - 1) (x + 1)$ ; and $x^{4} + x^{2} + 1 = (x^{2} + 1)^{2} - x^{2} = (x^{2} + x + 1) (x^{2} - x + 1)$ .

The same result.

CostAlong the way you meet $x^{4} + x^{2} + 1$ , which needs an added term to get any further.

Both roads end in the same place, because the complete factorization is unique. But the effort can be very different. As a rule, squares first, cubes second: the degree drops faster.

Stop 3 · Grouping needs the big picture too

$x^{5} + x^{4} + x^{3} + x^{2} + x + 1$ has six terms. You can group them in threes, or in pairs:

x^{4} (x + 1) + x^{2} (x + 1) + (x + 1) = (x + 1) (x^{4} + x^{2} + 1) .

The $x^{4} + x^{2} + 1$ that is left is exactly the expression from Stop 2. Problems you have solved become tools: recognize it, and you do not need to think it through again.

Practice

Exercise 1Basic

Factor $3 x^{3} - 27 x$ .

Exercise 2Advanced

Factor $x^{4} - 10 x^{2} + 9$ .

Exercise 3Advanced

Factor $(x^{2} + 2 x)^{2} - 2 (x^{2} + 2 x) - 3$ .

Exercise 4Challenge

Factor $x^{5} + x^{4} + x^{3} + x^{2} + x + 1$ .

In one sentence

One expression, three rounds of questions

Stop 1 · Treat a block as one letter

Example: factor (x² + x)² − 8(x² + x) + 12

Stop 2 · More than one road

Start with a difference of squares

Start with a difference of cubes

Stop 3 · Grouping needs the big picture too

Practice

Mistake Museum · Exhibits from this stop

Using a formula before taking out the common factor

Not substituting back after a substitution, or not factoring completely