Splitting and Adding Terms · Little Blue Book, Visualized

What this move does

The earlier methods all depend on seeing a common factor or a formula. But some expressions are one step short of a formula. Take $x^{4} + 4$ : there is $(x^{2})^{2}$ and there is $2^{2}$ , but the middle term $2 \cdot x^{2} \cdot 2 = 4 x^{2}$ is missing.

So add it yourself, and take it away again: something out of nothing.

x^{4} + 4 = x^{4} + 4 x^{2} + 4 - 4 x^{2} = (x^{2} + 2)^{2} - (2 x)^{2} = (x^{2} + 2 x + 2) (x^{2} - 2 x + 2)

The two gray terms add up to $0$ , so the value has not changed at all, but the shape has: the first three terms are a perfect square, and the $4 x^{2}$ being subtracted is exactly $(2 x)^{2}$ . The result is a difference of squares.

See what is missing: which term is needed to make a perfect square (or another formula)?
Add and subtract the same term (adding a term), or split an existing term into two (splitting a term).
Regroup and use a formula or take out a common factor. Finally, multiply back to check.

Predict first

To turn $x^{4} + x^{2} + 1$ into $(x^{2} + 1)^{2}$ , how many $x^{2}$ should you add (and subtract)?

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

Work through one

Example: factor x⁴ + 4

Step 1 / 6

The middle term is missing
There is $x^{4}$ and there is $4$ . To make a square of side $x^{2} + 2$ , $4$ strips of $x^{2}$ are still missing (the empty places in the picture).

$x^{4} + 4$
Try adding two
Add four
What is owed is a square too
Use the difference of squares
Does the other direction work?

Splitting a term

Splitting a term means breaking a term that is already there into two, again so that grouping works.

Example: factor x³ − 3x + 2

Step 1 / 5

Three terms, no common factor, no formula
Try splitting the middle term $- 3 x$ . Into what? Into pieces that let the first two terms share a factor and show a familiar formula.

$x^{3} - 3 x + 2$
Split it into −x and −2x
Group in pairs
A common factor appears
Factor completely

When it fails

You cannot split a term just any way. Split the middle term of $x^{2} + 5 x + 6$ into two terms and group them: which splits work?

Break it

Counterexample lab: remove one condition

Conclusion: After splitting, the groups fit together

$5 x = 2 x + 3 x$ :

Practice

Exercise 1Basic

Factor $x^{4} + 64$ .

Exercise 2Advanced

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

Exercise 3Advanced

Factor $x^{4} - 7 x^{2} + 9$ .

Exercise 4Challenge

Factor $x^{3} - 7 x + 6$ by splitting a term.

Mistake Museum · Exhibits from this stop

Before you leave, look at the traps other people fell into here.See every exhibit in the museum →

Exhibit

Adding without subtracting

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

What went wrong? Think first, then open

Diagnosis An extra $4 x^{2}$ appeared out of nowhere, so the value changed. The first equals sign is already false: at $x = 1$ the left side is $5$ and the right side is $9$ .

When you add

4 x^{2}

you must also subtract

4 x^{2}

x^{4} + 4 = (x^{2} + 2)^{2} - 4 x^{2} = (x^{2} + 2 x + 2) (x^{2} - 2 x + 2)

“Whatever you add, you must subtract”

Exhibit

Giving up as soon as you see a sum of squares

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

is a sum of squares, so it cannot be factored.

What went wrong? Think first, then open

Diagnosis “ $a^{2} + b^{2}$ cannot be factored” is about $a$ and $b$ treated as letters. Here $2 ab = 4 x^{2}$ happens to be a perfect square, so adding it and subtracting it produces a difference of squares.

First ask: is

2 ab

the square of something? If so, add a term.

x^{4} + 4 = (x^{2} + 2)^{2} - (2 x)^{2}

“A sum of squares can hide a difference of squares”