The Cross Method · Little Blue Book, Visualized

What this move does

Multiply out two linear factors and look:

(x + p) (x + q) = x^{2} + (p + q) x + pq

The constant term is the product of two numbers, and the coefficient of the middle term is the sum of the same two numbers. The cross method runs this backwards:

Look at the constant $c$ and list every pair of integers whose product is $c$ .
Among those pairs, find the one whose sum is exactly the middle coefficient $b$ .
Write down $(x + p) (x + q)$ , then multiply it back to check.

The rectangle on the right of the workbench says the same thing: its length is $x + p$ and its width is $x + q$ , so its area is made of $1$ big $x^{2}$ square, $p + q$ strips of size $x$ and $pq$ small unit squares. Only the right pair makes the pieces fit exactly.

Predict first

Suppose $x^{2} + b x + 12$ factors as $(x + p) (x + q)$ , where $p$ and $q$ are both positive integers. How many possible values of $b$ are there?

2

3

4

6

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

Work through one

When the constant is negative there is one extra check: the two numbers must have opposite signs.

Example: factor x² − x − 12

Step 1 / 5

Check the sign of the constant
The constant is $- 12$ , which is negative, so the two numbers we want have opposite signs. Put any pair up to start with.

$x^{2} - x - 12$
Now look at the middle term
Narrow it down
Got it
Write the answer and multiply back

When the leading coefficient is not 1

Now the first term has to be split as well. To get $a x^{2} + b x + c = (m x + p) (n x + q)$ we need $mn = a$ and $pq = c$ , and the middle term comes from multiplying across and adding:

m \cdot q + n \cdot p = b

So the same $p$ and $q$ give a different result if you swap them top to bottom. Try $3 \times 1$ and $1 \times 3$ in the machine below and compare.

Factor

2 x^{2} + 7 x + 3

. Which two numbers does the solid line join, and which does the dashed line join?

When no pair works

Break it

Counterexample lab: remove one condition

Conclusion: $x^{2} + b x + c$ factors with integers by the cross method

$x^{2} + 7 x + 10$ : $7^{2} - 4 \times 10 = 9 = 3^{2}$ is a perfect square. Click through the pairs: one of them is bound to work.

Practice

Exercise 1Basic

Factor $x^{2} + 9 x + 14$ .

Exercise 2Basic

Factor $x^{2} - 3 x - 28$ .

Exercise 3Advanced

Factor $3 x^{2} - 10 x + 8$ .

Exercise 4Challenge

Factor $x^{4} - 13 x^{2} + 36$ completely (until no factor can be split any further).

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

Matching the sum, forgetting the sign of the product

x^{2} - 5 x - 6 = (x - 2) (x - 3)

, because

- 2 - 3 = - 5

What went wrong? Think first, then open

Diagnosis The sum really is $- 5$ , but $(- 2) \times (- 3) = + 6$ : the constant has the wrong sign. The constant term is negative, so the two numbers must have opposite signs.

Look at the constant

- 6

first: opposite signs. Then find a pair that differs by

5

- 6

and

+ 1

. So

x^{2} - 5 x - 6 = (x - 6) (x + 1)

“Check the sign of the product before you hunt for the sum”

Exhibit

Adding straight across when the leading coefficient is not 1

2 x^{2} + 5 x + 3

: split as

2 \cdot 1

and

1 \cdot 3

and write

(2 x + 1) (x + 3)

What went wrong? Think first, then open

Diagnosis Multiply back and the middle term is $2 x \cdot 3 + 1 \cdot x = 7 x$ , not $5 x$ . When the leading coefficient is not $1$ , where each number sits changes what the cross products add up to.

Swap the

1

and the

3

(2 x + 3) (x + 1)

, whose cross products add to

2 \cdot 1 + 1 \cdot 3 = 5

. Always multiply back to check.

“It is a cross, so the positions are part of the answer”