Taking Out a Common Factor · Little Blue Book, Visualized

What this move does

It is the distributive law, used backwards:

m a + m b + m c = m (a + b + c)

The workbench above draws each term as the area of a rectangle. The common factor $m$ is their shared height. Rectangles of the same height can be put side by side to make one big rectangle: its height is the common factor, and its width is the sum in the bracket.

Coefficients: take the greatest common divisor of the coefficients.
Letters: take each letter that appears in every term, to its lowest power.
Put their product outside the bracket. Inside, write each term divided by the common factor. The number of terms stays the same.

Predict first

After taking the common factor $2 x$ out of $4 x^{2} + 2 x$ , how many terms are in the bracket?

1

2

3

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

Work through one

Example: factor 12x³ + 18x²

Step 1 / 5

Two areas
Think of $12 x^{3}$ and $18 x^{2}$ as two areas. Before taking anything out, the height is just $1$ , so both pieces are long and flat.

$12 x^{3} + 18 x^{2}$
Coefficients first
Then the letters
Join them
Check two things

A common factor can also be a whole bracket. For example,

x (a - b) + y (a - b) = (a - b) (x + y) .

Treat $(a - b)$ as a single letter $m$ , and it is exactly the same as above.

When it fails

Break it

Counterexample lab: remove one condition

Conclusion: You can take out a common factor

$6 x^{2} + 9 x$ : the coefficients share the factor $3$ , and every term contains $x$ . Plenty to take out.

Practice

Exercise 1Basic

Factor $8 x^{3} + 12 x$ .

Exercise 2Basic

Factor $6 x^{3} - 4 x^{2} + 2 x$ .

Exercise 3Advanced

Factor $6 a^{2} b - 9 a b^{2} + 3 ab$ .

Exercise 4Challenge

Factor $x (a - b) + y (b - a)$ .

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

Losing the 1

3 x^{2} + 3 x = 3 x (x)

3 x^{2} + 3 x = 3 x \cdot x

What went wrong? Think first, then open

Diagnosis The second term $3 x$ is taken out completely, but what is left is not “nothing”: it is $3 x \div 3 x = 1$ . Drop it, and multiplying back gives only $3 x^{2}$ : a term is missing.

3 x^{2} + 3 x = 3 x (x + 1)

. The bracket has as many terms as the original.

“When a whole term is taken out, a 1 stays behind”

Exhibit

Not taking everything out

12 x^{3} + 18 x^{2} = 6 x (2 x^{2} + 3 x)

What went wrong? Think first, then open

Diagnosis In the bracket, $2 x^{2}$ and $3 x$ still share the factor $x$ . Take out the largest common factor: the gcd of the coefficients, and each letter to its lowest power.

12 x^{3} + 18 x^{2} = 6 x^{2} (2 x + 3)

. When you are done, look inside the bracket again: is there still a common factor?

“After taking it out, look inside the bracket once more”

Exhibit

Taking out a minus sign and forgetting to change signs

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

What went wrong? Think first, then open

Diagnosis Taking out $- 2 x$ means dividing every term by $- 2 x$ : $4 x \div (- 2 x) = - 2$ , not $+ 2$ . Multiply back: $- 2 x \cdot 2 = - 4 x$ , the wrong sign.

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

. When you take out a minus sign, every term in the bracket changes sign.

“Take out a minus sign, and every sign in the bracket flips”