How to Factor Out a Common Factor

The distributive property turns $a (b + c)$ into $a b + a c$ . This lesson runs that rule backward. Reading $a b + a c$ as $a (b + c)$ is called factoring, and the first problems are pure arithmetic, with no letters in them at all.

Problem

Compute

57 \cdot 69 + 57 \cdot 31

without working out either product. Both terms contain the factor

57

, so pull it out front and see what the leftover sum becomes.

Show a hint

The first term is $57$ copies of $69$ , and the second is $57$ more copies of $31$ . Count copies before multiplying anything.

Show the full solution

Both terms share

57

, so pull it out front.

57 \cdot 69 + 57 \cdot 31 = 57 (69 + 31) = 57 \cdot 100 = 5700 .

Counting copies beats multiplying here, since the leftovers add to a round

100

Problem

Compute

58 \cdot 91 - 58 \cdot 81

the same way. The shared factor comes out front, and this time a subtraction is left behind it. What is the value of the whole expression?

Show a hint

The first product is $58$ copies of $91$ and the second is $58$ copies of $81$ , so subtracting leaves $58$ copies of some difference.

Show the full solution

Pull the shared

58

out front and the subtraction stays inside.

58 \cdot 91 - 58 \cdot 81 = 58 (91 - 81) = 58 \cdot 10 = 580 .

The long way,

5278 - 4698

, reaches the same

580

with far more work.

Two readings of one equality. Expanding opens the parentheses and factoring rebuilds them, so

3 (y + 8)

and

3 y + 24

are the same value written two ways. Below, twenty loose tiles become four equal groups of

2 + 3

, which is what

4 (2 + 3)

records.

Problem

Letters change nothing. The expression

4 w + 36

factors as

4 (w + c)

for one constant

c

. Divide each term by the shared

4

, then enter

c

Show a hint

Both terms get divided by the $4$ , not just the first one.

Show the full solution

Divide both terms by

4

. Since

4 w \div 4 = w

and

36 \div 4 = 9

, the factoring is

4 (w + 9)

, so

c = 9

. Expanding

4 (w + 9)

lands back on

4 w + 36

. Dividing only the first term gives

4 (w + 36)

, which expands to

4 w + 144

instead.

Problem

Dana factors

12 h + 30

2 (6 h + 15)

, and expanding it back checks out. But inside the parentheses,

6 h

and

15

still share a factor, so the job is unfinished. Enter the largest number that divides both

12 h

and

30

Show a hint

What is the largest number that divides both $12$ and $30$ ?

Show the full solution

Since

\gcd (12, 30) = 6

, the largest number dividing both terms is

6

, and the complete factoring is

6 (2 h + 5)

. Dana's

2 (6 h + 15)

expands back correctly, so passing the round trip does not mean the job is finished. Check whether the leftovers still share a factor.

Problem

A letter can be the shared factor. Since

y^{2}

means

y \cdot y

(1.5), both terms of

y^{2} + 10 y

contain

y

, so the expression factors as

y (y + c)

. Enter

c

Show a hint

Divide each term by $y$ and see what each one leaves behind.

Show the full solution

Divide each term by

y

. Since

y^{2} \div y = y

and

10 y \div y = 10

, the factoring is

y (y + 10)

, and the round trip

y (y + 10) = y^{2} + 10 y

confirms

c = 10

. The terms share no number bigger than

1

, but a shared letter is a common factor too.

Problem

Number and letter combine. Enter the greatest common factor of the two terms of

8 t^{2} + 20 t

, typed like a term, for example

6 z

Show a hint

Find the number part and the letter part separately, then put them together.

Show the full solution

The number part is

\gcd (8, 20) = 4

, and both terms contain one

t

, so the greatest common factor is

4 t

. Pulling it out gives

4 t (2 t + 5)

. Answering just

4

or just

t

stops halfway, since the shared factor carries a number and a letter.

Problem

A student factors

8 n + 36

4 (2 n + 8)

. Test it the 1.4 way, evaluate the original and the claimed factoring at

n = 1

, each on its own. How far apart are their values?

Show a hint

Work out each expression at $n = 1$ separately, then subtract the results.

Show the full solution

n = 1

the original gives

8 + 36 = 44

and the claimed factoring gives

4 (2 + 8) = 40

, so the two are

4

apart. Expanding catches it faster, since

4 (2 n + 8) = 8 n + 32

rather than

8 n + 36

. The

4

does divide both terms, but the leftover constant is wrong.

Problem

Three terms this time. The expression

9 a + 27 b - 18

factors as

9 (a + 3 b + c)

, and each term keeps its sign as the

9

comes out. Enter

c

, sign included.

Show a hint

What is $- 18$ divided by $9$ , sign and all?

Show the full solution

The third term is

- 18

, and

- 18 \div 9 = - 2

, so

9 a + 27 b - 18 = 9 (a + 3 b - 2)

and

c = - 2

. Each leftover keeps the sign of the term it came from, so dropping the minus would break the expansion.

Problem

m (m - 6) + 8 (m - 6)

, both terms contain the whole factor

m - 6

. Pull the chunk out front exactly the way the

57

came out in this lesson's first problem. The result is

(m + a) (m - 6)

. Enter

a

Show a hint

Treat $m - 6$ as a single object, like the shared $57$ in the hook problem.
The first term is $m$ copies of $m - 6$ , and the second term adds some more copies of the same thing. Count the total number of copies.

Show the full solution

Treat

m - 6

as one object. The first term is

m

copies of it and the second is

8

copies, which is

m + 8

copies in all, so the expression equals

(m + 8) (m - 6)

and

a = 8

. Expanding the two products into four separate terms first would erase the shared chunk, and chapter 9 factors quadratics with this same move.

Factoring reads distribution backward, the round trip checks it, completeness finishes it, and even a whole parenthesized chunk can be the shared factor. Next, 2.4 adds and subtracts entire expressions, and the full sign flip from distributing $- 1$ in 2.1 becomes the main skill.

Practice these ideas

Practice

Compute

34 \cdot 67 + 34 \cdot 33

without working out either product. Factor out the shared

34

first and see what sum is left behind it.

Show the solution

Both terms contain

34

, so the sum is

34 (67 + 33) = 34 \cdot 100 = 3400

. Grinding out

2278 + 1122

gets there too, just slowly.

Practice

Compute

76 \cdot 53 - 76 \cdot 43

by factoring first. Both products are ugly, but the difference left behind the shared factor is not.

Show the solution

Pull the shared

76

out front, since

76 (53 - 43) = 76 \cdot 10 = 760

. Subtracting inside the parentheses is far less work than computing

4028 - 3268

Practice

The expression

5 q + 60

factors as

5 (q + c)

for one constant

c

. Divide each term by the shared

5

, then enter

c

Show the solution

Divide each term by

5

. Since

5 q \div 5 = q

and

60 \div 5 = 12

, the factoring is

5 (q + 12)

, which expands back to

5 q + 60

, so

c = 12

. Writing

5 (q + 60)

divides only the first term and fails the round trip.

Practice

Enter the largest number that divides both terms of

38 x + 95

. Neither number is small, so factor each one before searching for what they share.

Show the solution

Since

38 = 2 \cdot 19

and

95 = 5 \cdot 19

, the largest common factor is

19

and the complete form is

19 (2 x + 5)

. Guessing small divisors stalls here, since

2

misses

95

and

5

misses

38

. Factor each number first instead.

Practice

Ravi factors

44 h + 66

11 (4 h + 6)

, and the expansion checks out. Even so,

4 h

and

6

still share a factor, so the factoring is not complete. Enter the largest common factor of the two original terms.

Show the solution

Since

\gcd (44, 66) = 22

, the largest common factor is

22

, and the complete form

22 (2 h + 3)

expands back to

44 h + 66

. Ravi's

11 (4 h + 6)

also expands correctly, but its leftovers

4 h

and

6

still share

2

, so it is not finished.

Practice

The shared factor is a letter this time. Since

n^{2}

n \cdot n

, both terms of

n^{2} + 13 n

contain

n

, so the expression factors as

n (n + c)

. Enter

c

Show the solution

The terms share no number, but both contain

n

. Dividing each by

n

leaves

n + 13

, and the round trip

n (n + 13) = n^{2} + 13 n

confirms

c = 13

Practice

Number and letter both matter here. Enter the greatest common factor of the two terms of

15 u^{2} + 25 u

, typed like a term, for example

8 w

Show the solution

The number part is

\gcd (15, 25) = 5

, and both terms contain

u

, so the greatest common factor is

5 u

. Pulling it out gives

5 u (3 u + 5)

. Entering only

5

or only

u

stops halfway.

Practice

The expression

26 x + 39 y - 52

factors as

13 (2 x + 3 y + e)

. Enter

e

, sign included.

Show the solution

Since

- 52 \div 13 = - 4

, the factoring is

13 (2 x + 3 y - 4)

, which expands back to

26 x + 39 y - 52

, so

e = - 4

. Each leftover keeps the sign of its own term, so dropping the minus changes the expression.

Practice

Evaluate

\frac{7 \cdot 93 + 7 \cdot 7}{10}

without any long multiplication. The top is a sum of two products with a shared factor, so factor it before dividing anything.

Show the solution

Factor the top first, since

7 \cdot 93 + 7 \cdot 7 = 7 (93 + 7) = 7 \cdot 100 = 700

, and

700 \div 10 = 70

. The shared

7

turns two ugly products into one round number.

Practice

Both terms of

d (d + 5) + 3 (d + 5)

contain the whole chunk

d + 5

, so it can come out front like any shared factor. Factoring it out gives

(d + a) (d + 5)

. Enter

a

Show the solution

Count copies of the chunk

d + 5

. The first term is

d

copies and the second is

3

more, so there are

d + 3

copies in all, giving

(d + 3) (d + 5)

and

a = 3

. The tempting answer

5

sits inside the shared chunk, not in the leftover factor.

Practice

One more full scan, this one on

6 t^{2} + 9 t

. Enter the greatest common factor of its two terms, typed like a term. Check that the leftovers share nothing before you commit.

Show the solution

Since

\gcd (6, 9) = 3

and both terms contain

t

, the greatest common factor is

3 t

and the complete form is

3 t (2 t + 3)

. The leftovers

2 t

and

3

share only

1

, so nothing is left to pull. Stopping at

3

would leave the shared

t

behind.

Practice

Evaluate

4 1^{2} + 41

by factoring before you compute anything. Done right, one clean product replaces a square plus an addition.

Show the solution

Both terms contain

41

, so

4 1^{2} + 41 = 41 (41 + 1) = 41 \cdot 42 = 1722

. The shape is general, since

n^{2} + n

always factors as

n (n + 1)

, a product of two consecutive whole numbers.

Factoring Out a Common Factor

Practice these ideas