Factoring x² + bx + c

Every product $(x + r) (x + s)$ expands to $x^{2} + (r + s) x + r s$ , so factoring $x^{2} + b x + c$ means finding two numbers with sum $b$ and product $c$ . Those are the two clues, and the product is the sharper one, since $c$ has only finitely many factor pairs, so the search starts there.

Problem

Two numbers have sum

15

and product

56

. The product is the clue to search on, so list the factor pairs of

56

and check each pair's sum. Enter the larger of the two numbers.

Show a hint

Starting from the product gives a shorter search. Only four pairs of whole numbers multiply to $56$ , so list those pairs.
The pairs are $1 \cdot 56$ , $2 \cdot 28$ , $4 \cdot 14$ and $7 \cdot 8$ . Check each sum against $15$ .

Show the full solution

The factor pairs of

56

are

1 \cdot 56

2 \cdot 28

4 \cdot 14

and

7 \cdot 8

, with sums

57

30

18

and

15

, so the numbers are

7

and

8

, and the larger is

8

. Starting from the product kept the search short, since only four pairs needed checking.

Problem

Factor

x^{2} + 8 x + 12

(x + r) (x + s)

. Both constants are positive, since the product and the sum are both positive. Enter the larger of the two.

Show a hint

Two numbers with sum $8$ and product $12$ . Run the factor pairs of $12$ .
The pairs of $12$ are $1 \cdot 12$ , $2 \cdot 6$ and $3 \cdot 4$ . One sum is $8$ .

Show the full solution

The pairs of

12

have sums

13

8

and

7

, so the numbers are

2

and

6

, giving

x^{2} + 8 x + 12 = (x + 2) (x + 6)

, and the larger constant is

6

. Expanding back gives

x^{2} + 8 x + 12

, the check from 9.1.

Problem

Factor

x^{2} + 3 x - 28

. The product of the two constants is

- 28

, so they have opposite signs, and since the sum is

+ 3

, the positive one is larger in size. Search the pairs of

28

with opposite signs. Enter the negative constant.

Show a hint

You need a factor pair of $28$ whose difference is $3$ , with signs chosen so the sum comes out positive.
The pairs of $28$ are $1 \cdot 28$ , $2 \cdot 14$ and $4 \cdot 7$ . Which pair has difference $3$ ?

Show the full solution

The pair with difference

3

4

and

7

, and the sum must be

+ 3

, so the positive number is

7

, giving

x^{2} + 3 x - 28 = (x + 7) (x - 4)

, and the negative constant is

- 4

. A quick expand gives

7 x - 4 x = 3 x

and

7 \cdot (- 4) = - 28

, matching the middle term and the constant.

Problem

Factor

x^{2} - 13 x + 22

. The product

+ 22

is positive, so the two constants share a sign, and since the sum

- 13

is negative, that shared sign is negative. Search the factor pairs of

22

, both taken negative. Enter the smaller of the two constants.

Show a hint

$22$ has only two factor pairs, $1 \cdot 22$ and $2 \cdot 11$ , and both numbers are negative here.
Which pair, made negative, sums to $- 13$ ? Then take the one farther below zero.

Show the full solution

The negative pairs are

- 1, - 22

with sum

- 23

and

- 2, - 11

with sum

- 13

, so

x^{2} - 13 x + 22 = (x - 2) (x - 11)

, and the smaller constant is

- 11

. Writing

(x + 2) (x + 11)

is the standard slip, since its expansion has

+ 13 x

, not

- 13 x

The rectangle from 9.1, read the other way. Its four cells hold the expansion,

x^{2}

r x

s x

and

r s

, and the gold side lengths

x + r

and

x + s

are what factoring recovers. The inside of the rectangle is the expanded form, the sides are the factors, and both clues are visible here, the cross cells add up to

b x

and the corner cell is

c

The expansion $x^{2} + (r + s) x + r s$ has leading coefficient $1$ , so the two clues are only trustworthy on a quadratic whose $x^{2}$ coefficient is also $1$ . Before any search, pull out the greatest common numeric factor and write it in front, where it stays. When there is no constant term, $x$ itself is a common factor.

Problem

Every term of

4 x^{2} + 52 x + 120

is divisible by the same number. Pull out that greatest common numeric factor, leaving a quadratic whose

x^{2}

coefficient is

1

, then finish the factoring. Enter the numeric factor you pulled out.

Show a hint

Find the largest number that divides $4$ , $52$ and $120$ evenly.
After it comes out, the search runs on $x^{2} + 13 x + 30$ .

Show the full solution

The greatest common factor of

4

52

and

120

4

. It comes out first, leaving

x^{2} + 13 x + 30

, which factors as

(x + 3) (x + 10)

, so the full form is

4 (x + 3) (x + 10)

, with the

4

kept in front.

Problem

x^{2} - 14 x

has no constant term, so both terms share the factor

x

itself. Factor the expression as

x

times a linear factor. Enter the constant inside the linear factor.

Show a hint

Write each term with an $x$ pulled out, $x \cdot x$ and $x \cdot (- 14)$ .
The factorization is $x (x + s)$ for one value of $s$ , and expanding it must give back $- 14 x$ .

Show the full solution

x^{2} - 14 x = x (x - 14)

, so the constant inside the linear factor is

- 14

. Expanding back gives

x \cdot x - 14 \cdot x

, the original expression.

When $b = 0$ the sum clue is $r + s = 0$ , so $s = - r$ and the product is $- r^{2}$ . That is why $x^{2} - 49$ factors as $(x + 7) (x - 7)$ , the conjugate identity from 9.1 run backwards, with the cross terms cancelling. The marks to recognize are a missing middle term and a negative perfect square constant.

Problem

Compute

5 3^{2} - 4 7^{2}

without working out either square. The difference of squares pattern applies to plain numbers too, and it turns the subtraction into one easy multiplication. Enter the value.

Show a hint

The conjugate identity from 9.1, $a^{2} - b^{2} = (a + b) (a - b)$ , applies with $a = 53$ and $b = 47$ .
$(53 - 47) (53 + 47) = 6 \cdot 100$ .

Show the full solution

5 3^{2} - 4 7^{2} = (53 - 47) (53 + 47) = 6 \cdot 100 = 600

. Running the identity in reverse replaced two squarings and a subtraction with one easy product, and the direct arithmetic agrees,

2809 - 2209 = 600

Problem

The quadratic

x^{2} + 18 x + 81

is the search's

r = s

case, since

81 = 9^{2}

and

18

is twice

9

, so it factors as

(x + 9)^{2}

. Now change the constant to

82

. Run the full search for

x^{2} + 18 x + 82

and answer yes or no. Does it factor with integer constants?

Show a hint

Both numbers would have to be positive, since the sum and the product are both positive. List every factor pair of $82$ .
$82 = 1 \cdot 82 = 2 \cdot 41$ , and those are all the pairs. Check both sums against $18$ .

Show the full solution

The pairs of

82

are

1 \cdot 82

and

2 \cdot 41

, with sums

83

and

43

, and negative pairs have negative sums, so the search comes up empty and the answer is

no

. An empty search is a proof, since the pairs are all there is to check.

Once the left side of $x^{2} + b x + c = 0$ is factored, it is a product equal to zero, and by the zero product property from 9.1 at least one of the factors is zero. Each factor set to zero is a linear equation. The property applies to a product equal to zero and nothing else, so put zero on one side before splitting anything.

Problem

Solve

x^{2} - 4 x - 45 = 0

. Factor the left side, then the zero product property splits the equation into two linear ones. Enter the positive root.

Show a hint

Two numbers with product $- 45$ and sum $- 4$ . The signs differ, and the negative one is larger in size.
The pair is $- 9$ and $5$ , so the factors are $x - 9$ and $x + 5$ .

Show the full solution

x^{2} - 4 x - 45 = (x - 9) (x + 5)

, so

x = 9

x = - 5

, and the positive root is

9

. Checking,

81 - 36 - 45 = 0

Problem

Solve

x (x + 5) = 24

. The left side is already factored, but the right side is not zero, so neither factor is forced to equal anything. Expand, move every term to one side, and factor again before splitting. Enter the positive root.

Show a hint

The zero product property only works on a product equal to zero. Expand and subtract $24$ first.
$x^{2} + 5 x - 24 = 0$ needs two numbers with sum $5$ and product $- 24$ .

Show the full solution

Expanding gives

x^{2} + 5 x - 24 = 0

, which factors as

(x + 8) (x - 3)

, so

x = - 8

x = 3

, and the positive root is

3

. The check runs against the original equation,

3 \cdot 8 = 24

. Splitting

x (x + 5) = 24

into

x = 24

x + 5 = 24

is exactly the trap 9.1 named.

You can now factor any quadratic that turns into $x^{2} + b x + c$ once a numeric factor comes out, whenever an integer pair exists, and solve the equations built from those factorizations. Two things are still out of reach, a leading coefficient that does not divide out, and a quadratic whose search comes up empty.

Problem

A poster is

6

inches taller than it is wide, and its area is

187

square inches. Writing

w

for the width gives

w (w + 6) = 187

. Rearrange, factor and split. Both roots solve the equation, but only one can be a width. Enter the width in inches.

Show a hint

Expand and move $187$ over, then search for two numbers with sum $6$ and product $- 187$ .
$187 = 11 \cdot 17$ , and $17 - 11 = 6$ .

Show the full solution

w^{2} + 6 w - 187 = 0

factors as

(w + 17) (w - 11)

, so

w = - 17

w = 11

, and a width is a positive length, so the width in inches is

11

. The check is the original statement,

11 \cdot 17 = 187

, an

11

17

inch poster.

Problem

Factor

3 x^{2} - 45 x + 150

completely. The common numeric factor comes out first. The quadratic left behind has a positive constant and a negative middle term, so both constants in its factored form are negative. Enter the smaller of the two.

Show a hint

Divide $3$ out of every term, then search $x^{2} - 15 x + 50$ among negative pairs only.
Which factor pair of $50$ , both taken negative, sums to $- 15$ ?

Show the full solution

Pulling out

3

leaves

x^{2} - 15 x + 50

, and the negative pair with sum

- 15

and product

50

- 5

and

- 10

, so

3 x^{2} - 45 x + 150 = 3 (x - 5) (x - 10)

, and the smaller constant is

- 10

. Dropping the

3

when writing the complete factorization is the standard slip, it stays in front.

Every quadratic here began monic or became monic after a numeric factor came out. When the leading coefficient stays, as in $2 x^{2} + 7 x + 3$ , the sum and product clues are no longer enough on their own. Lesson 9.3, Factoring $a x^{2} + b x + c$ , extends the search to that case.

Practice these ideas

Practice

Two numbers have sum

16

and product

39

. Enter the larger one.

Show the solution

The factor pairs of

39

have sums

40

and

16

, so the numbers are

3

and

13

, and the larger is

13

Practice

Factor

x^{2} + 10 x + 16

. Enter the smaller of the two constants in the factored form.

Show the solution

The pair with sum

10

2

and

8

, so

x^{2} + 10 x + 16 = (x + 2) (x + 8)

, and the smaller constant is

2

Practice

Factor

x^{2} + 15 x + 36

. Enter the larger of the two constants in the factored form.

Show the solution

The pair with sum

15

3

and

12

, so

x^{2} + 15 x + 36 = (x + 3) (x + 12)

, and the larger constant is

12

Practice

Factor

x^{2} + 2 x - 48

. The two constants in the factors differ in sign. Enter the negative one.

Show the solution

The pair of

48

with difference

2

6

and

8

, with the

8

positive, so

x^{2} + 2 x - 48 = (x + 8) (x - 6)

, and the negative constant is

- 6

Practice

Factor

x^{2} - 12 x + 35

. Both constants are negative. Enter the smaller of the two.

Show the solution

The negative pair with sum

- 12

- 5

and

- 7

, so

x^{2} - 12 x + 35 = (x - 5) (x - 7)

, and the smaller constant is

- 7

Practice

Factor

7 x^{2} + 42 x + 35

completely, pulling out the greatest common numeric factor first. Enter the larger of the two constants in the factored form.

Show the solution

The common factor is

7

, and pulling it out leaves

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

, so

7 x^{2} + 42 x + 35 = 7 (x + 1) (x + 5)

and the larger constant is

5

. Without the pull the search on

7 x^{2}

never even starts, since the pattern needs a plain

x^{2}

Practice

Factor

x^{2} - 22 x

x

times a linear factor. Enter the constant inside the linear factor.

Show the solution

x^{2} - 22 x = x (x - 22)

, so the constant inside the linear factor is

- 22

Practice

x^{2} - 225

fits the form

x^{2} + b x + c

with

b = 0

and

c = - 225

. Factor it as

(x + k) (x - k)

and enter the positive integer

k

Show the solution

The pair must multiply to

- 225

and add to

0

, so it is

15

and

- 15

, giving

x^{2} - 225 = (x + 15) (x - 15)

and

k = 15

Practice

Compute

2 9^{2} - 2 1^{2}

in your head, as a difference times a sum. Enter the value.

Show the solution

2 9^{2} - 2 1^{2} = (29 - 21) (29 + 21) = 8 \cdot 50 = 400

. The sum does not have to be a power of ten for the shortcut to be quick.

Practice

x^{2} - 32 x + 256

is a perfect square trinomial,

(x + r)^{2}

for one value of

r

. Enter

r

, sign included.

Show the solution

256 = 1 6^{2}

and

- 32

is twice

- 16

, so

x^{2} - 32 x + 256 = (x - 16)^{2}

and

r = - 16

. This is the search's

r = s

case, the pair

- 16

and

- 16

Practice

Does

x^{2} - 7 x - 44

factor with integer constants? Run the factor pairs of

44

with opposite signs and answer yes or no.

Show the solution

The pair with difference

7

4

and

11

, with the

11

negative, so

x^{2} - 7 x - 44 = (x - 11) (x + 4)

, and the answer is

yes

Practice

Solve

x^{2} - 3 x - 70 = 0

. Enter the larger root.

Show the solution

x^{2} - 3 x - 70 = (x - 10) (x + 7)

, so

x = 10

x = - 7

, and the larger root is

10

. Checking,

100 - 30 - 70 = 0

Practice

Solve

x^{2} = 8 x + 65

. Move every term to one side before factoring. Enter the negative root.

Show the solution

Rearranging gives

x^{2} - 8 x - 65 = 0

, which factors as

(x - 13) (x + 5) = 0

, so

x = 13

x = - 5

, and the negative root is

- 5

. Checking in the original,

25 = - 40 + 65

Practice

Solve

x^{2} = 19 x

. Do not divide both sides by

x

, since a root is lost that way. Enter the smaller root.

Show the solution

x^{2} - 19 x = 0

factors as

x (x - 19) = 0

, so

x = 0

x = 19

, and the smaller root is

0

. Dividing by

x

at the start loses exactly this root.

Practice

Solve

3 x^{2} + 27 x + 42 = 0

. Pull out the common factor first. Both roots are negative. Enter the larger one.

Show the solution

Dividing out

3

gives

x^{2} + 9 x + 14 = (x + 2) (x + 7) = 0

, so

x = - 2

x = - 7

, and the larger root is

- 2

. On the number line

- 2

is to the right of

- 7

Practice

Two pages of a book have consecutive page numbers that multiply to

306

. Consecutive numbers give

n (n + 1) = 306

. Solve it, reject the root that cannot be a page number, and enter the smaller page number.

Show the solution

n^{2} + n - 306 = 0

factors as

(n + 18) (n - 17)

, so

n = 17

n = - 18

, and a page number is positive, so the pages are

17

and

18

and the smaller is

17

. The check is the original statement,

17 \cdot 18 = 306

Practice

Solve

2 x^{2} - 338 = 0

. The common factor comes out first, and what remains is a difference of squares. Enter the negative root.

Show the solution

Dividing out

2

gives

x^{2} - 169 = (x + 13) (x - 13) = 0

, so

x = 13

x = - 13

, and the negative root is

- 13

. Checking,

2 \cdot 169 - 338 = 0