Factoring ax² + bx + c

The quadratic $2 x^{2} + 7 x + 3$ factors as $(2 x + 1) (x + 3)$ . Expand to check, $2 x^{2} + 6 x + x + 3 = 2 x^{2} + 7 x + 3$ . What is new is the leading term. $2 x^{2}$ is the product of the two leading terms, so the leading coefficient must also be split between the factors, and the middle coefficient depends on that split as well as on the constants.

Problem

Factor

5 x^{2} + 29 x + 20

. Since

5

is prime, the leading terms can only be

5 x

and

x

, so the factorization has the shape

(5 x + q) (x + s)

with

q

and

s

positive whole numbers. Enter

q

, the constant beside

5 x

Show a hint

The leading split is forced, so only the two constants are unknown. List the positive factor pairs of $20$ , keeping both orders of each pair, since swapping the constants changes the middle term.
Check each candidate on its cross sum $5 s + q$ , which must equal $29$ . The pair with $q = 20$ beside $5 x$ gives $25$ , close but wrong, so keep going.

Show the full solution

The cross sum

5 s + q

must equal

29

. Among the factor pairs of

20

, the pair

q = 4

s = 5

gives

5 \cdot 5 + 4 = 29

, so

5 x^{2} + 29 x + 20 = (5 x + 4) (x + 5)

and

q = 4

. The swapped order

q = 5

s = 4

gives

25

instead, which is why each pair is checked in both orders.

Problem

Factor

11 x^{2} + 39 x - 20

. The split of

11

is forced, so the shape is

(11 x + q) (x + s)

. The product

q s = - 20

is negative, so

q

and

s

have opposite signs. Enter the negative constant.

Show a hint

The cross sum $11 s + q$ must equal $+ 39$ , a large positive number, so the positive term must be $11 s$ . That makes $s$ positive and $q$ negative.
Test the factor pairs of $20$ with $s$ positive and $q$ negative, checking the cross sum $11 s + q$ against $39$ each time.

Show the full solution

With

s

positive and

q

negative, the pair

q = - 5

s = 4

gives the cross sum

11 \cdot 4 - 5 = 39

, so

11 x^{2} + 39 x - 20 = (11 x - 5) (x + 4)

and the negative constant is

- 5

. Fixing the signs first cut the candidate list in half, since a cross sum as large as

+ 39

is out of reach when

s

is negative.

Problem

Factor

25 x^{2} + 75 x + 54

. Now the leading terms can split as

25 x

and

x

or as

5 x

and

5 x

, and both splits have to be searched. Both constants are positive. Enter the larger of the two constants in the factorization.

Show a hint

The constant pairs to test are $1 \cdot 54$ , $2 \cdot 27$ , $3 \cdot 18$ and $6 \cdot 9$ , each one usable in either order. Check every candidate on its cross sum against $75$ .
In the $(25 x + q) (x + s)$ rows the cross sum is $25 s + q$ , and the smallest one those pairs give is $77$ , which is already above $75$ . In the $(5 x + q) (5 x + s)$ rows it is $5 (q + s)$ , so $q + s = 15$ with $q s = 54$ .

Show the full solution

Pairing the larger constant with

25 x

gives cross sums

79

77

93

and

159

, and swapping each pair only raises the sum, so the

25 x

and

x

split is out. In the

(5 x + q) (5 x + s)

split the cross sum is

5 (q + s) = 75

, so

q + s = 15

with

q s = 54

, and the pair is

6

and

9

. So

25 x^{2} + 75 x + 54 = (5 x + 6) (5 x + 9)

and the larger constant is

9

. When the two leading terms are equal, the condition on the constants is exactly 9.2's sum and product search.

Problem

Every candidate for

5 x^{2} + 53 x + 72

has the shape

(5 x + q) (x + s)

with

q

and

s

positive and

q s = 72

, and counting order there are twelve such pairs. The middle coefficient

53

is odd. How many of the twelve candidates survive that one observation?

Show a hint

The cross sum is $5 s + q$ . Work out its parity when $q$ and $s$ are both even, using the fact that $5 s$ has the same parity as $s$ .
When $q$ and $s$ are both even the cross sum is even, and $53$ is odd. Count the ordered pairs of $72$ with both entries even, then subtract from twelve.

Show the full solution

Since

5 s

has the parity of

s

, the cross sum

5 s + q

is even whenever

q

and

s

are both even, and an even sum cannot equal

53

. Both odd is impossible, since

q s = 72

is even. The all-even ordered pairs are

2 \cdot 36

4 \cdot 18

6 \cdot 12

and their swaps, six in all, and removing them from the twelve leaves

6

. One survivor checks,

5 \cdot 9 + 8 = 53

, so

5 x^{2} + 53 x + 72 = (5 x + 8) (x + 9)

The candidate check for

a x^{2} + b x + c

. Stack the candidate factors,

p x + q

over

r x + s

. The left column product

p r

must equal

a

, and the right column product

q s

must equal

c

. The gold diagonals mark

p s

and

q r

, and their sum must equal

b

. Every candidate matches

a

and

c

by construction, so the middle term is the only place one can fail.

The first step has not changed. Before any lists get written, pull out the greatest common numeric factor and keep it in front. Sometimes what remains is monic, and the whole problem is back in 9.2's shorter search. Sometimes the remaining leading coefficient is smaller but still above $1$ , and both lists are shorter for it.

Problem

Factor

30 x^{2} - 156 x - 144

completely. Written as

g (5 x + q) (x + s)

, where

g

is the greatest common numeric factor, enter

s

Show a hint

All three coefficients are divisible by $6$ . Take the $6$ out first. The quadratic that remains has leading coefficient $5$ , smaller but above $1$ , so a search is still needed.
Factor $5 x^{2} - 26 x - 24$ as $(5 x + q) (x + s)$ . The constant is negative, so $q$ and $s$ have opposite signs, and the cross sum $5 s + q$ must equal $- 26$ .

Show the full solution

The greatest common factor is

6

, leaving

6 (5 x^{2} - 26 x - 24)

. The constant is negative, so

q

and

s

have opposite signs, and

q = 4

s = - 6

gives the cross sum

5 (- 6) + 4 = - 26

, so

30 x^{2} - 156 x - 144 = 6 (5 x + 4) (x - 6)

and

s = - 6

. Taking the

6

out first made every number in the search smaller. The leading coefficient stayed above

1

, so a search was still needed, just on easier numbers.

Problem

Factor

81 x^{2} - 100

into two binomials with integer coefficients. There is no middle term to search on, and both end terms are perfect squares, so this is one of 9.1's identities rather than a candidate list. The two constants are opposites. Enter the positive one.

Show a hint

No candidate search is needed here. Expand $(9 x + q) (9 x - q)$ with the double distribution and see which terms cancel.
The expansion leaves $81 x^{2} - q^{2}$ , so $q^{2} = 100$ , and $q$ is the positive choice.

Show the full solution

Since

81 x^{2} = (9 x)^{2}

and

100 = 1 0^{2}

, the conjugate identity gives

81 x^{2} - 100 = (9 x + 10) (9 x - 10)

, so

q = 10

. Expanding, the cross products

90 x

and

- 90 x

sum to zero, which is why the middle term is missing.

An empty search is still a proof. In $3 x^{2} + 4 x + 5$ the only leading split is $3 x$ and $x$ , and the constants must be the positive pair $1$ and $5$ . $(3 x + 1) (x + 5)$ has cross sum $16 x$ , $(3 x + 5) (x + 1)$ has cross sum $8 x$ , neither is $4 x$ , and no integer factorization exists. Quadratics like this are handled in chapter 11.

Problem

Does

4 x^{2} + 13 x + 6

factor into two binomials with integer coefficients? Run the full search, every leading split against every constant pair, before answering yes or no.

Show a hint

Both constants have to be positive, since their product $6$ and the middle coefficient $13$ are both positive. Then $4$ splits only as $4 \cdot 1$ or $2 \cdot 2$ , so the whole list has six candidates.
The middle coefficient $13$ is odd, so a candidate whose two cross products are both even is impossible. That removes the $(2 x + q) (2 x + s)$ row, and checking the rest settles it.

Show the full solution

The

(4 x + q) (x + s)

row has cross sums

25

10

14

11

for

(q, s) = (1, 6)

(6, 1)

(2, 3)

(3, 2)

, and the

(2 x + q) (2 x + s)

row has

14

and

10

. Both constants had to be positive and

4

splits only as

4 \cdot 1

2 \cdot 2

, so those six are the whole list, and none has cross sum

13

, making the answer

no

Problem

Solve

4 x^{2} + 27 x - 40 = 0

. Factor the left side, then split with the zero product property. Enter the positive root, as a fraction in lowest terms.

Show a hint

The leading $4$ splits as $4 \cdot 1$ or $2 \cdot 2$ , and a $2 x$ paired with $2 x$ always has an even cross sum, while $27$ is odd.
So the factors have the form $(4 x + q) (x + s)$ , with $q s = - 40$ and cross sum $4 s + q = 27$ . The constant is negative, so $q$ and $s$ have opposite signs. Once the factoring is done, set each factor equal to zero on its own.

Show the full solution

The left side factors as

4 x^{2} + 27 x - 40 = (4 x - 5) (x + 8)

, since the cross products

32 x

and

- 5 x

add to

27 x

. By the zero product property either

4 x - 5 = 0

x + 8 = 0

, so the roots are

\frac{5}{4}

and

- 8

, and the positive root is

\frac{5}{4}

. The root is a fraction because

4 x - 5 = 0

is finished by dividing by

4

The zero product property applies to a product equal to zero and nothing else, so the first move has not changed. When a quadratic equation has terms on both sides, subtract until one side is zero, collect like terms, and only then factor and split. A factored left side proves nothing while the right side is nonzero.

Problem

Solve

5 x^{2} + 22 x = 48

. The right side is not zero, and the zero product property applies only to a product that equals zero, so rearrange first, then factor and split. Enter the positive root as a fraction in lowest terms.

Show a hint

Move $48$ across so the right side is zero. The leading coefficient $5$ is prime, so the split of the leading terms is forced.
Search $(5 x + q) (x + s)$ with $q s = - 48$ , and keep the candidate whose cross products add to $22 x$ .

Show the full solution

Moving

48

across gives

5 x^{2} + 22 x - 48 = 0

. Since

5

is prime the leading terms split as

5 x

and

x

, and

- 8

with

6

gives cross products

30 x

and

- 8 x

, which add to

22 x

, so

5 x^{2} + 22 x - 48 = (5 x - 8) (x + 6)

. Then

5 x - 8 = 0

x + 6 = 0

, so

x = \frac{8}{5}

x = - 6

, and the positive root is

\frac{8}{5}

. Checking in the original equation,

5 \cdot \frac{64}{25} + 22 \cdot \frac{8}{5} = \frac{64}{5} + \frac{176}{5} = 48

Problem

The first and last terms of

144 x^{2} + b x + 49

are both perfect squares. Find the positive value of

b

that makes the whole trinomial the square of a binomial.

Show a hint

Start by writing those two terms as squares, $144 x^{2} = (12 x)^{2}$ and $49 = 7^{2}$ .
Expanding $(p x + q)^{2}$ gives $p^{2} x^{2} + 2 p q x + q^{2}$ , so the middle coefficient is twice the product of the two terms being squared, not just their product.

Show the full solution

The first and last terms are

(12 x)^{2}

and

7^{2}

, and with

b

positive the trinomial is

(12 x + 7)^{2} = 144 x^{2} + 168 x + 49

, so

b = 168

. A common slip is

84

, the product

12 \cdot 7

without the doubling. The middle term of a square is twice that product, one copy from each cross product.

You can now take any quadratic with integer coefficients through one finite process, common factor out, identities checked, candidates listed, narrowed, and tested on their cross sums. When an integer factorization exists the process ends on it, and an exhausted list is a proof that none exists. The two closing problems draw on most of it at once.

Problem

Factor

24 x^{2} - 34 x - 45

into two binomials with integer coefficients, writing both leading coefficients as positive. Since the constant is

- 45

, the two constants in the factors have opposite signs, so exactly one of them is negative. Enter that negative constant.

Show a hint

Both constants in the factors are odd, since their product is $- 45$ . The cross products must add to the even number $- 34$ , and that happens only when the two leading coefficients have the same parity.
Try the split $4 \cdot 6$ of $24$ against the odd factor pairs of $45$ , with signs chosen so the cross products add to $- 34$ .

Show the full solution

Both constants are odd, since their product is

- 45

, so each cross product has the parity of its leading coefficient, and those add to the even

- 34

. Both leading coefficients are therefore even, leaving the splits

2 \cdot 12

and

4 \cdot 6

24

. No odd pair of

45

fits

2 \cdot 12

, and

4 \cdot 6

with constants

- 9

and

5

gives cross products

- 54 x

and

20 x

, so

24 x^{2} - 34 x - 45 = (4 x - 9) (6 x + 5)

and the negative constant is

- 9

. With an odd constant and an even middle coefficient, only the even splits of

a

are possible, which is a fast first cut.

Problem

Solve

36 x^{2} + 13 x - 40 = 0

. There are five splits of

36

and four factor pairs of

40

, so narrow the list with sign and parity reasoning before trying candidates. Enter the negative root as a fraction in lowest terms.

Show a hint

The middle coefficient $13$ is odd, so one cross product is odd and one is even. An all even split of $36$ or an all even factor pair of $40$ makes both of them even.
After the parity cut the splits still open are $1 \cdot 36$ , $3 \cdot 12$ , $4 \cdot 9$ and the pairs $1 \cdot 40$ , $5 \cdot 8$ . The constant is negative, so the two cross product sizes differ by $13$ .

Show the full solution

Factoring gives

36 x^{2} + 13 x - 40 = (4 x + 5) (9 x - 8)

, so

4 x + 5 = 0

9 x - 8 = 0

, and the roots are

x = - \frac{5}{4}

and

x = \frac{8}{9}

, making the negative root

- \frac{5}{4}

. The search is short because

13

is odd, which rules out every all even split of

36

and every all even factor pair of

40

So far every root in this chapter was found the same way, factor first, then read each root off its factor. Lesson 9.4, Roots, Sums, and Products, works without that step. The sum and the product of a quadratic's roots can be read straight from its coefficients, so questions about the roots can be answered without ever finding them. That is new.

Practice these ideas

Practice

Factor

5 x^{2} + 31 x + 44

. The leading coefficient is prime, so the two factors can only begin with

5 x

and

x

. Enter the constant in the factor whose leading term is

5 x

Show the solution

The shape is

(5 x + q) (x + s)

with

q s = 44

and cross sum

5 s + q = 31

. The pair

q = 11

s = 4

gives

5 \cdot 4 + 11 = 31

, so

5 x^{2} + 31 x + 44 = (5 x + 11) (x + 4)

and the constant beside

5 x

11

. The swapped order

q = 4

s = 11

gives

59

, which is why order matters.

Practice

Factor

11 x^{2} - 38 x - 24

. The leading coefficient is prime, so the linear terms can only be

11 x

and

x

. Enter the negative one of the two constants in the factored form.

Show the solution

(11 x + q) (x + s)

with

q s = - 24

, the middle coefficient is

11 s + q

, and

s = - 4

with

q = 6

gives

- 44 + 6 = - 38

. So

11 x^{2} - 38 x - 24 = (11 x + 6) (x - 4)

, and the negative constant is

- 4

. Flipping both signs gives

(11 x - 6) (x + 4)

, whose middle coefficient is

+ 38

, the right size and the wrong sign.

Practice

Factor

6 x^{2} + 41 x + 30

into two binomials with positive integer coefficients. The leading coefficient is not prime, so there is more than one way to split it between the two factors. Enter the smaller of the two constants.

Show the solution

Splitting the leading coefficient as

6 \cdot 1

gives

6 x^{2} + 41 x + 30 = (6 x + 5) (x + 6)

, with middle coefficient

6 \cdot 6 + 5 = 41

and constants multiplying to

30

, so the smaller constant is

5

. The balanced split

(2 x + q) (3 x + s)

is a dead end, since no factor pair of

30

makes

2 s + 3 q = 41

Practice

Factor

9 x^{2} - 31 x - 70

into two binomials with integer coefficients. The middle coefficient is not a multiple of

3

, which rules out the split with

3 x

in both factors, since that split makes every cross sum a multiple of

3

. Enter the positive constant.

Show the solution

In the

(3 x + q) (3 x + s)

split every cross sum is

3 (q + s)

, a multiple of

3

, and

- 31

is not, so the shape is

(9 x + q) (x + s)

with

9 s + q = - 31

and

q s = - 70

. The pair

q = 14

s = - 5

gives

9 (- 5) + 14 = - 31

, so

9 x^{2} - 31 x - 70 = (9 x + 14) (x - 5)

and the positive constant is

14

. Ruling out a whole split by divisibility is cheaper than testing its rows one at a time.

Practice

Factor

4 x^{2} + 31 x + 60

. The middle coefficient is odd, so the two factors cannot both have an even coefficient on

x

, since both cross products would then be even. Enter the constant in the factor whose leading term is

4 x

Show the solution

The all-even split is out, so the linear terms are

4 x

and

x

. In

(4 x + q) (x + s)

with

q s = 60

, the middle coefficient is

4 s + q

, and

q = 15

with

s = 4

gives

16 + 15 = 31

. So

4 x^{2} + 31 x + 60 = (4 x + 15) (x + 4)

, and the constant asked for is

15

. Checking parity first rules out the

2 x

times

2 x

split before any candidate is expanded.

Practice

Factor

5 x^{2} + 60 x + 160

completely, pulling out the greatest common numeric factor before any candidate search. Enter the larger of the two constants in the binomials that remain.

Show the solution

Every coefficient is a multiple of

5

, so

5 x^{2} + 60 x + 160 = 5 (x^{2} + 12 x + 32)

. The pair with sum

12

and product

32

4

and

8

, giving

5 (x + 4) (x + 8)

, so the larger constant is

8

. Pulling the

5

out first leaves a monic quadratic, so the sum and product search finishes it with no candidate list to check.

Practice

Each end term of

196 x^{2} - 169

is a perfect square and there is no middle term, so this is the conjugate identity rather than a search. Factor it as

(14 x + q) (14 x - q)

with

q

positive. What is

q

Show the solution

Since

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

and

169 = 1 3^{2}

, the identity gives

196 x^{2} - 169 = (14 x + 13) (14 x - 13)

, so

q = 13

. The two cross products are

- 182 x

and

+ 182 x

, which cancel, and that cancellation is why the middle term is missing.

Practice

Does

6 x^{2} + 5 x + 4

factor into two binomials with integer coefficients? Both the constant and the middle coefficient are positive, so both constants would have to be positive. Run every leading split against every constant pair before answering yes or no.

Show the solution

For

(6 x + q) (x + s)

the cross sum

6 s + q

gives

25

14

and

10

, and for

(2 x + q) (3 x + s)

the cross sum

2 s + 3 q

gives

11

10

and

14

. None of the six equals

5

, so the answer is

no

. The search is finite, so coming up empty is a complete answer, and quadratics like this one get their treatment in chapter 11.

Practice

Factor

6 x^{2} - 17 x - 114

(6 x + q) (x + s)

with integers

q

and

s

. The constant term is negative, so

q

and

s

have opposite signs. Enter

q

Show the solution

The conditions are

q s = - 114

and

6 s + q = - 17

. Among the factor pairs of

114

, only

6

and

19

fit, with

s = - 6

and

q = 19

giving

6 (- 6) + 19 = - 17

, so

6 x^{2} - 17 x - 114 = (6 x + 19) (x - 6)

and

q = 19

. The order inside the pair matters, since

s = - 19

with

q = 6

gives

6 (- 19) + 6 = - 108

instead.

Practice

Factor

4 x^{2} + 9 x - 100

(4 x + q) (x + s)

with integers

q

and

s

. The constant is large while the middle coefficient is small, so the two cross products very nearly cancel. Enter

q

Show the solution

The conditions are

q s = - 100

and

4 s + q = 9

. The pair

s = - 4

q = 25

meets both, since

25 \cdot (- 4) = - 100

and

4 (- 4) + 25 = 9

, so

4 x^{2} + 9 x - 100 = (4 x + 25) (x - 4)

and

q = 25

. A

2 \cdot 2

split of the leading coefficient is impossible here, since

(2 x + m) (2 x + n)

has even middle coefficient

2 m + 2 n

while

9

is odd.

Practice

Each end term of

25 x^{2} - k x + 81

is a perfect square. There is exactly one positive number

k

that makes the whole quadratic the square of a binomial. What is

k

Show the solution

The end terms are

(5 x)^{2}

and

9^{2}

, and

(5 x - 9)^{2} = 25 x^{2} - 90 x + 81

, so

k = 90

. The middle term of a squared binomial is twice the product of the two terms being squared, so its size here is

2 \cdot 5 \cdot 9 = 90

, and it is negative because of the minus in

5 x - 9

Practice

Solve

9 x^{2} + 16 x - 80 = 0

. One root is negative and one is a positive fraction. Enter the positive root as an exact fraction in lowest terms.

Show the solution

The conditions

q s = - 80

and

9 s + q = 16

hold for

q = - 20

s = 4

, so

9 x^{2} + 16 x - 80 = (9 x - 20) (x + 4)

. The zero product property gives

9 x - 20 = 0

x + 4 = 0

, so

x = \frac{20}{9}

x = - 4

, and the positive root is

x = \frac{20}{9}

. Checking the middle term,

36 x - 20 x = 16 x

Practice

Solve

20 x^{2} + 45 x - 275 = 0

. A numeric factor comes out of all three terms first, and what remains still has a leading coefficient above

1

. Enter the positive root as a fraction in lowest terms.

Show the solution

Dividing by

5

leaves

4 x^{2} + 9 x - 55 = 0

. A

2 x

times

2 x

split would make the middle coefficient even, so the leading terms are

4 x

and

x

, and

(4 x - 11) (x + 5)

has cross sum

20 - 11 = 9

with constant

- 55

. The roots of

5 (4 x - 11) (x + 5) = 0

are

\frac{11}{4}

and

- 5

, so the positive root is

\frac{11}{4}

. The factor

5

is never zero, so only the binomials give roots.

Practice

Factor

20 x^{2} - 89 x + 80

into two binomials with integer coefficients. Enter the constant in the factor whose leading term is

5 x

, sign included.

Show the solution

The split

(4 x + q) (5 x + s)

needs

q s = 80

and

4 s + 5 q = - 89

. The pair

q = - 5

s = - 16

fits, since

4 (- 16) + 5 (- 5) = - 64 - 25 = - 89

, so

20 x^{2} - 89 x + 80 = (4 x - 5) (5 x - 16)

and the constant in the

5 x

factor is

- 16

. An odd middle coefficient needs one odd constant, and the only odd divisors of

80

are

1

and

5

, so only two candidates ever need checking.

Practice

Factor

36 x^{2} - 60 x - 119

into two binomials with integer coefficients, writing both leading coefficients as positive. The constant is negative, so exactly one of the two constants in the factors is negative. Enter that negative constant.

Show the solution

With the split

6 \cdot 6

the cross sum is

6 (q + s) = - 60

, giving

q + s = - 10

with

q s = - 119

, which is the pair

7

and

- 17

. Then

36 x^{2} - 60 x - 119 = (6 x + 7) (6 x - 17)

and the negative constant is

- 17

. A middle coefficient divisible by

6

makes the split with two equal leading terms worth trying first, since equal leading terms make the cross sum a multiple of them.

Practice

Solve

20 x^{2} + 51 x - 90 = 0

. Neither root is a whole number here, which is ordinary once the leading coefficient is above

1

. Enter the positive root as a fraction in lowest terms.

Show the solution

The split

(5 x + q) (4 x + s)

with

q = - 6

and

s = 15

has cross sum

5 \cdot 15 + 4 \cdot (- 6) = 51

, so

(5 x - 6) (4 x + 15) = 0

and the roots are

\frac{6}{5}

and

- \frac{15}{4}

, with positive root

\frac{6}{5}

. The root from

5 x - 6 = 0

\frac{6}{5}

, not

- \frac{6}{5}

, since

5 x = 6

Practice

Solve

48 x^{2} + 26 x = 55

. The right side is not zero, and the zero product property needs it to be, so rearrange before factoring. Enter the negative root as a fraction in lowest terms.

Show the solution

Rearranged,

48 x^{2} + 26 x - 55 = 0

. The split

(8 x + q) (6 x + s)

with

q = 11

s = - 5

gives

8 (- 5) + 6 \cdot 11 = 26

, so

(8 x + 11) (6 x - 5) = 0

with roots

- \frac{11}{8}

and

\frac{5}{6}

. The negative root is

- \frac{11}{8}

. Each factor

p x + q

contributes the root

- q / p

, so a leading coefficient that does not divide its constant leaves a fraction.