Factoring in Action

The equation $x^{4} - 20 x^{2} + 64 = 0$ has degree four, so it is not a quadratic. But $x^{4} = (x^{2})^{2}$ , so the only powers of $x$ in it are $x^{2}$ and the square of $x^{2}$ . Write $u$ for $x^{2}$ . The equation is then $u^{2} - 20 u + 64 = 0$ , which 9.2's search factors as $(u - 4) (u - 16) = 0$ , so $x^{2} = 4$ or $x^{2} = 16$ , and $x = \pm 2$ or $x = \pm 4$ .

Problem

Solve

x^{4} - 34 x^{2} + 225 = 0

. Write

u

for

x^{2}

, factor the quadratic in

u

, then go back to

x

. Enter the largest of the four solutions.

Show a hint

Since $x^{4} = (x^{2})^{2}$ , writing $u$ for $x^{2}$ turns the equation into $u^{2} - 34 u + 225 = 0$ .
Two numbers with product $225$ and sum $- 34$ are $- 9$ and $- 25$ , so $x^{2} = 9$ or $x^{2} = 25$ . Each of those has two solutions.

Show the full solution

With

u = x^{2}

the equation is

u^{2} - 34 u + 225 = 0

, and two numbers with product

225

and sum

- 34

are

- 9

and

- 25

, so it factors as

(u - 9) (u - 25) = 0

. Then

x^{2} = 9

gives

x = \pm 3

and

x^{2} = 25

gives

x = \pm 5

, so the largest of the four is

5

. The values

9

and

25

are values of

x^{2}

, so neither is a solution of the original equation.

Problem

(2 x + 3)^{2} - 14 (2 x + 3) + 40 = 0

the expression

2 x + 3

appears squared and again on its own. Write

u

for

2 x + 3

, solve for

u

, then solve for

x

. Enter the larger value of

x

as a fraction in lowest terms.

Show a hint

The repeated piece does not have to be a power of $x$ . Here it is $2 x + 3$ , so the equation is $u^{2} - 14 u + 40 = 0$ .
$u^{2} - 14 u + 40 = (u - 4) (u - 10)$ , so $2 x + 3 = 4$ or $2 x + 3 = 10$ . Subtract $3$ , then divide by $2$ .

Show the full solution

Writing

u

for

2 x + 3

gives

u^{2} - 14 u + 40 = 0

, which factors as

(u - 4) (u - 10) = 0

. Then

2 x + 3 = 4

gives

x = \frac{1}{2}

, and

2 x + 3 = 10

gives

x = \frac{7}{2}

, so the larger value is

\frac{7}{2}

. Both

4

and

10

are values of

u

, not of

x

, so neither one is a solution until the substitution is undone.

Problem

Solve

(x - 6)^{2} - 2 (x - 6) - 24 = 0

and enter the larger solution. The expression

x - 6

repeats, so give that repeated piece a single letter before you factor.

Show a hint

Write $u = x - 6$ . The equation becomes $u^{2} - 2 u - 24 = 0$ , which needs two numbers with product $- 24$ and sum $- 2$ .
The factoring gives $(u - 6) (u + 4) = 0$ , so $u = 6$ or $u = - 4$ . Now recover $x$ from each value of $u$ , then take the larger.

Show the full solution

With

u = x - 6

the equation is

u^{2} - 2 u - 24 = (u - 6) (u + 4) = 0

, so

u = 6

u = - 4

. Those are values of

x - 6

, so

x = 12

x = 2

, and the larger is

12

. Reporting

6

is the common mistake. It is the larger value of

u

, and

u

stands for

x - 6

Problem

Solve

x^{4} - 13 x^{2} - 48 = 0

by writing

u

for

x^{2}

. How many real solutions does the equation have? Enter a single number.

Show a hint

The equation in $u$ is $u^{2} - 13 u - 48 = 0$ . Solve that first and read off both values of $u$ .
The values are $u = 16$ and $u = - 3$ . Ask which real numbers satisfy $x^{2} = 16$ , and which satisfy $x^{2} = - 3$ .

Show the full solution

u

the equation is

u^{2} - 13 u - 48 = 0

, and two numbers with product

- 48

and sum

- 13

are

- 16

and

3

, so it factors as

(u - 16) (u + 3) = 0

. Then

x^{2} = 16

gives

x = 4

and

x = - 4

, while

x^{2} = - 3

has no real solution, since no real number squares to a negative. The count is

2

. Two values of

u

do not automatically mean four values of

x

The same equation on two rows. The top row has

x^{2}

in both boxes, once squared and once multiplied by

b

, so the equation has degree four. The bottom row has a single letter in those same two boxes and nothing outside them changes, so what is left is

a u^{2} + b u + c = 0

. The quadratic was there in the top row too. Naming the repeated piece is what makes it visible, and a binomial in the boxes works the same way.

The repeated piece is not always obvious. In $\frac{1}{x^{2}} - \frac{11}{x} + 24 = 0$ it is $\frac{1}{x}$ , since $\frac{1}{x^{2}} = {(\frac{1}{x})}^{2}$ , and $u^{2} - 11 u + 24 = (u - 3) (u - 8)$ gives $\frac{1}{x} = 3$ or $8$ , so $x = \frac{1}{3}$ or $x = \frac{1}{8}$ . Multiplying the original by $x^{2}$ gives $24 x^{2} - 11 x + 1 = 0$ , with the same two roots.

Problem

Solve

x + \frac{56}{x} = 18

. Multiply both sides by

x

, collect everything on one side, and factor. Enter the larger solution.

Show a hint

Multiplying by $x$ turns $\frac{56}{x}$ into $56$ and $x$ into $x^{2}$ , so no fraction is left.
The cleared equation is $x^{2} - 18 x + 56 = 0$ . Two numbers with product $56$ and sum $- 18$ are $- 4$ and $- 14$ .

Show the full solution

Multiplying by

x

gives

x^{2} + 56 = 18 x

, so

x^{2} - 18 x + 56 = 0

, which factors as

(x - 4) (x - 14) = 0

. The solutions are

4

and

14

, so the larger is

14

. Both check in the original equation, and

x = 0

is not a candidate, since

\frac{56}{0}

is not a number.

Problem

Solve

\frac{4}{x - 2} + \frac{x}{x + 2} = \frac{16}{x^{2} - 4}

, where

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

. Multiplying through by

(x - 2) (x + 2)

produces two candidates, and only one of them solves the original equation. Enter that one.

Show a hint

Multiply every term by $(x - 2) (x + 2)$ . In the first term the $x - 2$ divides out, in the second the $x + 2$ does, and on the right both do.
You get $4 (x + 2) + x (x - 2) = 16$ , so $x^{2} + 2 x - 8 = 0$ . Factor it, then put each candidate into $\frac{4}{x - 2}$ before deciding.

Show the full solution

Multiplying every term by

(x - 2) (x + 2)

gives

4 (x + 2) + x (x - 2) = 16

, so

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

and

(x + 4) (x - 2) = 0

. The candidates are

- 4

and

2

. At

x = 2

the denominator

x - 2

is zero, so

2

is not a solution, and the answer is

- 4

. Checking,

\frac{4}{- 6} + \frac{- 4}{- 2} = \frac{4}{3}

, which matches

\frac{16}{12}

A numerator and a denominator can share a factor. Since $x^{2} + 3 x - 18 = (x + 6) (x - 3)$ , the expression $\frac{x^{2} + 3 x - 18}{x - 3}$ equals $x + 6$ at every value of $x$ except $3$ . At $x = 3$ the original is $\frac{0}{0}$ and has no value, while $x + 6$ is $9$ . Cancelling a factor leaves the excluded value excluded.

Problem

Solve

\frac{x^{2} - 3 x - 10}{x + 2} = - 7

. The numerator factors, and one of its factors is the denominator. How many solutions does the equation have? Enter a single number.

Show a hint

$x^{2} - 3 x - 10 = (x - 5) (x + 2)$ , so for every $x$ except $- 2$ the left side equals $x - 5$ .
That leaves $x - 5 = - 7$ , so the only candidate is $x = - 2$ . Check that value against the original denominator.

Show the full solution

Since

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

, the left side equals

x - 5

wherever it is defined, and

x - 5 = - 7

gives

x = - 2

. But

x = - 2

makes the denominator zero, so it is not a solution and the count is

0

. Multiplying through instead gives

x^{2} - 3 x - 10 = - 7 x - 14

, so

x^{2} + 4 x + 4 = 0

and

(x + 2) (x + 2) = 0

, the same excluded candidate again.

Problem

The numbers

a

and

b

satisfy

3 a^{2} - 11 a b + 6 b^{2} = 0

, and

b

is not zero. Read the left side as a quadratic in

a

, with

b

held as a constant, and factor it. Enter the smaller of the two possible values of

\frac{a}{b}

as a fraction in lowest terms.

Show a hint

As a quadratic in $a$ the three coefficients are $3$ , $- 11 b$ and $6 b^{2}$ . Each factor has an $a$ term and a $b$ term.
Use the middle-term split from 9.3. Multiply $3 a^{2}$ by $6 b^{2}$ to get $18 a^{2} b^{2}$ , break $- 11 a b$ into two terms with that product, then group in pairs. Set each factor to zero and divide by $b$ .

Show the full solution

As a quadratic in

a

3 a^{2} - 11 a b + 6 b^{2} = (3 a - 2 b) (a - 3 b) = 0

, so

3 a = 2 b

a = 3 b

. Dividing by

b

, which is not zero, gives

\frac{a}{b} = \frac{2}{3}

\frac{a}{b} = 3

, and the smaller is

\frac{2}{3}

. Dividing the original equation by

b^{2}

is the second route, since

3 t^{2} - 11 t + 6 = 0

with

t = \frac{a}{b}

has the same two solutions.

Several routes are open now. If the equation is already $a x^{2} + b x + c = 0$ , factor it and take a root from each factor. If one piece repeats, name it, solve, then undo the name. If the variable sits under a fraction bar, clear the denominators first. If the question asks only for the sum or the product, read it off the coefficients with 9.4's two facts.

Problem

A cyclist covers

60

kilometres at a steady speed. At

5

kilometres per hour faster the ride would have taken one hour less, so her speed

v

satisfies

\frac{60}{v} - \frac{60}{v + 5} = 1

. Find

v

in kilometres per hour.

Show a hint

Multiply both sides by $v (v + 5)$ . On the left the first term becomes $60 (v + 5)$ and the second becomes $60 v$ , and the right side becomes $v (v + 5)$ .
That gives $300 = v^{2} + 5 v$ , so $v^{2} + 5 v - 300 = 0$ . Factor it, then ask which of the two candidates can be a speed.

Show the full solution

Multiplying by

v (v + 5)

gives

60 (v + 5) - 60 v = v (v + 5)

, so

300 = v^{2} + 5 v

and

v^{2} + 5 v - 300 = 0

, which factors as

(v - 15) (v + 20) = 0

. The roots are

15

and

- 20

, and a speed cannot be negative, so

v = 15

. The root

- 20

solves the equation exactly, so it is ruled out by what

v

measures and not by the algebra. Checking,

\frac{60}{15} - \frac{60}{20} = 4 - 3 = 1

Problem

Clearing the denominators in

\frac{5}{x} + \frac{5}{x - 4} = 3

gives

3 x^{2} - 22 x + 20 = 0

, which has no integer factorization, so no method in this chapter produces either solution. Find the sum of the two solutions of the original equation as a fraction in lowest terms.

Show a hint

9.4 gives the sum of the roots as $- \frac{b}{a}$ , read straight off the coefficients with no factoring.
Here $a = 3$ and $b = - 22$ . Before using them, check that neither excluded value, $0$ or $4$ , is a root of the cleared quadratic.

Show the full solution

For

3 x^{2} - 22 x + 20

the sum of the roots is

- \frac{b}{a} = - \frac{- 22}{3} = \frac{22}{3}

. Neither excluded value is a root, since

3 x^{2} - 22 x + 20

equals

20

x = 0

and

- 20

x = 4

, so both roots of the quadratic solve the original equation. Factoring is not needed here, since the question asks about the two solutions together.

Substitution helps only when the quadratic it produces factors, and plenty do not. In $x^{4} - 7 x^{2} + 5 = 0$ the substitution gives $u^{2} - 7 u + 5 = 0$ , and the only integer pairs with product $5$ sum to $6$ or $- 6$ , never $- 7$ . Chapter 10 adds factorizations chapter 9 has no name for, and chapter 11 gives a method that solves every quadratic.

Problem

Solve

(x^{2} + 2 x)^{2} = 2 x^{2} + 4 x + 3

. The right side contains a copy of the piece that is squared on the left. Enter the smallest real solution.

Show a hint

$2 x^{2} + 4 x = 2 (x^{2} + 2 x)$ , so the right side is $2 (x^{2} + 2 x) + 3$ .
With $u = x^{2} + 2 x$ the equation is $u^{2} = 2 u + 3$ , so $u^{2} - 2 u - 3 = 0$ and $u = 3$ or $u = - 1$ . Each value gives its own quadratic in $x$ .

Show the full solution

Since

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

, the equation is

u^{2} = 2 u + 3

with

u = x^{2} + 2 x

, so

u^{2} - 2 u - 3 = (u - 3) (u + 1) = 0

and

u = 3

u = - 1

. Then

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

gives

x = 1

and

x = - 3

, and

x^{2} + 2 x + 1 = 0

gives

x = - 1

, so the smallest is

- 3

. There are three distinct solutions, not four, because the second value of

u

gives a repeated factor.

Problem

Solve

{(\frac{2 x}{x - 1})}^{2} - 16 (\frac{2 x}{x - 1}) + 28 = 0

. One of the two values of the repeated piece produces no value of

x

at all. Enter the solution as a fraction in lowest terms.

Show a hint

Write $u$ for $\frac{2 x}{x - 1}$ . The equation is $u^{2} - 16 u + 28 = 0$ .
$u^{2} - 16 u + 28 = (u - 2) (u - 14)$ . For $u = 2$ , multiply out $2 x = 2 (x - 1)$ and see what is left. Then do $u = 14$ .

Show the full solution

With

u = \frac{2 x}{x - 1}

the equation is

u^{2} - 16 u + 28 = (u - 2) (u - 14) = 0

, so

u = 2

u = 14

. For

u = 2

2 x = 2 (x - 1)

gives

0 = - 2

, which no

x

satisfies. For

u = 14

2 x = 14 (x - 1)

gives

12 x = 14

, so

x = \frac{7}{6}

. Two values of

u

can still leave one solution, and here that is because

\frac{2 x}{x - 1}

is never equal to

2

Every equation in this lesson was a quadratic in something, a renamed piece or $x$ itself once the denominators were cleared, and chapter 9's tools were enough. Chapter 10, Special Factorizations, adds factorizations this chapter has no name for, sums and differences of cubes among them, recognized by shape rather than found by a search.

Practice these ideas

Practice

The expression

\frac{x^{2} + x - 56}{x + 8}

equals

x - 7

at every value of

x

but one. Enter the value of

x

where the expression is undefined.

Show the solution

The denominator is zero at

x = - 8

. Since

x^{2} + x - 56 = (x + 8) (x - 7)

, cancelling

x + 8

leaves

x - 7

, which is

- 15

there, while the original is

\frac{0}{0}

and has no value. Cancelling is legal only where

x + 8

is not zero, so that one value stays excluded.

Practice

Solve

x^{4} - 53 x^{2} + 196 = 0

by writing

u

for

x^{2}

. Enter the largest value of

x

that solves it.

Show the solution

With

u = x^{2}

the equation is

u^{2} - 53 u + 196 = 0

, which factors as

(u - 4) (u - 49) = 0

, so

x^{2} = 4

x^{2} = 49

and the largest value of

x

7

. The four solutions are

- 7

- 2

2

and

7

, so stopping at

u = 49

reports the square of the answer.

Practice

Solve

(x + 5)^{2} - 5 (x + 5) + 4 = 0

, where the repeated piece is

x + 5

. Enter the larger value of

x

Show the solution

With

u = x + 5

(u - 1) (u - 4) = 0

gives

u = 1

u = 4

, so

x = - 4

x = - 1

, and the larger is

- 1

. Both values of

u

are positive while both solutions are negative, which is why the undo step cannot be skipped.

Practice

Solve

x^{4} - 8 x^{2} - 9 = 0

. Enter the positive solution.

Show the solution

With

u = x^{2}

the equation is

u^{2} - 8 u - 9 = 0

, which factors as

(u - 9) (u + 1) = 0

, so

u = 9

u = - 1

. No real number squares to

- 1

, and

x^{2} = 9

gives

x = \pm 3

, so the positive solution is

3

. The value

9

x^{2}

rather than

x

, so it is not itself a solution.

Practice

Solve

\frac{54}{x} = x + 3

by multiplying both sides by

x

. Enter the positive solution.

Show the solution

Multiplying by

x

gives

54 = x^{2} + 3 x

, so

x^{2} + 3 x - 54 = 0

and

(x + 9) (x - 6) = 0

. The solutions are

- 9

and

6

, so the positive one is

6

. Neither is

0

, so both are legal values of

x

in the original equation.

Practice

Solve

\frac{x^{2} - 12 x + 27}{x - 9} = 5

. The numerator shares a factor with the denominator. Enter the solution.

Show the solution

Since

x^{2} - 12 x + 27 = (x - 3) (x - 9)

, the left side equals

x - 3

wherever the expression is defined, so

x - 3 = 5

and

x = 8

. The excluded value is

9

, and

8

is not it, so this candidate survives.

Practice

How many different real values of

x

satisfy

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

? Enter a single number.

Show the solution

With

u = x^{2}

the equation is

u^{2} + 9 u = 0

, so

u (u + 9) = 0

and

u = 0

u = - 9

. Then

x^{2} = 0

gives only

x = 0

, and

x^{2} = - 9

gives no real

x

, so the count is

1

. A

u

of zero is the one case that yields a single

x

, since a positive

u

yields two and a negative

u

none.

Practice

The square of a positive integer plus twice that integer is

288

. Enter the integer.

Show the solution

With

n

the integer,

n^{2} + 2 n = 288

, so

n^{2} + 2 n - 288 = 0

and

(n - 16) (n + 18) = 0

. The roots are

16

and

- 18

, and the integer has to be positive, so it is

16

. Checking,

256 + 32 = 288

Practice

Clearing the denominators in

\frac{x}{x + 5} + \frac{2}{x} = 3

gives

2 x^{2} + 13 x - 10 = 0

, which has no integer factorization. Neither

0

nor

- 5

is a root of it. Enter the sum of the two solutions as a fraction in lowest terms.

Show the solution

For

2 x^{2} + 13 x - 10

the sum of the roots is

- \frac{b}{a} = - \frac{13}{2}

, so the sum of the two solutions is

- \frac{13}{2}

. The excluded values

0

and

- 5

are not roots, so nothing is discarded and no factorization is needed.

Practice

Solve

(4 x + 3)^{2} - 7 (4 x + 3) - 8 = 0

. Enter the larger solution as a fraction in lowest terms.

Show the solution

With

u = 4 x + 3

the equation is

(u - 8) (u + 1) = 0

, so

4 x + 3 = 8

4 x + 3 = - 1

, which give

x = \frac{5}{4}

and

x = - 1

, and the larger is

\frac{5}{4}

. Stopping at

u = 8

is the usual slip, since

8

is a value of the repeated piece and not a value of

x

Practice

Solve

4 x^{4} - 17 x^{2} + 4 = 0

by writing

u

for

x^{2}

. Enter the smallest positive solution as a fraction in lowest terms.

Show the solution

With

u = x^{2}

4 u^{2} - 17 u + 4 = (4 u - 1) (u - 4) = 0

, so

u = \frac{1}{4}

u = 4

. Undoing the substitution,

x^{2} = \frac{1}{4}

gives

x = \pm \frac{1}{2}

and

x^{2} = 4

gives

x = \pm 2

, so the smallest positive solution is

\frac{1}{2}

. Each positive

u

gives two values of

x

, which is why a quadratic in

x^{2}

can have four solutions.

Practice

Solve

\frac{x + 5}{x - 6} = \frac{66}{x^{2} - 6 x}

. The denominator on the right factors as

x (x - 6)

, and one of the two candidates is extraneous. Enter the single value of

x

that solves the equation.

Show the solution

Multiplying by

x (x - 6)

gives

x (x + 5) = 66

, so

x^{2} + 5 x - 66 = 0

and

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

. The candidates are

- 11

and

6

, and

x = 6

makes both original denominators zero, so the solution is

- 11

. Checking,

\frac{- 6}{- 17} = \frac{6}{17}

and

\frac{66}{187} = \frac{6}{17}

Practice

Solve

\frac{6}{x - 1} + \frac{x}{x + 1} = \frac{12}{x^{2} - 1}

. The denominator on the right factors as

(x - 1) (x + 1)

, and one candidate makes a denominator zero. Enter the solution.

Show the solution

Multiplying by

(x - 1) (x + 1)

gives

6 (x + 1) + x (x - 1) = 12

, so

x^{2} + 5 x - 6 = 0

and

(x + 6) (x - 1) = 0

. The candidates are

- 6

and

1

, and

x = 1

makes

x - 1

zero, so the solution is

- 6

Practice

The numbers

a

and

b

satisfy

2 a^{2} - 9 a b + 10 b^{2} = 0

, and

b

is not zero. Enter the larger of the two possible values of

\frac{a}{b}

as a fraction in lowest terms.

Show the solution

As a quadratic in

a

(2 a - 5 b) (a - 2 b) = 0

, so

2 a = 5 b

a = 2 b

. Dividing by

b

gives

\frac{a}{b} = \frac{5}{2}

\frac{a}{b} = 2

, and the larger is

\frac{5}{2}

. Both are genuinely possible, since

(a, b) = (5, 2)

and

(a, b) = (2, 1)

each satisfy the equation.

Practice

The numbers

a

and

b

satisfy

4 a^{2} + 4 a b - 3 b^{2} = 0

, and

b

is not zero. Enter the smaller of the two possible values of

\frac{a}{b}

as a fraction in lowest terms.

Show the solution

As a quadratic in

a

, the left side factors as

(2 a - b) (2 a + 3 b)

, so

2 a = b

2 a = - 3 b

. Dividing each by

b

gives

\frac{a}{b} = \frac{1}{2}

\frac{a}{b} = - \frac{3}{2}

, and the smaller of the two is

- \frac{3}{2}

. Dividing the original equation by

b^{2}

is the same route, since with

t = \frac{a}{b}

it reads

4 t^{2} + 4 t - 3 = 0

Practice

Solve

(x^{2} + 8 x)^{2} + 6 (x^{2} + 8 x) - 135 = 0

. Each value of the repeated piece gives a quadratic that factors over the integers. Enter the sum of all four solutions.

Show the solution

With

u = x^{2} + 8 x

(u - 9) (u + 15) = 0

gives

u = 9

u = - 15

. Then

x^{2} + 8 x - 9 = 0

gives

x = 1

and

x = - 9

, and

x^{2} + 8 x + 15 = 0

gives

x = - 3

and

x = - 5

, so the sum is

1 - 9 - 3 - 5 = - 16

. Both quadratics have the form

x^{2} + 8 x + k

and both have real roots, so by 9.4 each pair of roots sums to

- 8

and the total is twice

- 8

Practice

Solve

{(\frac{x + 1}{x - 1})}^{2} - 11 (\frac{x + 1}{x - 1}) + 10 = 0

. One value of the repeated piece produces no

x

at all. Enter the solution as a fraction in lowest terms.

Show the solution

With

u = \frac{x + 1}{x - 1}

(u - 1) (u - 10) = 0

gives

u = 1

u = 10

. For

u = 1

x + 1 = x - 1

reduces to

1 = - 1

, so no

x

works. For

u = 10

x + 1 = 10 x - 10

gives

9 x = 11

, so

x = \frac{11}{9}