Meet the Quadratic

Chapter 8 ended with a promise, that $x^{2} + 7 x + 12$ equals $(x + 3) (x + 4)$ . Multiply to check. $(x + 3) (x + 4) = x (x + 4) + 3 (x + 4) = x^{2} + 4 x + 3 x + 12 = x^{2} + 7 x + 12$ , so the promise holds. An expression $a x^{2} + b x + c$ with $a$ not zero is a quadratic. This lesson multiplies linear factors into quadratics. Reversing that is the next lesson's job.

Problem

Multiply

(x + 5) (x + 6)

the way the opener multiplied its product, splitting the first factor and distributing each part over the second. Collect like terms and enter the coefficient of

x

Show a hint

Write the product as $x (x + 6) + 5 (x + 6)$ . That is chapter 2's distributive property with the whole factor $(x + 6)$ as the multiplier.
Each part distributes again, four products in all. Two of them contain $x$ to the first power, and those two collect.

Show the full solution

(x + 5) (x + 6) = x (x + 6) + 5 (x + 6) = x^{2} + 6 x + 5 x + 30 = x^{2} + 11 x + 30

, so the coefficient of

x

11

. The two cross products

6 x

and

5 x

are like terms, and collecting them is where the middle term of every such product comes from.

Problem

Expand

(x + 10) (x - 2)

, keeping the minus sign with the

2

in both products that use it. Enter the constant term of the result.

Show a hint

Split as $x (x - 2) + 10 (x - 2)$ and distribute each part. Two of the four products come out negative.
The constant term is the product of the two constants, $+ 10$ and $- 2$ , sign included.

Show the full solution

(x + 10) (x - 2) = x^{2} - 2 x + 10 x - 20 = x^{2} + 8 x - 20

, so the constant term is

- 20

. The only constant is

10 \cdot (- 2)

, negative because exactly one of the two constants is negative.

Problem

Expand

(x - 5) (x - 8)

. Both constants are negative this time, so all four sign decisions in the four products matter. Enter the coefficient of

x

Show a hint

Split as $x (x - 8) - 5 (x - 8)$ and distribute each part, keeping every sign.
The two cross products are $- 8 x$ and $- 5 x$ . The last product, $(- 5) (- 8)$ , is positive.

Show the full solution

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

, so the coefficient of

x

- 13

. Both cross products are negative and add, while the constant

(- 5) (- 8) = 40

comes out positive.

Problem

Expand

(8 x - 5) (2 x + 9)

. The

x^{2}

coefficient is now a product of two numbers, and two of the four products contain

x

to the first power. Enter the coefficient of

x

Show a hint

Distribute twice as before, $8 x (2 x + 9) - 5 (2 x + 9)$ , and write all four products before collecting anything.
The two $x$ products come from $8 x \cdot 9$ and $- 5 \cdot 2 x$ . Compute both, signs included, then add.

Show the full solution

(8 x - 5) (2 x + 9) = 16 x^{2} + 72 x - 10 x - 45 = 16 x^{2} + 62 x - 45

, so the coefficient of

x

62

. With leading coefficients the middle is no longer a plain sum of the two constants, it is

72 - 10

, one cross product at a time.

The rectangle's width is

a + b

and its height is

c + d

, so its area is

(a + b) (c + d)

. The four cells have areas

a c

b c

a d

and

b d

, and they tile the rectangle exactly, so

(a + b) (c + d) = a c + a d + b c + b d

. Every term of one factor multiplies every term of the other, the same four products that appear in the double distribution.

A few products come up so often they are worth knowing on sight, a sum squared, a difference squared, and a sum times the matching difference. Squaring means multiplying by itself, so $(x + k)^{2}$ means $(x + k) (x + k)$ , four products again, and it is not $x^{2} + k^{2}$ . The next two problems work out two of these, and the third follows the same way.

Problem

Write

(x + 8)^{2}

(x + 8) (x + 8)

, the square written out as the product it abbreviates, and expand it with the same double distribution as every product so far. Enter the coefficient of

x

Show a hint

Squaring means multiplying by itself, so the same double distribution applies, four products again.
The cross product is $8 x$ both times, and the two copies are like terms.

Show the full solution

(x + 8)^{2} = (x + 8) (x + 8) = x^{2} + 8 x + 8 x + 64 = x^{2} + 16 x + 64

, so the coefficient of

x

16

. The cross product appears twice, which is why the answer is

16

and not

8

, and why

(x + 8)^{2}

is not

x^{2} + 64

Problem

Compute

61 \times 59

without a calculator by writing it as

(60 + 1) (60 - 1)

and expanding all four products. Enter the product.

Show a hint

The four products are $60 \cdot 60$ , $60 \cdot (- 1)$ , $1 \cdot 60$ , and $1 \cdot (- 1)$ . Two of them cancel.
The cross products are $- 60$ and $+ 60$ , so they add to zero. You are left with one square minus a much smaller square.

Show the full solution

61 \times 59 = (60 + 1) (60 - 1) = 3600 - 60 + 60 - 1 = 3600 - 1 = 3599

. The cross products cancel whenever the two factors are the same distance from a round number, one above and one below, which is what makes this a mental calculation.

An expansion is a claim that the two sides are equal at every value of $x$ , so they must agree at any value you try. At $x = 10$ , $(x + 1) (x + 8)$ is $11 \cdot 18 = 198$ , and $x^{2} + 9 x + 8$ is $100 + 90 + 8 = 198$ . At $x = 0$ only the constant term is left, so pick a nonzero value. Agreement at one value is strong evidence. A disagreement proves an error.

Problem

A student expands

(x + 6) (x + 9)

and writes

x^{2} + 14 x + 54

. Substitute

x = 10

into the product and into the claim to confirm they disagree, then enter the correct coefficient of

x

Show a hint

At $x = 10$ the product is $16 \cdot 19$ . Evaluate the claim at $10$ as well and compare.
The constant $54$ is correct, since $6 \cdot 9 = 54$ . Recompute the two cross products, $9 x$ and $6 x$ .

Show the full solution

x = 10

the product is

16 \cdot 19 = 304

while the claim gives

100 + 140 + 54 = 294

, so the claim is wrong. Expanding correctly,

(x + 6) (x + 9) = x^{2} + 9 x + 6 x + 54 = x^{2} + 15 x + 54

, and the coefficient of

x

15

. The gap between

304

and

294

10

, exactly one missing

x

x = 10

Problem

Expand

(x - 10 y) (x + 8 y)

. The same rule applies, every term of the first factor times every term of the second, with two letters to track. Enter the coefficient of

x y

Show a hint

Four products again, $x \cdot x$ , $x \cdot 8 y$ , $- 10 y \cdot x$ , and $- 10 y \cdot 8 y$ . Two of them contain both letters.
Collect $8 x y$ and $- 10 x y$ . The $y^{2}$ product is separate and is not the answer.

Show the full solution

(x - 10 y) (x + 8 y) = x^{2} + 8 x y - 10 x y - 80 y^{2} = x^{2} - 2 x y - 80 y^{2}

, so the coefficient of

x y

- 2

. The

x y

coefficient is the collected sum

8 - 10

, while

- 80

is the

y^{2}

coefficient, the product.

The payoff is immediate. In $(x - 6) (x + 10) = 0$ a product equals zero, so at least one factor is zero, and each factor gives a chapter 4 equation. $x - 6 = 0$ gives $x = 6$ , and $x + 10 = 0$ gives $x = - 10$ , the sign flipped. Of the two, $6$ is the larger, since $- 10$ is negative. A factor can also be $x$ itself, and $x = 0$ is already solved.

Problem

Solve

(x - 8) (x + 15) = 0

. The left side is zero exactly when one of the factors is zero. Enter the larger of the two solutions.

Show a hint

Set each factor equal to zero and solve the two linear equations separately.
$x + 15 = 0$ is solved by a negative number. Compare the two solutions on the number line before choosing the larger.

Show the full solution

x - 8 = 0

gives

x = 8

and

x + 15 = 0

gives

x = - 15

, so the larger solution is

8

. The factor

x + 15

is zero at

- 15

, not

15

, and

8 > - 15

even though

15

is bigger than

8

Problem

The left side of

x (x - 16) = 0

has

x

itself as a factor. Count every value of

x

that makes the left side zero, then enter the number of solutions of the equation.

Show a hint

The zero product property applies to any two factors. Set each of the two factors equal to zero.
The equation $x = 0$ is already solved, and its solution counts.

Show the full solution

The factors are

x

and

x - 16

, so

x = 0

x = 16

, and the number of solutions is

2

. Dropping

x = 0

is the common error, but zero is a solution, since substituting it back gives

0 \cdot (0 - 16) = 0

Multiplying two binomials is now mechanical, four products and a collection, the same every time. The chapter's prize is the reverse direction, starting from $x^{2} + b x + c$ and recovering the factors, because the zero product property applies only to factored form. That reversal starts in the next lesson.

Problem

Solve

(10 x - 25) (2 x + 1) = 0

. Each factor is zero at its own value of

x

, and neither value is an integer this time. Enter the positive solution as a fraction in lowest terms.

Show a hint

Each factor gives one linear equation, $10 x - 25 = 0$ and $2 x + 1 = 0$ . Solve both.
One solution is negative. The other comes out as a fraction that is not yet in lowest terms, so reduce it before entering.

Show the full solution

10 x - 25 = 0

gives

x = \frac{25}{10} = \frac{5}{2}

, and

2 x + 1 = 0

gives

x = - \frac{1}{2}

, so the positive solution is

\frac{5}{2}

. Lowest terms means dividing out the common factor of

25

and

10

, which is

5

Problem

There is one value of

b

for which

36 x^{2} + b x + 121

is exactly the square

(6 x - 11)^{2}

. Expand the square and compare the two expressions term by term. Enter

b

Show a hint

Use the square-of-a-difference identity with $6 x$ as the first term and $11$ as the second.
The middle term of the expansion is $- 2 \cdot 6 x \cdot 11$ . Match it against $b x$ .

Show the full solution

(6 x - 11)^{2} = 36 x^{2} - 2 \cdot 6 \cdot 11 x + 121 = 36 x^{2} - 132 x + 121

, so matching middle terms gives

b = - 132

. The standard misses are

- 66

, forgetting the

2

2 a b

, and

+ 132

, dropping the sign of the subtraction.

Every product $(x + r) (x + s)$ expands to $x^{2} + (r + s) x + r s$ , the sum in the middle and the product at the end. Lesson 9.2, Factoring $x^{2} + b x + c$ , starts from the expanded form and recovers $r$ and $s$ , a search with exactly two clues, the sum and the product.

Practice these ideas

Practice

Expand

(x + 1) (x + 9)

. Enter the coefficient of

x

Show the solution

(x + 1) (x + 9) = x^{2} + 9 x + x + 9 = x^{2} + 10 x + 9

, so the coefficient of

x

10

Practice

Expand

(x - 1) (x + 14)

. Enter the constant term.

Show the solution

(x - 1) (x + 14) = x^{2} + 14 x - x - 14 = x^{2} + 13 x - 14

, so the constant term is

- 14

. The sign comes from

(- 1) \cdot 14

, one negative constant.

Practice

Expand

(x - 2) (x - 9)

. Enter the coefficient of

x

Show the solution

(x - 2) (x - 9) = x^{2} - 9 x - 2 x + 18 = x^{2} - 11 x + 18

, so the coefficient of

x

- 11

. Two negative constants give a negative middle and a positive constant.

Practice

Expand

(2 x + 1) (5 x + 8)

. Enter the coefficient of

x

Show the solution

(2 x + 1) (5 x + 8) = 10 x^{2} + 16 x + 5 x + 8 = 10 x^{2} + 21 x + 8

, so the coefficient of

x

21

Practice

Expand

(x + 11)^{2}

. Enter the coefficient of

x

Show the solution

(x + 11)^{2} = x^{2} + 2 \cdot 11 x + 121 = x^{2} + 22 x + 121

, so the coefficient of

x

22

. The middle is

2 a b

, not

a b

, because the cross product appears twice.

Practice

Expand

(x - 10)^{2}

. Enter the constant term.

Show the solution

(x - 10)^{2} = x^{2} - 20 x + 100

, so the constant term is

100

. The constant of a square is always positive, only the middle term carries the minus.

Practice

Compute

85 \times 75

in your head by writing it as

(80 + 5) (80 - 5)

. Enter the product.

Show the solution

85 \times 75 = (80 + 5) (80 - 5) = 6400 - 25 = 6375

. The cross products,

+ 400

and

- 400

, cancel, so only

8 0^{2} - 5^{2}

is left.

Practice

Expand

(x + 20) (x - 20)

. Enter the coefficient of

x

Show the solution

(x + 20) (x - 20) = x^{2} - 20 x + 20 x - 400 = x^{2} - 400

, so the coefficient of

x

0

. The expansion has no

x

term at all, which is the difference of squares pattern.

Practice

Expand

(2 x - 9) (8 x - 1)

. Enter the coefficient of

x

Show the solution

(2 x - 9) (8 x - 1) = 16 x^{2} - 2 x - 72 x + 9 = 16 x^{2} - 74 x + 9

, so the coefficient of

x

- 74

. The constant

(- 9) (- 1) = 9

is positive while both cross products are negative.

Practice

A student expands

(x + 6) (x + 10)

and writes

x^{2} + 16 x + 66

. Substitute

x = 5

into the product and into the claim to see they disagree, then enter the correct constant term.

Show the solution

x = 5

the product is

11 \cdot 15 = 165

while the claim gives

25 + 80 + 66 = 171

, so the claim is wrong. The constant is

6 \cdot 10 = 60

, so the correct constant term is

60

. A check at

x = 0

would have caught this even faster, since substituting

x = 0

leaves only the constant term.

Practice

Expand

(x - 6 y) (x + 15 y)

. Enter the coefficient of

x y

Show the solution

(x - 6 y) (x + 15 y) = x^{2} + 15 x y - 6 x y - 90 y^{2} = x^{2} + 9 x y - 90 y^{2}

, so the coefficient of

x y

9

Practice

Solve

(x - 13) (x + 6) = 0

. Enter the larger solution.

Show the solution

x - 13 = 0

gives

x = 13

and

x + 6 = 0

gives

x = - 6

, so the larger solution is

13

Practice

Solve

(x + 16) (x + 21) = 0

. Both solutions are negative. Enter the larger one.

Show the solution

The solutions are

x = - 16

and

x = - 21

, and

- 16 > - 21

, so the larger solution is

- 16

. Larger means further right on the number line, not bigger digits.

Practice

Solve

x (2 x - 15) = 0

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

Show the solution

x = 0

2 x - 15 = 0

, which gives

x = \frac{15}{2}

, so the positive solution is

\frac{15}{2}

. Zero is a solution too, just not the positive one.

Practice

How many solutions does

(x - 17)^{2} = 0

have? Enter the count.

Show the solution

Both factors are

x - 17

, so the only solution is

x = 17

, and the count is

1

. A repeated factor gives one solution, not two, since both linear equations are the same.

Practice

To solve

(x - 1) (x + 13) = 10

, Mara sets

x - 1 = 10

and answers

x = 11

. Evaluate

(x - 1) (x + 13)

x = 11

. Enter the value you get.

Show the solution

x = 11

the factors are

10

and

24

, so

(x - 1) (x + 13) = 10 \cdot 24 = 240

. Many pairs of numbers multiply to

10

, so

x - 1 = 10

was only a guess. Splitting an equation factor by factor is valid only when the other side is

0

Practice

Find the value of

b

that makes

81 x^{2} + b x + 25

exactly

(9 x - 5)^{2}

Show the solution

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

, so

b = - 90

. Forgetting the factor

2

gives

- 45

, and dropping the subtraction's sign gives

+ 90

, the two standard misses.