Chapter Eleven — Polynomials

The Humongous Book of Algebra Problems

246

11.26 Calculate the product and simplify: (x – y)(x

2

+ 6xy – 9y

2

).

Distribute each term of the left binomial through the right trinomial.

11.27 Calculate the product and simplify: (x + 3y – 1)(2x – y + 7).

Distribute each term of the left trinomial through the right trinomial.

11.28 Calculate the product and simplify: (a – 3b + 2c)(4a – b + 5c).

Distribute each term of the left trinomial through the right trinomial.

Long Division of Polynomials

A lot like long dividing integers

11.29 Calculate (x + 4) ÷ x using long division.

Rewrite the expression as a long division problem, with the divisor outside the

division symbol and the dividend within it.

Even though

both polynomials

have three terms,

it doesn’t change

the way you multiply.

Distribute x, then 3y,

and then –1 through

the right-hand

polynomial and add

everything together:

x(2x – y + 7) + 3y(2x

– y + 7) – 1(2x – y

+ 7).

Before you

try to perform

long division on

polynomials, make

sure you remember

how the process

works with whole

numbers—ip back to

Problems 2.3 and 2.5.

The two techniques

are basically the same,

except with whole

numbers you divide

one DIGIT at a time,

and with polynomials

you divide one

TERM at a time.

The divisor is

what you’re dividing BY and

the dividend is what you’re

dividing INTO.

Chapter Eleven — Polynomials

The Humongous Book of Algebra Problems

247

Divide the leftmost term of the dividend by the divisor: x ÷ x = 1. Write this

number above its like term in the dividend.

Multiply the newly placed number by the divisor: 1 · x = x. Write the opposite of

the result beneath the dividend, once again lining up like terms.

Combine like terms: x – x = 0.

Add the next term of the dividend (+4).

Combine like terms: 0 + 4 = 4.

The degree of the divisor x is greater than the degree of the polynomial below

the horizontal line, so the process of long division is complete.

The number above the division symbol (1) is the quotient and the number

below the horizontal line (4) is the remainder. Write the solution using the

format below.

The number

4 in the dividend

x + 4 has the same

variable as 1 (because

neither of them have

any variables). That

means 1 and 4 are like

terms. When you write

numbers above the

division symbol, line

them up with their

like terms.

Whenever

you multiply a

number on top of the

division symbol by the

number out front, write

the OPPOSITE of that

number (in this case –x

instead of x) beneath

the dividend.

x = x

1

, so x has

degree one. 4 is a

constant and doesn’t

have any variables,

so it has degree zero.

Because 1 > 0, you’re

done dividing.

Chapter Eleven — Polynomials

The Humongous Book of Algebra Problems

248

Note: Problems 11.30–11.31 refer to the quotient (x

2

+ 6x – 9) ÷ (x – 2).

11.30 Calculate the quotient and remainder using long division.

Express the quotient as a long division problem.

Divide the leftmost term of the dividend by the leftmost term of the divisor:

x

2

÷ x = x. Write the answer above the like term 6x in the dividend.

Multiply the newly placed x by each term of the divisor and write the opposites of

the products beneath the dividend.

Combine like terms (6x + 2x = 8x) and add the next term of the dividend (–9).

Divide the leftmost term below the horizontal line by the leftmost term of the

divisor: 8x ÷ x = 8. Write the answer above the division symbol, multiply both

terms of the divisor by 8, and write the opposites of the products below 8x – 9.

Therefore, .

In other words,

x (from the divisor)

times what equals x

2

(from the dividend)?

The answer’s x:

x

.

x = x

2

. If you write

the polynomials from

highest to lowest powers

of x, you’ll always be

dealing with the terms

on the left: “The left

term of the divisor

times what equals

the left term of the

dividend?”

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