Chapter Seventeen — Calculating Roots of Functions

The Humongous Book of Algebra Problems

380

Identifying Rational Roots

Factoring polynomials given a head start

17.1 Given a polynomial of degree n that has real number coefﬁcients, what is

guaranteed by the Fundamental Theorem of Algebra?

The Fundamental Theorem of Algebra states that the polynomial will have

exactly n complex roots. It is an existence theorem, merely guaranteeing that

those roots exist but providing no means by which to calculate them. Neither

can the theorem be used to further classify the roots, such as to determine how

many of the complex roots are real numbers or how many are rational.

17.2 According to the remainder theorem, if the quotient (ax

3

+ bx

2

+ cx + d) ÷

(x – m) has remainder r, what conclusion can be drawn? Assume that a, b, c, d,

and m are real numbers.

The remainder theorem states that substituting x = m into a polynomial written

in terms of x produces a value exactly equal to the remainder when that poly-

nomial is divided by x – m. In this example, dividing ax

3

+ bx

2

+ cx + d by x – m

produces remainder r, so the remainder theorem states that substituting x = m

into the expression results in the same value.

a(m)

3

+ b(m)

2

+ c(m) + d = r

Note: Problems 17.3–17.4 refer to the polynomial x

3

– 3x

2

+ 7x – 4.

17.3 Evaluate the expression for x = 5 using the remainder theorem. Verify your

answer.

According to the remainder theorem, the remainder of (x

3

– 3x

2

+ 7x – 4) ÷

(x – 5) is equal to the expression x

3

– 3x

2

+ 7x – 4 evaluated for x = 5. Calculate

the quotient using synthetic division.

Therefore, x

3

– 3x

2

+ 7x – 4 = 81 when x = 5. Verify this by actually substituting

x = 5 into the expression and simplifying.

A “polynomial

written in terms

of x” is a polynomial

containing x’s and no

other variables.

The number

left over when

you divide a

polynomial by

(x – m) is equal to the

number you get out

when you plug

x = m into the

polynomial.

If you

need synthetic

division practice,

check out Problems

11.38–11.45.

Chapter Seventeen — Calculating Roots of Functions

The Humongous Book of Algebra Problems

381

Note: Problems 17.3–17.4 refer to the polynomial x

3

– 3x

2

+ 7x – 4.

17.4 Evaluate the expression for x = –2 using the remainder theorem.

Calculate the remainder of (x

3

– 3x

2

+ 7x – 4) ÷ (x + 2) using synthetic division.

Therefore, (–2)

3

– 3(–2)

2

+ 7(–2) – 4 = –38.

Note: Problems 17.5–17.9 refer to the function f(x) = x

3

+ 13x

2

+ 31x – 45.

17.5 Evaluate f(1) using the remainder theorem.

Apply synthetic division.

The remainder is 0, so f(1) = 0.

Note: Problems 17.5–17.9 refer to the function f(x) = x

3

+ 13x

2

+ 31x – 45.

17.6 Explain why x = 1 is a root of f(x).

If f(c) = 0, then c is a root of function f(x). According to Problem 17.5, f(1) = 0,

so x = 1 is a root of the function.

Note: Problems 17.5–17.9 refer to the function f(x) = x

3

+ 13x

2

+ 31x – 45.

17.7 Explain why x – 1 is factor of x

3

+ 13x

2

+ 31x – 45.

According to Problem 17.5, f(x) ÷ (x – 1) has remainder 0. Because f(x) is evenly

divisible by x – 1, by deﬁnition, x – 1 is a factor of f(x).

Instead

of plugging

x = –2 into the

expression, plug

it into a synthetic

division box—the

signs match. Use the

opposite of –2 to

write the divisor:

x + 2.

Roots are also

called “zeroes”

because plugging

them into the

function makes the

function equal 0.

C is a

factor of

D if C divides

evenly into D.

In other words,

D ÷ C has no

remainder.

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