Chapter Seventeen — Calculating Roots of Functions

The Humongous Book of Algebra Problems

394

Synthesizing Root Identiﬁcation Strategies

Factoring big polynomials from the ground up

Note: Problems 17.35–17.37 refer to the function f(x) = 2x

3

– 12x

2

+ 15x – 25.

17.35 Apply Descartes’ rule of signs to predict the total number of positive and

negative roots of f(x).

Adjacent terms of f(x) are opposites, so there are three sign changes in f(x).

According to Descartes’ rule of signs, f(x) has either three roots or one root. To

determine the possible number of negative roots, substitute –x into f(x).

Every term in f(–x) is negative, so there are no sign changes. Therefore, f(x) has

no negative roots.

Note: Problems 17.35–17.37 refer to the function f(x) = 2x

3

– 12x

2

+ 15x – 25.

17.36 Apply the rational root test to list the possible roots of f(x). Modify the list in

light of the information provided by Problem 17.35.

The constant of h(x) is –25, which has factors 1, 5, and 25; the leading

coefﬁcient is 2, which has factors 1 and 2. Apply the rational root test.

There are only six possible rational roots. Note that the list does not consider

negative roots, because Problem 17.35 concluded that h(x) had no negative

roots. Of the numbers in the list of possible rational roots, only 5 is a root of

f(x), as f(5) = 0.

In other

words, every

term has the

opposite sign of the

terms beside it. The

maximum number of

sign changes you can

have is one fewer

than the number

of terms in the

polynomial.

If you’re

wondering why

there are no ± signs

in this list, it’s not

a typo—keep

reading.

Chapter Seventeen — Calculating Roots of Functions

The Humongous Book of Algebra Problems

395

Note: Problems 17.35–17.37 refer to the function f(x) = 2x

3

– 12x

2

+ 15x – 25.

17.37 According to the Fundamental Theorem of Algebra, a polynomial of degree n

has exactly n complex roots. The degree of f(x) is 3, indicating that it should

have exactly three complex roots. Problem 17.36 identiﬁed only one root;

identify the two remaining roots.

According to Problem 17.36, 5 is a root of f(x). Therefore, f(x) is evenly divisible

by (x – 5). Calculate the quotient using synthetic division.

Thus f(x) = (x – 5)(2x

2

– 2x + 5). Identify the remaining roots by solving the

equation 2x

2

– 2x + 5 = 0 via the quadratic formula.

The roots of f(x) are 5, 1 – 3i, and 1 + 3i.

Note: Problems 17.38–17.39 refer to the function g(x) = x

4

– 2x

3

– 7x

2

+ 8x + 12.

17.38 Apply Descartes’ rule of signs to predict the total number of positive and

negative roots of g(x).

Consecutive terms of g(x) change sign twice, so g(x) has either 2 or 0 positive

roots. The terms of g(–x) = x

4

+ 2x

3

– 7x

2

– 8x + 12 change signs twice as well, so

g(x) has either 2 or 0 negative roots.

Complex

roots include

real numbers

and imaginary

numbers.

g(–x) =

(–x)

4

– 2(–x)

3

– 7(–x)

2

+ 8(–x) + 12

= x

4

– 2(–x

3

) – 7x

2

– 8x

+ 12.

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