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The Humongous Book of Algebra Problems by W. Michael Kelley

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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 youre
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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