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.
Chapter Seventeen — Calculating Roots of Functions
The Humongous Book of Algebra Problems
396
Note: Problems 17.38–17.39 refer to the function g(x) = x
4
– 2x
3
– 7x
2
+ 8x + 12.
17.39 Identify all four roots of g(x).
According to the rational root test, the possible roots of g(x) are ±1, ±2, ±3, ±4,
±6, and ±12. Notice that –1 is a root of g(x).
Therefore, g(x) = (x + 1)(x
3
– 3x
2
– 4x + 12). Factor the cubic by identifying
another root of g(x), such as 3.
Thus g(x) = (x + 1)(x
3
– 3x
2
– 4x + 12) = (x + 1)(x – 3)(x
2
– 4). Factor the
remaining quadratic, a difference of perfect squares: x
2
– 4 = (x + 2)(x – 2).
Therefore, g(x) = (x + 1)(x – 3)(x + 2)(x – 2). Set all four factors equal to zero
and solve to identify the roots of g(x): –2, –1, 2, and 3.
Note: Problems 17.40–17.43 refer to the function h(x) = c(x – 4)
2
+ 1. Assume c is a rational
number and c < 0.
17.40 Use the leading coefficient test to explain why h(x) has exactly two real roots.
As explained in Problem 17.20, the graph of h(x) is the graph of y = x
2
after
three transformations are performed. Each point on the graph of h(x) is c times
further away from the x-axis than the graph of y = x
2
. Furthermore, the graph of
h(x) is shifted four units to the right and one unit up on the coordinate plane.
Whereas the graph of y = x
2
has vertex (0,0), the vertex of h(x) is (4,1).
The problem states that c < 0, so the graph of y = x
2
must also be reflected across
the x-axis. According to the leading coefficient test, both ends of the graph
of h(x) go down, because h(x) has an even degree (2) and a negative leading
coefficient (c).
The graph of h(x) is a downward-pointing parabola whose maximum value
(1) occurs at its vertex (4,1). The domain of the function is all real numbers,
because no value of x causes h(x) to be undefined. Its ends point down, so
the graph must intersect the x-axis twice. Those two x-intercepts are the roots
of h(x).
Instead
of picking
random numbers
from this list, you may
want to use a graphing
calculator to sketch
the graph of g(x). Pay
attention to its x-
intercepts, because the
roots of a function are
also the x-intercepts
of its graph. The graph
of g(x) APPEARS to pass
through x = –2, x = –1,
x = 2, and x = 3.
So 3 is a
root of g(x)
because 3
4
– 2(3)
3
– 7(3)
2
+ 8(3) + 12
= 0. When you just
synthetically divided,
you got x
3
– 3x
2
– 4x –
12, and 3 has to be a
root of that polynomial
as well. This time,
synthetically divide
the cubic by the
root instead of
the original
function g(x).
There’s no
x term because
the quotient has an
x-coefcient of 0.
Here’s the answer
in a nutshell. The graph of h(x) is a parabola (which is
sort of U-shaped). The ends of the parabola point down, which means
it’s an upside-down U-shaped graph. The vertex is (4,1), so the graph reaches
no higher than y = 1. The function only goes DOWN from there, so the x-axis
cuts through that U twice (as will all horizontal lines below y = 1).
Chapter Seventeen — Calculating Roots of Functions
The Humongous Book of Algebra Problems
397
Note: Problems 17.40–17.43 refer to the function h(x) = c(x – 4)
2
+ 1. Assume c is a rational
number and c < 0.
17.41 For what values of c might h(x) have two or zero positive roots, according to
Descartes’ rule of signs?
Expand h(x) by squaring (x – 4) and distributing c.
The function will have either two or zero positive roots when there are exactly
two sign changes in h(x). Consider the coefficients of h(x) one at a time. The
first term has a negative coefficient (c < 0,) and the coefficient of –8cx is positive,
so there is a sign change between the first two terms. The constant term is
represented by 16c + 1.
For two sign changes to occur, 16c + 1 must be a negative number, as the
preceding term –8cx is positive. Solve the equation 16c + 1 < 0 for c.
When , the consecutive terms of h(x) exhibit two sign changes, which
indicates that h(x) has either two or zero positive roots according to Descartes’
rule of signs.
Note: Problems 17.40–17.43 refer to the function h(x) = c(x – 4)
2
+ 1. Assume c is a rational
number and c < 0.
17.42 For what values of c does h(x) have exactly one negative root, according to
Descartes’ rule of signs?
Substitute –x into h(x) and expand the function.
Because c < 0, the first term of h(–x) is negative, as is the second term. If h(x)
has one negative root, then h(–x) should exhibit one sign change—the constant
term 16c + 1 must be positive. According to Problem 17.41, the constant term
is negative when . Therefore, the constant is positive when .
–8 is negative
and so is c. Multiplying
two negatives gives
you a positive.
16c and
1 look like
different terms
but they should
be considered a
single, constant term,
because neither
contains x. It’s sort of
like the polynomial
x
2
+ 6x + 7 + 1, which is
the sum of four things,
but there’s actually
only three terms
after you add
the constants
7 and 1.
The constant
16c + 1 equals
zero when
.
Chapter Seventeen — Calculating Roots of Functions
The Humongous Book of Algebra Problems
398
Note: Problems 17.40–17.43 refer to the function h(x) = c(x – 4)
2
+ 1. Assume c is a rational
number and c < 0.
17.43 Apply the rational root test to identify all possible rational roots of h(x).
As explained in Problems 17.40–17.41, the constant of h(x) is 16c + 1, and the
leading coefficient is c. According to the rational root test, all rational roots of
h(x) have the form , such that m is a factor of 16c + 1 and n is a factor of c.
A more specific answer cannot be given, because the value of c is unknown.
..................Content has been hidden....................
You can't read the all page of ebook, please click here login for view all page.