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.
Chapter Seventeen — Calculating Roots of Functions
The Humongous Book of Algebra Problems
382
Note: Problems 17.5–17.9 refer to the function f(x) = x
3
+ 13x
2
+ 31x – 45.
17.8 Factor f(x).
According to Problem 17.5, (x
3
+ 13x
2
+ 31x – 45) ÷ (x – 1) = x
2
+ 14x + 45.
Therefore, x
3
+ 13x
2
+ 31x – 45 = (x – 1)(x
2
+ 14x + 45). Factor the quadratic
expression.
(x – 1)(x
2
+ 14x + 45) = (x – 1)(x + 9)(x + 5)
Therefore, the factored form of x
3
+ 13x
2
+ 31x – 45 is (x – 1)(x + 9)(x + 5).
Note: Problems 17.5–17.9 refer to the function f(x) = x
3
+ 13x
2
+ 31x – 45.
17.9 Identify all three roots of f(x).
According to Problem 17.8, f(x) = (x – 1)(x + 9)(x + 5). Substitute f(x) = 0 into
the equation, use the zero product property to set each factor equal to 0, and
solve the individual equations.
The roots of f(x) are x = –9, x = –5, and x = 1.
Note: Problems 17.10–17.11 refer to the function g(x) = 12x
3
+ 25x
2
– 38x – 15.
17.10 Factor g(x) given that –3 is one of its roots.
If –3 is a root of g(x), then (12x
3
+ 25x
2
– 38x – 15) ÷ (x + 3) has remainder 0.
Apply synthetic division.
Therefore, (12x
3
+ 25x
2
– 38x – 15) ÷ (x + 3) = 12x
2
– 11x – 5. Factor the quadratic
by decomposition.
Therefore, g(x) = (x + 3)(4x – 5)(3x + 1).
To factor the
function, take the
root you’re given (in
this case 1), divide it
out synthetically, make
a quadratic polynomial
out of the numbers you
get (in this case
x
2
+ 14x + 45), and then
try to factor that
quadratic.
Problems
17.28–17.34
explain how to
factor a cubic (or a
function with a higher
degree) when you’re
not given a root (like
–3) to start
with.
The number
in the box
matches the root
(–3), and the number
in the corresponding
factor always has
the opposite sign
(x + 3).
Chapter Seventeen — Calculating Roots of Functions
The Humongous Book of Algebra Problems
383
Note: Problems 17.10–17.11 refer to the function g(x) = 12x
3
+ 25x
2
– 38x – 15.
17.11 Identify all three roots of g(x).
According to Problem 17.10, g(x) = (x + 3)(4x – 5)(3x + 1). Let g(x) = 0, set each
factor equal to zero, and solve the resulting equations.
The roots of g(x) are x = –3, , and .
Note: Problems 17.12–17.13 refer to the function h(x) = 7x
3
– 30x
2
+ 9x – 4.
17.12 Factor h(x) given that 4 is one of its roots.
Calculate (7x
3
– 30x
2
+ 9x – 4) ÷ (x – 4) using synthetic division.
The quotient is 7x
2
– 2x + 1, so h(x) = (x – 4)(7x
2
– 2x + 1). The quadratic
expression is prime.
Note: Problems 17.12–17.13 refer to the function h(x) = 7x
3
– 30x
2
+ 9x – 4.
17.13 Identify all three roots of h(x).
According to Problem 17.12, h(x) = (x – 4)(7x
2
– 2x + 1). Let h(x) = 0 and set
both factors equal to zero.
x – 4 = 0 7x
2
– 2x + 1 = 0
The solution to the left equation is x = 4. Solving the right equation requires the
quadratic formula.
The roots of
g(x) are the same
as the solutions to
the equation
12x
3
+ 25x
2
– 38x – 15
= 0, which is the
equation h(x) = 0.
Prime = “unfactorable,”
so you can’t factor
7x
2
– 2x + 1.
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