Sullivan Algebra and Trigonometry: Section 5.2

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Sullivan Algebra and Trigonometry Section 5.2

• Objectives
• Use the Remainder and Factor Theorems
• Use Descartes Rule of Signs
• Use the Rational Zeros Theorem
• Find the Real Zeros of a Polynomial Function
• Solve Polynomial Equations
• Use the Theorem for Bounds on Zeros
• Use the Intermediate Value Theorem

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Division Algorithm for Polynomials If f(x) and
g(x) denote polynomial functions and if g(x) is
not the zero polynomial, then there are unique
polynomial functions q(x) and r(x) such that
dividend
quotient
divisor
remainder
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Remainder Theorem Let f be a polynomial function.
If f (x) is divided by x - c, then the remainder
is f (c).
x 3 x - (-3)
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Factor Theorem Let f be a polynomial function.
Then x - c is a factor of f (x) if and only if f
(c) 0.
In other words, if f(c) 0, then the remainder
found if f(x) is divided by x - c is zero.
Hence, since x - c divided into f(x) evenly
(remainder 0), x - c is a factor of f(x).
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a.) f(-3) 30 Therefore, x 3 does not divide
into f. So, x 3 is not a factor of f.
b.) f(4) 0 Therefore, x 4 does divide into
f. So, x 4 is a factor of f.
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Theorem Number of Zeros A polynomial function
cannot have more zeros than its degree.
Theorem Descartes Rule of Signs Let f denote of
polynomial function. The number of positive real
zeros of f either equals the number of variations
in sign of the nonzero coefficients of f (x) or
else equals that number less an even integer. The
number of negative real zeros of f either equals
the number of variations in sign of the nonzero
coefficients of f (-x) or else equals that number
less an even integer.
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Discuss the real zeros of
There are at most three zeros, since the function
is a polynomial of degree three.
Using Descartes Rule of Signs, f(x) has one sign
change. So, there is one positive real zero.
Using Descartes Rule of Signs, f (-x) has two
sign changes. So, there are two or zero negative
real zeros.
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Rational Zeros Theorem Let f be a polynomial
function of degree 1 or higher of the form
where each coefficient is an integer. If p/q, in
lowest terms, is a rational zero of f, then p
must be a factor of a0 and q must be a factor of
an.
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List the potential rational zeros of
According the the theorem, the numerator of
potential rational zeros will be factors of p -
12 and the denominator will be factors of q 2
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Find the real zeros of
Factor f over the reals.
First, determine the nature of the zeros. Since
the polynomial is degree 5, there are at most
five zeros.
Using Descartes Rule of Signs, there are three
or one positive real zero(s).
Using Descartes Rule of Signs again, there are
two or no negative real zeros.
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Now, list all possible rational zeros p/q by
factoring the first and last coefficients of the
function.
Now, begin testing each potential zero using
synthetic division. If a potential zero k is in
fact a zero, then x - k divides into f (remainder
will be zero) and is a factor of f.
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Thus, -3 is a zero of f and x 3 is a factor of
f.
Thus, -2 is a zero of f and x 2 is a factor of
f.
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We know from Descartes Rule of Signs that there
are no more negative real zeros.
Thus, 1 is a zero of f and x - 1 is a factor of f.
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Theorem Bounds on Zeros
Let f denote a polynomial function whose leading
coefficient is 1.
A bound M on the zeros of f is the smaller of the
two numbers
where Max means choose the largest entry in

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Find a bound on the zeros of
Every zero of f lies between -21 and 21.
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Intermediate Value Theorem Let f denote a
continuous function. If a lt b and if f(a) and
f(b) are of opposite sign, then the graph of f
has at least one x-intercept between a and b.
y
f(b)
x-intercept
a
x
b
f(a)
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Use the intermediate value theorem to show that
the graph of the function
has an x - intercept in the interval -3, -2
f (- 3) -11.2 lt 0
f (-2) 1.8 gt 0
Therefore, since f (-3) lt 0 and f (-2) gt 0, there
exists a number between - 3 and -2 where f (x)
0. So, the function has an x - intercept in the
interval -3,-2