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Math 2 Precalculus Algebra
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Real Zeros of Polynomials — 5.5
Division Algorithm for Polynomials
If f ( x ) and g ( x ) are polynomials and g ( x ) ≠ 0 then there exists unique polynomials q ( x ) and r ( x ) such that
f ( x) = q ( x) ⋅ g ( x) + r ( x)
or
f ( x=
) quotient ⋅ g ( x) + remainder ,
where the degree of r ( x ) is less than the degree of g ( x ) .
Find the quotient and remainder if f ( x ) is divided by g ( x ) .
1.
2.
f ( x ) = x 4 + 3x 2 + 6 x − 5
g ( x )= x − 2
Find f ( 2 ) for f ( x ) = x 4 + 3 x 2 + 6 x − 5
Remainder Theorem
If a polynomial f ( x ) is divided by x − c (a linear factor), then the remainder is f ( c ) .
Factor Theorem
A polynomial f ( x ) has a linear factor x − c , if and only if f ( c ) = 0 .
3.
Show that x − 2 is a factor of f ( x ) = x3 − 4 x 2 + 3 x + 2
Find the quotient and remainder if f ( x ) is divided by g ( x ) . Use this information to write f ( x ) as a product of linear and
quadratic factors irreducible over  .
4.
f ( x) =
− 2 x3 − 8 x 2 − 2 x + 1 2
g ( x )= x − 1
5.
f ( x=
) x6 − 6 4
p ( x ) = x2 − 2 x + 4
Maximum Number of Real Zeros
A polynomial function cannot have more real zeros than its degree.
Descartes’ Rule of Signs
Let f denote a polynomial function written in standard form.
• The number of positive real zeros of f either equals the number of variations in the 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 the sign of the nonzero coefficients of
f ( − x ) or else equals that number less an even integer.
State the maximum number of real zeros that each polynomial may have, then use Descartes’ Rule of Signs to determine how
many positive and how many negative zeros each polynomial function may have. Do not attempt to find the zeros.
6.
f ( x ) = 5x4 + 2 x2 − 6 x − 5
Theorem on Rational Zeros of a Polynomial
If the polynomial f ( x )= an x n + an −1 x n −1 +  + a1 x + a0 has integer coefficients and if
c
is a rational zero of f ( x ) , such
d
that conjugate c and d have no common prime factor, then
1. The numerator, c , of the zero is a factor of the constant term a0 .
2. The denominator, d , of the zero is a factor of the leading coefficient an .
In other words:
f a c t o r s o f t h e c o n s t a n t t e r m a0
Possible rational zeros
=
f a c t o r s o f t h e l e a d i n g c o e f f i c i e n t an
List all possible rational zeros.
7.
3 x5 − 1 0 x 4 − 6 x3 + 2 4 x 2 + 1 1x − 6 =
0
Factors of -6:
Factors of 3:
Possible Rational zeros:
Use the Rational Zeros Theorem to find all the real zeros of the polynomial. Use the zeros to factor f over the real numbers.
8.
3 x5 − 1 0 x 4 − 6 x3 + 2 4 x 2 + 1 1x − 6 =
0
Use synthetic division to test a possible rational zero. Try:
Use synthetic division to test a possible rational zero. Try:
Rewrite equation:
Note: The constant and leading coefficient are still the same as the original, so the possible rational zeros are still
the same, and what didn’t work previously won’t work now.
Use synthetic division to test a possible rational zero for the quotient. Try:
Rewrite equation:
Note: The constant and leading coefficient are still the same as the original, so the possible rational zeros are still
the same, and what didn’t work previously won’t work now.
Use synthetic division to test a possible rational zero for the quotient. Try:
Use synthetic division to test a possible rational zero for the quotient. Try:
Rewrite equation:
Note: The quotient is now a quadratic that can be factored.
Write the equation as a product of linear and quadratic factors irreducible over  .
Solutions to 3 x5 − 1 0 x 4 − 6 x3 + 2 4 x 2 + 1 1x − 6 =
0 are:
Solve each equation in the real number system.
9.
x 4 − 2 x3 + 1 0 x 2 − 1 8 x + 9 =
0
10.
x3 +
3 2
x + 3x − 2 =
0
2
Graph the polynomial.
11.
f ( x ) = x3 +
3 2
x + 3x − 2
2