Respuesta :
Answer:
If
€
p(x) is a polynomial, the solutions to the equation
€
p(x) = 0 are called the zeros of the
polynomial. Sometimes the zeros of a polynomial can be determined by factoring or by using the
Quadratic Formula, but frequently the zeros must be approximated. The real zeros of a polynomial
p(x) are the x-intercepts of the graph of
€
y = p(x).
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The Factor Theorem: If
€
(x − k) is a factor of a polynomial, then
€
x = k is a zero of the polynomial.
Conversely, if
€
x = k is a zero of a polynomial, then
€
(x − k) is a factor of the polynomial.
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Example 1: Find the zeros and x-intercepts of the graph of
€
p(x) =x
4−5x
2 + 4.
€
x
4−5x
2 + 4 = 0
(x
2 − 4)(x
2 −1) = 0
(x + 2)(x − 2)(x +1)(x −1) = 0
x + 2 = 0 or x − 2 = 0 or x +1= 0 or x −1= 0
x = −2 or x = 2 or x = −1 or x =1
So the zeros are –2, 2, –1, and 1 and the x-intercepts are (–2,0), (2,0), (–1,0), and (1,0).
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The number of times a factor occurs in a polynomial is called the multiplicity of the factor. The
corresponding zero is said to have the same multiplicity. For example, if the factor
€
(x − 3) occurs to
the fifth power in a polynomial, then
€
(x − 3) is said to be a factor of multiplicity 5 and the
corresponding zero, x=3, is said to have multiplicity 5. A factor or zero with multiplicity two is
sometimes said to be a double factor or a double zero. Similarly, a factor or zero with multiplicity
three is sometimes said to be a triple factor or a triple zero.
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Example 2: Determine the equation, in factored form, of a polynomial
€
p(x) that has 5 as double
zero, –2 as a zero with multiplicity 1, and 0 as a zero with multiplicity 4.
€
p(x) = (x − 5)
2(x + 2)x
4
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Example 3: Give the zeros and their multiplicities for
€
p(x) = −12x
4 + 36x3 − 21x
2.
€
−12x
4 + 36x3 − 21x
2 = 0
−3x
2(4x
2 −12x + 7) = 0
−3x
2 = 0 or 4x
2 −12x + 7 = 0
x
2 = 0 or x = −(−12)± (−12)
2−4(4)(7)
2(4)
x = 0 or x = 12± 144−112
8 = 12± 32
8 = 12±4 2
8 = 12
8 ± 4 2
8 = 3
2 ± 2
2
So 0 is a zero with multiplicity 2,
€
x = 3
2 − 2
2 is a zero with multiplicity 1, and
€
x = 3
2 + 2
2 is a zero
with multiplicity 1.
(Thomason - Fall 2008)
Because the graph of a polynomial is connected, if the polynomial is positive at one value of x and
negative at another value of x, then there must be a zero of the polynomial between those two values
of x.
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Example 4: Show that
€
p(x) = 2x3 − 5x
2 + 4 x − 7 must have a zero between
€
x =1 and
€
x = 2.
€
p(1) = 2(1)
3 − 5(1)
2 + 4(1) − 7 = 2(1) − 5(1) + 4 − 7 = 2 − 5 + 4 − 7 = −6
and
€
p(2) = 2(2)3 − 5(2)
2 + 4(2) − 7 = 2(8) − 5(2) + 8 − 7 =16 −10 + 8 − 7 = 7.
Because
€
p(1) is negative and
€
p(2) is positive and because the graph of
€
p(x) is connected,
€
p(x)
must equal 0 for a value of x between 1 and 2.
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If a factor of a polynomial occurs to an odd power, then the graph of the polynomial actually goes
across the x-axis at the corresponding x-intercept. An x-intercept of this type is sometimes called an
odd x-intercept. If a factor of a polynomial occurs to an even power, then the graph of the
polynomial "bounces" against the x-axis at the corresponding x-intercept, but not does not go across
the x-axis there. An x-intercept of this type is sometimes called an even x-intercept.
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Example 5: Use a graphing calculator or a computer program to graph
€
y = 0.01x
2(x + 2)3(x − 2)(x − 4)
4 .
x
y
–2 2 4
5
Because the factors
€
(x + 2) and
€
(x − 2) appear to odd
powers, the graph crosses the x-axis at
€
x = −2
and
€
x = 2.
Because the factors x and
€
(x − 4) appear to even
powers, the graph bounces against the x-axis at
€
x = 0
and
€
x = 4.
Note that if the factors of the polynomial were
multipled out, the leading term would be
€
0.01x10.
This accounts for the fact that both tails of the graph
go up; in other words, as
€
x → −∞,
€
y
Step-by-step explanation: