Transcript ppt

3.3 Zeros of polynomial functions
Rational Zero Test: If a polynomial f(x) has
integer coefficients, every rational zero of f has the form
p factors of the constant term
rational zero = 
q factors of the L. coefficint
where p and q have no common factors other than 1.
• p is a factor of the constant term.
• q is a factor of the leading coefficient.
Example: List all possible rational zeros of f(x) = 4x3 + 3x2 – x – 6.
q=4
p=–6
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Find the zeros of
f(x) = 2x3 + x2 – 5x + 2.
f(x) = 2x3 – 9 x2 + 2.
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2 2
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f ( x)  x  3x  x  9 x  4
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Upper and Lower Bounds Theorem
• When P(x)÷(x-c) by syn. Div and P(x) with real coefficients.
Upper Bound: a) c>0 and the leading coefficient > 0
x = c is an upper bound for the real zeros if all
numbers in the bottom row are either +ve or zeros
b) c>0 and the leading coefficient < 0
x = c is an upper bound for the real zeros if all
numbers in the bottom row are either -ve or zeros
Lower Bound: c<0
x = c is a lower bound for the real zeros if all
numbers in the bottom row are alternate in sign
( the zero can be considered +ve or –ve)
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Example: According to the Bounds Theorem find both the smallest
positive integer that is upper bound and the largest negative integer that is
lower bound for the real zeros
• f (x) = – 4 x4 +12 x3+3x2 – 12x + 7.
• f (x) = 4 x3 – 11x2 – 3x + 15.
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Descartes’s Rule of Signs: If f(x) is a polynomial with real
coefficients and a nonzero constant term,
1. The number of positive real zeros of f is equal to the number
of variations in sign of f(x) or less than that number by an
even integer.
2. The number of negative real zeros of f is equal to the
number of variations in sign of f(–x) or less than that number
by an even integer.
A variation in sign means that two consecutive, nonzero
coefficients have opposite signs.
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Example: Use Descartes’s Rule of Signs to determine the
possible number of positive and negative real zeros of
f(x) = 2x4 – 17x3 + 35x2 + 9x – 45.
The polynomial has three variations in sign.
+ to –
+ to –
f(x) = 2x4 – 17x3 + 35x2 + 9x – 45
– to +
f(x) has either three positive real zeros or one positive real zero.
f(– x) = 2(– x)4 – 17(– x)3 + 35(– x)2 + 9(– x) – 45
=2x4 + 17x3 + 35x2 – 9x – 45
f(x) has one negative real zero.
One change in sign
f(x) = 2x4 – 17x3 + 35x2 + 9x – 45 = (x + 1)(2x – 3)(x – 3)(x – 5).
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