Use Descartes’ Rule of Signs
Describe the possible real zeros of
f(x) = x 4 – 3x 3 – 5x 2 + 2x + 7.
Examine the variations of sign for f(x) and for f(–x).
f(x) = x4 – 3x3 – 5x2 + 2x + 7
f(–x) = (–x)4 – 3(–x)3 – 5(–x)2 + 2(–x) + 7
= x 4 + 3x 3 – 5x 2 – 2x + 7
Use Descartes’ Rule of Signs
The original function f(x) has two variations in sign,
while f(–x) also has two variations in sign. By
Descartes' Rule of Signs, you know that f(x) has either
2 or 0 positive real zeros and either 2 or 0 negative
real zeros.
Answer: 2 or 0 positive real zeros, 2 or 0 negative
real zeros
Describe the possible real zeros of
g(x) = –x 3 + 8x 2 – 7x + 9.
A. 3 or 1 positive real zeros, 1 negative real zero
B. 3 or 1 positive real zeros, 0 negative real zeros
C. 2 or 0 positive real zeros, 0 negative real zeros
D. 2 or 0 positive real zeros, 1 negative real zero
Find a Polynomial Function Given Its Zeros
Write a polynomial function of least degree with real
coefficients in standard form that has –1, 2, and
2 – i as zeros.
Because 2 – i is a zero and the polynomial is to have
real coefficients, you know that 2 + i must also be a zero.
Using the Linear Factorization Theorem and the zeros
–1, 2, 2 – i, and 2 + i, you can write f(x) as follows:
f(x) = a[x – (–1)](x – 2)[x – (2 – i)](x – (2 + i)]
Find a Polynomial Function Given Its Zeros
While a can be any nonzero real number, it is simplest to
let a = 1. Then write the function in standard form.
f(x) = (1)(x + 1)(x – 2)[x – (2 – i)][x – (2 + i)]
Answer:
Let a = 1.
Write a polynomial function of least degree with
real coefficients in standard form that has
–2 (multiplicity 2), 0, and 3i as zeros.
A. f(x) = x 5 + 4x 4 + 13x 3 + 36x 2 + 36x
B. f(x) = x 5 + 4x 4 + 9x 3 + 18x
C. f(x) = x 3 + 2x 2 – 3ix 2 – 6xi
D. f(x) = x 4 + 4x 3 – 5x 2 – 36x – 36