1. 2.5
Real Zeros of
Polynomial Functions
Descartes Rule of Signs
The number of positive real zeros of f is
either equal to the number of sign changes of
f(x) or is less by an even integer.
The number of negative real zeros of f is
either equal to the number of sign changes of
f(-x) or is less by an even integer.

2. Ex. Apply Descartes’s Rule of Signs to
f(x) = 2x4 + 7x3 – 4x2 – 27x - 18
How many sign variations are there in f(x)?
Only one sign change, so f must have exactly
one positive real zero.
How about f(-x)
f(-x) = 2x4 – 7x3 – 4x2 + 27x -18
3 variations in sign means f has either 3 or 1
negative real zeros.

3. Apply Descartes’s Rule to f(x) = x3 – x + 1
2 or 0 (+)
1 (-)
The Rational Zero Test
Rational Zero =
p = a factor of the constant term
q = a factor of the leading coefficient

4. Find the rational zeros of
Descartes Rule
3 or 1 (+)
Ex. f(x) = x4 – x3 + x2 – 3x - 6
1 (-)
factors of 6
Possible rational zeros =
factors of 1
Start with 1 1 1 1 -2 -8
did not = 0, so not a zero 2 1 1 3 3
Works! (x – 2) is a factor -1 1 3
Works! (x + 1) is a factor
(x2 + 3)(x – 2)(x + 1) are the factors & -1 and 2 are zeros

5. Ex. Find the rational zeros of f(x) = 2x3 + 3x2 – 8x + 3
Desc.
2 or 0 (+)
Poss. zeros
1 (-) 1 2 5 -3
It works!
(x – 1)(2x2 + 5x –3)
(x – 1)(2x – 1)(x + 3) ½

6. Ex. f(x) = 10x3 – 15x2 – 16x + 12
Desc.
2 or 0 (+)
Poss. Zeros
1 (-) 1 10 -5
-21 -9 2 10 5 -6
Works!
(x – 2)(10x2 + 5x – 6)
Quad. form

7. Use the given zero to find all the zeros of the function.
Given: 3i
f(x) = x3 + x2 + 9x + 9
If 3i is a zero, then -3i, it’s conjugate, is a zero.
Multiply these zeros in factored form.
(x - 3i)(x + 3i)
= x2 + 9
Zeros are 3i, -3i, and -1