lesson 5.6 rational zeros of polynomial functions

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LESSON 5.6 Rational Zeros of Polynomial Functions

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Page 1: LESSON 5.6 Rational Zeros of Polynomial Functions

LESSON 5 .6

Rational Zeros of Polynomial Functions

Page 2: LESSON 5.6 Rational Zeros of Polynomial Functions

Solve x3 – 5x2 + 8x – 4= 0. How? Can you factor this? Can you use the quadratic formula?

Now what if I tell you that one root is x =1?Then x3 – 5x2 + 8x – 4 = (x – 1)( ). Use synthetic division to find the missing factor.

The missing factor is x2 – 4x + 4, which can be factored. So solve and learn that x = 2.

So the solutions are 1 and 2 (double root).

Page 3: LESSON 5.6 Rational Zeros of Polynomial Functions

But what if you don’t know any zeros and cannot factor a polynomial?

Rational Zero TheoremTo find the possible rational zeros of a polynomial divide the factors of the constant by the factors of the leading coefficient.

List the possible rational zeros for f(x) = 2x3 + 3x2 – 4x + 6.

Answer: ±1,±2,±3,±6,±1/2,±3/2

Page 4: LESSON 5.6 Rational Zeros of Polynomial Functions

Now try this one….

a) Solve x3 + 4x2 + x – 6 = 0

Answer: x = 1, x = -2, x = -3

Do they all check?

Why do you think there were 3 solutions?

Page 5: LESSON 5.6 Rational Zeros of Polynomial Functions

and this one…

b) Find all of the roots of f(x) = x4 + 2x3 – 7x2 – 20x – 12

Answer: x = 3, -2 (double root) and -1. Verify this using your graphing calculator.

Page 6: LESSON 5.6 Rational Zeros of Polynomial Functions

and one last problem!

c) Find the zeros of f(x) = 6x4 – 25x3 + 30x2 – 5x – 6

How many answers should there be?Answer: x = 1, x = 2, x = 3/2, x = -1/3