Polynomial function and one factor

Question

use the Factor Theorem to find all real zeros for the given polynomial function and one factor.

Answer 1

By remainder theorem remainder r = f(-3) = - 81 1 + 9 + 60 + 12 = 0, so (x + 3) is a factor and x = - 3 is real zero of f(x) Now, to get others zeros, let do synthetic division:

-3| 3 1 - 20 12

-9 24 -12

3 -8 4 0

So we have quotiont 3x² - 8x + 4 = 0; we can factor left side:(3x - 2)(x - 2) = 0; 3x - 2 = 0 or x - 2 = 0

and x = 3/2 and x = 2 and x = 2/3 all zeros of our equation!

Answer 2

We try x= - 3

-3 | 3 1 -20 12

-9 24 -12

3 -8 4 0

f(-3) = 0, so x+3 is a factor

(3x - 2)(x - 2) = 0 factors as 3x^2 - 8x + 4

If you wish, you may plug 3/2 and 2 into the original polynomial to verify via the factor theorem.

The real zeros are -3, 2/3, and 2

Answer 3

With the factor provided, I know one zero must be x=-3, use this to do synthetic division.

-3 | 3 1 -20 12

-9 24 -12

3 -8 4 0

The remaining polynomial is 3x²-8x+4=0 Factor

(3x-2)(x-2)=0

Set the factors equal to zero and solve

3x-2=0

3x=2

x=2/3

x-2=0

x=2