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

Question

f(x) = 2x³ − 15x² + 31x − 18;    x − 1

Answer 1

(2x³ – 15x² + 31x –18) ÷ (x – 1) = 2x² – 13x +18 is found by long division.

2x² – 13x +18 = 2x² – 4x – 9x +18 = 2x(x – 2) – 9(x – 2) = (2x – 9)(x – 2)

So we can completely factor 2x³ – 15x² + 31x –18:

2x³ – 15x² + 31x –18 = (x – 1)(2x – 9)(x – 2)

2x³ – 15x² + 31x –18 = (x – 1)(2x – 9)(x – 2) = 0 yields three zeroes, which arise from

x – 1 = 0, 2x – 9 = 0, and x – 2=0

Answer

The three zeroes of the polynomial are

x = 1

x = 9/2

x = 2

Answer 2

divide the cubic by (x-1)

=

2x^2-13+18=0

factor

(2x-9)(x-2)

set each of the 3 factors= 0, solve for x

the zeros are x=9/2, 2 and 1

possible zeros using Descartes' rule of signs were 1 or 3 positive real zeros

no possible negative real zeros, 0 or 2 possible imaginary

18,9,6,3,2 1 or

9/2, 3/2, 1/2