<u>Answer:</u>
(x - 2) is not a factor of f(x) = 3x^5 - 7x^3 - 11x^2 + 2
<u>Step-by-step explanation:</u>
According to the Remainder Theorem, when a polynomial f(x) is divided by a term (x - r), then the remainder must be f(r).
So first we will divide f(x) by (x - 2) using long division to get:
(3x^5 - 7x^3 - 11x^2 + 2) / (x - 2)
= 3x^4 + (6x^4 - 7x^3 - 11x^2 + 2) / (x-2)
= 3x^4 + 6x^3 + (5x^3-11x^2+2) / (x - 2)
= 3x^4 + 6x^3 + 5x^2 (-x^2 + 2) / (x - 2)
= 3x^4 + 6x^3 + 5x^2 -x -2 - (2) / (x -2)
=
Therefore the remainder is -2.
Now check x = 2 for 3x^5 - 7x^3 - 11x^2 + 2:
3(2)^5 - 7(2)^3 - 11(2)^2 + 2 = -2
The Remainder Factor theorem also states that if (x - r) is a factor of f(x) then f(r) must be 0.
So we found that f(2) = -2, therfore (x - 2) is not a factor of f(x) .
