The binomial x-2 is a factor of the polynomial below. f(x)=x3+x2+nx+10 What is the value of n?

Explanation:

Since x-2 is a factor of f(x), this means f(2) = 0.

More generally, if x-k is a factor of p(x), then p(k) = 0. This is a special case of the remainder theorem.

So if we plugged x = 2 into f(x), we'd get

f(x) = x^3+x^2+nx+10

f(2) = 2^3+2^2+n(2)+10

f(2) = 8+4+2n+10

f(2) = 2n+22

Set this equal to 0 and solve for n

2n+22 = 0

2n = -22

n = -22/2

n = -11 is the answer

Therefore, x-2 is a factor of f(x) = x^3+x^2-11x+10

Plug x = 2 into that updated f(x) function to find

f(x) = x^3+x^2-11x+10

f(2) = 2^3+2^2-11(2)+10

f(2) = 8+4-22+10

f(2) = 0

Which confirms our answer.