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.

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.