Express f(x) in the form f(x)=(x-k)q(x)+r for the given value of k.

f(x)=x^3 +6x^2 +10x+4,k=-2

Is f(x)= 0 ?

no, but r=0, as a little synthetic division will show.

To express f(x) in the form f(x) = (x - k)q(x) + r, where k is a given value, we need to perform polynomial long division.

In this case, we have f(x) = x^3 + 6x^2 + 10x + 4 and k = -2.

Performing polynomial long division, we divide f(x) by (x - k):

x^2 + 4x + 2
__________________
x - (-2) | x^3 + 6x^2 + 10x + 4
- (x^3 - 2x^2 )
_______________
8x^2 + 10x
- (8x^2 - 16x)
_______________
26x + 4
- (26x - 52)
______________
56

The remainder is 56.

Therefore, f(x) = (x - k)q(x) + r can be written as:

f(x) = (x - (-2))(x^2 + 4x + 2) + 56

Simplifying, we have:

f(x) = (x + 2)(x^2 + 4x + 2) + 56

So, f(x) ≠ 0, unless the value of x makes (x + 2)(x^2 + 4x + 2) + 56 = 0.