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 ?

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

First, let's divide the given function f(x) = x^3 + 6x^2 + 10x + 4 by x - k, where k = -2.

To perform the polynomial long division, we start by dividing the highest degree term of the dividend (x^3) by the first term of the divisor (x). The result is x^2.

x^2
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x - (-2)| x^3 + 6x^2 + 10x + 4

Next, we multiply x^2 by the entire divisor, x - (-2), obtaining x^3 - 2x^2. We then subtract this result from the dividend:

x^3 + 6x^2 + 10x + 4
- (x^3 - 2x^2 + 2x - 4)
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8x^2 + 8x + 8

Now, we bring down the next term of the dividend, which is 8x^2, and repeat the process:

x^2 + 8

x^2 + 8
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x - (-2) | x^3 + 6x^2 + 10x + 4

- (x^3 - 2x^2 + 2x - 4)
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8x^2 + 8x + 8

Since there are no more terms to bring down, the remainder is 8x^2 + 8x + 8, which is denoted as r.

Hence, we can express f(x) in the form f(x) = (x - k)q(x) + r, where k = -2, as:

f(x) = (x + 2)(x^2 + 8) + (8x^2 + 8x + 8)

Now, regarding whether f(x) = 0. To determine if f(x) equals zero, we substitute x = 0 into the expression:

f(0) = (0 + 2)(0^2 + 8) + (8(0)^2 + 8(0) + 8)
= (2)(0 + 8) + (0 + 0 + 8)
= 2(8) + 8
= 16 + 8
= 24

Since f(0) = 24, f(x) is not equal to zero.