Pure Maths - Polynomial

(1) f(x) = g(x) + x^3 + kx^2 + 8x +8
Both f(x) and g(x) are divisible by x+2, so
f(-2)=0; g(-2)=0
f(-2) = g(-2) +(-2)^3 +k(-2)^2 +8(-2) + 8
0 = 0 - 8 + 4k - 16 + 8
k=4

(b) g(x) = h(x)(x^2-1) + (4x - 1)
g(1) = h(1)(1-1) + 4(1) -1
g(1) = 3

(i) f(1) = g(1) + 1 + 4 + 8 + 8
= 3 + 21
= 24

(ii) By long division,
x^3 + 4x^2 + 8x +8 = (x+4)(x^2-1) + (9x+12)
so f(x) = g(x) + (x+4)(x^2-1) + (9x+12)
= h(x)(x^2-1) + (4x - 1) + (x+4)(x^2-1) + (9x+12)
= [h(x) + (x+4)](x^2-1) + 13x + 11

So the remainder when f(x) is divided by (x^2-1) is 13x + 11