A little synthetic division yields:
f(x) = (x+2)(x^2+2x+k-4) + (-4-2k)
f(x) = (x-2)(x^2+6x+k+12) + (2k+20)
-4-2k = 2k+20
k = -6
Consider the function f(x)=x^3+4x^2+kx-4. The remainder from f(x)/(x+2) is twice the remainder from f(x) / (x-2). Determine the value of k
2 answers
first, sub in both x values
first (x-2), x=2
f(2) = (2)^3 + 4(2)^2 + k(2) - 4
f(2) = 20 + 2k
same with (x+2), x= -2
f(-2) = (-2)^3 + 4(-2)^2 + k(-2) - 4
f(-2) = 4 - 2k
we know that f(2) (x - 2) remainder is 2 times larger than f(-2) (x + 2) remainder, so:
2(20 + 2k) = 4 - 2k
40 + 4k = 4 - 2k
36 = -6k
therefore, k = -6
first (x-2), x=2
f(2) = (2)^3 + 4(2)^2 + k(2) - 4
f(2) = 20 + 2k
same with (x+2), x= -2
f(-2) = (-2)^3 + 4(-2)^2 + k(-2) - 4
f(-2) = 4 - 2k
we know that f(2) (x - 2) remainder is 2 times larger than f(-2) (x + 2) remainder, so:
2(20 + 2k) = 4 - 2k
40 + 4k = 4 - 2k
36 = -6k
therefore, k = -6