Calculate the value of k for which the polynomial equation f(x)=x^3-kx+12 has the root of x=-4

If x = -4 is a root of f(x), then dividing f(x) by x + 4 leaves a remainder of 0.

Compute the quotient,

(x³ - kx + 12) / (x + 4)

x³ = x² • x, and

x² (x + 4) = x³ + 4x²

Subtract this from the dividend (f(x)) to get an initial remainder of

(x³ - kx + 12) - (x³ + 4x²) = -4x² - kx + 12

-4x² = -4x • x, and

-4x (x + 4) = -4x² - 16x

Subtract this from the previous remainder to get a new one of

(-4x² - kx + 12) - (-4x² - 16x) = (16 - k) x + 12

(16 - k) x = (16 - k) • x, and

(16 - k) (x + 4) = (16 - k) x + 64 - 4k

which gives the next remainder,

((16 - k) x + 12) - ((16 - k) x + 64 - 4k) = 4k - 52

4k - 52 does not divide x, so we're done and what we've shown is

f(x) / (x + 4) = x² - 4x + 16 - k + (4k - 52)/(x + 4)

(just gathering the bold terms above and the last remainder)

The remainder should be 0, so that

4k - 52 = 0

4k = 52

k = 13

4444444444444 i think