Question

if p(x) =×^3-2×^2+kx+5 is divisible by (×-2), the remainder is 11, find k, hence find all the zeroes of x^3+kx^2+3k+1

Answer 1

Hey there!


Let f(x) = x3 + 2 x2 + k x + 3
Remainder when f(x) is divided by (x - 30 is f(3).
⇒ f(3) = (3)3 + 2(3)2 + k(3) + 3 = 21
⇒ f(3) = 27 + 18 + 3k + 3 = 21
⇒ 48 + 3k = 21
⇒ 3k = -27
⇒ k = -9
By substituting k = -9 in x3 + 2x2 + kx – 18 we get x3 + 2x2 - 9x – 18.
Let f(x) = x3 + 2x2 - 9x – 18
f(-2) = (-2)3 + 2(2)2 - 9(2) - 18 = 0
Therefore, (x + 2) is a factor of x3 + 2x2 - 9x – 18.
f(-3) = (-3)3 + 2(-3)2 - 9(-3) - 18 = 0
Therefore, (x + 3) is a factor of x3 + 2x2 - 9x – 18.
f(3) = (3)3 + 2(3)2 - 9(3) - 18 = 0
Therefore, (x - 3) is a factor of x3 + 2x2 - 9x – 18.

The factors of x3 + 2x2 - 9x – 18 are (x - 3), (x + 3) and (x + 2).

Hope it help