POLYNOMIALS 1A

If f(x) = x⁴ + kx² + 6x + 7 has a remainder of 22 when divided by x+2; find k.

solution

f(x) = x⁴ + kx² + 6x + 7

x + 2 = 0

x = -2

f(x) = (-2)⁴ + k(-2)² + 6(-2) + 7 = 22

16 + 4k + (-12) + 7 = 22

16 + 4k - 12 + 7 = 22

16 + 4k - 5 = 22

4k + 16 - 5 = 22

4k + 11= 22

4k = 22 - 11

4k = 11

4k = 11

4      4

k = ¹¹/4

hence k=¹¹/4

TRY THIS......................

 If f(x) = x⁴ - kx² + 3x - 11 has a remainder of 16 when divided by x-3; find k.