Question

The polynomial f(x) = x³-x²+px+q, where p and q are constants, has remainder 8 when divided by (x-1). If (x+1) is factor of f(x), find the values of p and q​

Answer 1

Answer:p=-1 and q=1

Step-by-step explanation:

Answer 2

Answer:

p=3, q=5

Step-by-step explanation:

f(x) = x³-x²+px+q

factor: x+1 = 0   x = -1

(-1)³ - (-1)² + p*(-1) + q = 0

-2 - p + q = 0

p = q-2

(x-1) is factor of f(x) - 8, root x = 1 : f(x) = x³-x²+(q-2)x+q-8

(1)³ - (1)² + (q-2)(1) + q - 8 = 0

q - 2 + q - 8 = 0

2q = 10     q = 5

p = q-2 = 3

check: x³-x²+3x+5 = (x+1)(x²-2x+5)

x³-x²+3x+5 = (x²+3)(x-1) + 8