When f(x)= x^4+3x^3+px^2-2x+q is divided by (x-2) the remainder is 15, and (x+3) is a factor of f(x). find values of p and q
Division by (x-2). Remainder = 15.
(x-2) ) x^4+3x^3+px^2 - 2x + q( x^3 + 5x^2 + (10+p)x + (18+2p)
x^4-2x^3
5x^3+px^2
5x^3-10x^2
(10+p)x^2 – 2x
(10+p)x^2 – (20+2p)x
(18+2p)x + q
(18+2p)x – (36+4p)
(36+4p+q)
Remainder is R1 = 36 + 4p + q = 15
Division by (x+3). Remainder = 0, since (x+3) is a factor
(x+3) ) x^4+3x^3+px^2 - 2x + q( x^3 + px – (3p+2)
x^4+3x^3
0 + px^2 – 2x
px^2 + 3px
-(3p+2)x + q
-(3p+2)x – (9p+6)
9p+q+6
Remainder is R2 = 9p + q + 6 = 0
Using R1 and R2,
4p + q = -21
9p + q = -6
Subtracting,
5p = 15
p = 3
q = -33
Answer: the values of p and q are: p = 3, q = -33