Question

If p(x)=x³-ax²+bx+3 leaves a remainder -19 when divided by (x+2) and a remainder 17 When divided by (x-2).prove that a+b=6

Answer 1

p(x)=x³-ax²+bx+3

I case:- Remainder -19 when divided by (x+2)

⇒Remainder = p(-2) = (-2)³-a(-2)²+b(-2)+3

⇒-8-4a-2b+3

⇒-(4a+2b+5) = -19

i.e 4a + 2b = 14

or 2a+b = 7..............(1)

II case:- Remainder 17 when divided by (x-2)

⇒Remainder = p(2) = (2)³-a(2)²+b(2)+3

⇒8-4a+2b+3

⇒-4a+2b+11 = 17

i.e -4a +2b = 6

or -2a+b = 3............(2)

On solving (1) & (2)

a = 1 & b = 5