Question

if X+2 is the factor of P of X= ax3(read as a xcube) +bx2(read as bxsquare) +x -6 and when divided by x-2 ,leaves a reminder 4 prove that a=0, b=2

Answer 1

p(x)=ax³+bx²+x-6

since x+2 is a factor of p(x)
p(-2)=0
a(-2)³+b(-2)²+(-2)-6=0
-8a+4b-8=0
-8a+4b=8
-2a+b=2    ...(1)

Given p(x) leaves remainder 4 when divided by x-2

p(2)=4
a(2)³+b(2)²+2-6=4
8a+4b=8
2a+b=2     ...(2)

adding (1) and (2) we get 
2b=4  ⇒b=2

putting b=2 in (1) we get 
-2a+2=2⇒-2a=0
a=0