Question

Find the remainder when p(x) = x^3-ax+6 is divided by (a+x)​

Answer 1

Answer:

We have x

We have x 3

We have x 3 −ax

We have x 3 −ax 2

We have x 3 −ax 2 +6x−a

We have x 3 −ax 2 +6x−a Apply remainder theorem

We have x 3 −ax 2 +6x−a Apply remainder theorem x−a=0

We have x 3 −ax 2 +6x−a Apply remainder theorem x−a=0x=a

We have x 3 −ax 2 +6x−a Apply remainder theorem x−a=0x=aPut x=a in equation.

We have x 3 −ax 2 +6x−a Apply remainder theorem x−a=0x=aPut x=a in equation. (a)

We have x 3 −ax 2 +6x−a Apply remainder theorem x−a=0x=aPut x=a in equation. (a) 3

We have x 3 −ax 2 +6x−a Apply remainder theorem x−a=0x=aPut x=a in equation. (a) 3 −a(a)

We have x 3 −ax 2 +6x−a Apply remainder theorem x−a=0x=aPut x=a in equation. (a) 3 −a(a) 2

We have x 3 −ax 2 +6x−a Apply remainder theorem x−a=0x=aPut x=a in equation. (a) 3 −a(a) 2 +6a−a

We have x 3 −ax 2 +6x−a Apply remainder theorem x−a=0x=aPut x=a in equation. (a) 3 −a(a) 2 +6a−a=a

We have x 3 −ax 2 +6x−a Apply remainder theorem x−a=0x=aPut x=a in equation. (a) 3 −a(a) 2 +6a−a=a 3

We have x 3 −ax 2 +6x−a Apply remainder theorem x−a=0x=aPut x=a in equation. (a) 3 −a(a) 2 +6a−a=a 3 −a

We have x 3 −ax 2 +6x−a Apply remainder theorem x−a=0x=aPut x=a in equation. (a) 3 −a(a) 2 +6a−a=a 3 −a 3

We have x 3 −ax 2 +6x−a Apply remainder theorem x−a=0x=aPut x=a in equation. (a) 3 −a(a) 2 +6a−a=a 3 −a 3 +6a−a

We have x 3 −ax 2 +6x−a Apply remainder theorem x−a=0x=aPut x=a in equation. (a) 3 −a(a) 2 +6a−a=a 3 −a 3 +6a−a=6a−a

We have x 3 −ax 2 +6x−a Apply remainder theorem x−a=0x=aPut x=a in equation. (a) 3 −a(a) 2 +6a−a=a 3 −a 3 +6a−a=6a−a=5a

We have x 3 −ax 2 +6x−a Apply remainder theorem x−a=0x=aPut x=a in equation. (a) 3 −a(a) 2 +6a−a=a 3 −a 3 +6a−a=6a−a=5aThen reminder is 5a