Question

find the values of Q and R if x^3-8 =(x-2) (x^2+Qx+R)

Answer 1

Foil out the right side first. Then equate the coefficients of equal-ordered terms on the two sides.
Ex: x^2-1= (x+A)(x+B) = x^2 + (A+B)x+AB
So I have A+B=0 and AB=-1 => A=-1 and B = 1. Vice versa works too.

Answer 2

so after foiling out (x-2)(x^2+Qx+R) I would get x^3 + Qx^2 +Rx - 2x^2 -2Qx-2R?

Answer 3

Yes. Now group terms with x's that have the same exponent.

Answer 4

i grouped x^3 + 2Qx^2+Rx-2Qx-2R? is that right?

Answer 5

I mean factor them out. See what I did up there with the Ax+Bx=(A+B)x

Answer 6

factor them out as in factorise?

Answer 7

No I just meant to rewrite it as x^3 + Qx^2 +Rx - 2x^2 -2Qx-2R= x^3+(Q-2)x^2+(R-2Q)x-2R
Are you able to see how I got that?

Answer 8

how did you rewrite it as that?

Answer 9

I grouped the terms that have x^2 together and those that have x together.

Answer 10

but i dont see how you got (Q-2) (R-2Q)x

Answer 11

It's (Q-2)x^2+(R-2Q)x. Expand and you see you get something like part of the original foil