Question

What is the remainder of this long division problem?

3x-2 then long hand with 6x^3+2x^2+11x-10 in it.

Answer 1

when does 3x-2 = 0?

Answer 2

if P(x) = (3x-2) Q(x) + R(x) then 3x-2 = 0, P(x) = R(x)

Answer 3

So it would be a 0 remainder?

Answer 4

dunno, when does 3x-2 = 0?

Answer 5

since it is stated that:

6x^3+2x^2+11x-10 = (3x-2) Q(x) + R(x)

then the value of x that makes 3x-2 = 0, lets say its u, will zero out the Q(x) part and we are left with

6u^3+2u^2+11u-10 = 0 + R(u)

Answer 6

3x-2 = 0 when x=2/3 so the remainder has to be:

6(2/3)^3+2(2/3)^2+11(2/3)-10 = R(2/3)

Answer 7

or just work thru the longhand division stuff ....