when ax^ - 4x + 4 is divided by x - 2, the remainder is 4, what is the quotient?
The exponent of x in the first term is missing but here is the general procedure.
We have $\displaystyle ax^\alpha - 4x + 4 = P(x)(x-2)+4$, where $\displaystyle P(x)$ is a polynomial of degree $\displaystyle \alpha-1$
$\displaystyle \implies (ax^\alpha - 4x + 4) - 4 = P(x)(x-2) \implies ax^\alpha - 4x = P(x)(x-2) \implies $ 2 is a root of $\displaystyle ax^\alpha - 4x$ $\displaystyle \implies a2^\alpha = 8 \implies a = \frac{8}{2^\alpha}$
is that so?
$\displaystyle 4a-4$ is the remainder ... set that equal to 4 and solve for $\displaystyle a$Code:2] a ..... -4 ...... 4 .............2a ....(4a-8) -------------------------- .....a....(2a-4) ...(4a-4)
... note you could also use the remainder theorem to answer this question.