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The polynomial f(x) is divisible by (2x-3) and leaves a remainder of -2 when divided by (x-1). Find the remainder when f(x) is divided by (2x-3)(x-1).

I figured that since (2x-3) is a factor, I can ignore it since I am only interested in the remainder. So is the remainder just equals to -2?

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That reasoning can't possibly be right: $f$ is divisible by $(2x-3)$ so the remainder must also be divisible by $2x-3$ so it can't be a constant (other than $0$, perhaps).

The very general way to solve problems like this is: Write $f(x) = (2x-3)(x-1)g(x) + Ax + B$ and the reminder is $Ax+B$, with $A$ and $B$ constants that we must determine.

Now plug in $x = 1.5$ and $x = 1$. $f(1.5) = 0$, so $1.5A + B = 0$, and $f(1) = -2$, so $A+B = -2$. Now solve for $A$ and $B$ (which is trivial).

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Hint $\ f = (2x\!-\!3)g\ $ and $\,-2 = f(1) = -g(1),\,$ so $\,g = 2+(x\!-\!1)h$.

Thus $\ f = (2x\!-\!3)(2+(x\!-\!1)h)\, =\, 4x\!-\!6+ (2x\!-\!3)(x\!-\!1)h$

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