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My Precalc test has the following Synthetic Division question: if $F(x)=4x^3 +2x^2-2ax-4a^2$, and $F(a)=0$, find all possible values of $a$.

So, when I got to the residual part of the synthetic division, I get $-4a^2 + 4a^3 $ and I set this equal to $0 $ and get $a=1$ and $a=0$. All I get is a big red x thru my answer.

What is wrong with those answers?

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    Your answer is right. Maybe there was more to the question (or answer) than you posted. – 2017-02-19
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    I have the strong feeling that either you misread the question, mis-typed the answer, or your "answer machine" is a tad bit broken. At least I'm pretty sure you are right... – 2017-02-19

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When you divide your polynomial $f(x) = 4x^3+2x^2-2ax-4a^2$ by the linear polynomial $x-a$, you say you get the constant polynomial $r(x) = -4a^2+4a^3 = -4a^2(1-a)$, which is to say that there exists a polynomial $h(x)$ such that

$f(x) = h(x)(x-a) + r(x)$

But you know $0 = f(a) = h(a)(a-a)+r(a) = r(a)$. Thus $-4a^2(1-a) = 0$. For what $a$ can this be possible?

You are right by saying it must be that $a=0,1$. When $a=0$, the original polynomial becomes $f(x)=4x^3 +2x^2$, which clearly satisfies $0=4(0)^3+2(0)^2 =f(0)=f(a)$. Moreover, when $a=1$, the original polynomial becomes $f(x) = 4x^3+2x^2-2x-4$, which as you can see also satisfies $0 = 4(1)^3+2(1)^2-2(1)-4 = f(1) = f(a)$.

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    You are suggesting $a=-1$, which, through plug and chug into the original equation, is clearly false. – 2017-02-19
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    I was suggesting that a=1 or a=0 – 2017-02-19
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    I don't see the sign error. I was going to suggest that your version has a sign error. Can you show me, I just don't see it. If I compute F(1) and F(0), I do indeed get 0 – 2017-02-19
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    That was Joeb's work, not mine. I get $4a^2(a-1)=0$ which results in $a=0$ and $a=1$. – 2017-02-19
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    Let's move this to chat. – 2017-02-19
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53882/discussion-between-user163862-and-dxiv). – 2017-02-19
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    So, does that make my answer of a=1 and a=0 correct? – 2017-02-19
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    @user163862 Your answer *is* correct, as commented directly under your question. All the other comments here, below joeb's post, were about *his* answer, *not* yours. – 2017-02-19
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    I fixed the sign error. Sorry about that – 2017-02-19
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    I was just seeking another opinion other than my precalc teahcer's that my answer was correct of a=0 and a=1. When I calculate $F(a)$, I get $4a^3 -4a^2$, which to me certainly looks like $a=1$ and $a=0$, but I'm beginning to think there must be something more to this, since it has generated so much discussion here. – 2017-02-19