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How can I prove this:

Prove that, if the polynomial $p(x)=ax^3+bx^2+cx+d$ has roots $x_1,x_2,x_3$, then $d=ax_1 x_2 x_3$.

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    Hint: $p(x) = a(x-x_1)(x - x_2)(x - x_3)$. Note also, that $d = - ax_1x_2x_3$2011-11-28
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    Multiply out $a(x-x_1)(x-x_2)(x-x_3)$ and note the constant term...2011-11-28
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    $p(x)=a(x-x_1)(x-x_2)(x-x_3)$2011-11-28
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    I think there's a negative sign missing...2011-11-28
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    Just so everything's above the board: the image that originally accompanied the question had a minus sign missing. Check the edit history.2011-11-28
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    It's not true without further hypotheses, e.g. mod $6$ note $x^3-x$ has roots $x = -1,1,2$ but their product $\rm -2 \not\equiv 0$.2011-11-28
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    @BillDubuque um... The roots of $x^3-x$ are 0,1,-1 ... Or did you mean to write something else?2012-01-05
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    @BillDubuque Oh, nevermind, I misunderstood. I think though that we can safely assume that the OP meant polynomials over a field.2012-01-05
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    @Tim Even better, Bill's example is over a ring, not a field :-). In fact, the statement is true over all fields.2012-01-05

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As many have said in the comments ...

If a polynomial $p$ has a root $x_0$ then we can factor out that root, i.e. $p(x) = q(x) (x-x_0)$ where $q$ is a polynomial of degree one less than $p$.

This allows us to rewrite $p(x) = ax^3 + bx^2 + cx + d$ as $p(x) = k(x-x_1)(x-x_2)(x-x_3)$ where $k$ is some constant. Multiplying through, we see that the coefficient of the $x^3$ term is $k$, so $k = a$ ; and the constant term is $-kx_1x_2x_3$, so we see that $d = -ax_1x_2x_3$ , as desired.