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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 $d = - ax_1x_2x_3$ – Alexander Thumm Nov 28 '11 at 13:22
Multiply out $a(x-x_1)(x-x_2)(x-x_3)$ and note the constant term... – J. M. Nov 28 '11 at 13:23
$p(x)=a(x-x_1)(x-x_2)(x-x_3)$ – pedja Nov 28 '11 at 13:23
I think there's a negative sign missing... – David Mitra Nov 28 '11 at 13:26
Just so everything's above the board: the image that originally accompanied the question had a minus sign missing. Check the edit history. – J. M. Nov 28 '11 at 13:41

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.

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