I started off with:
$$f(x)= a(x-(-3)) (x-(2)) (x-(-1)) (x-(-2))$$ $$f(x)= a(x+3) (x-2) (x+1) (x+2)$$
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2Now develop and choose the right $a$. Coefficient of $x^2$ is $-a$. – Jean-Claude Arbaut Mar 22 '13 at 19:28
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You are right that $f(x)=a(x+3)(x+2)(x+1)(x-2)$
Expanding gives $$f(x)=a(x^{2}+3x+2x+6)(x^{2}+x-2x-2)=a(x^{2}+5x+6)(x^{2}-x-2)$$
Multiplying the quadratics, $$f(x)=a(x^{4}+5x^{3}+6x^{2}-x^{3}-5x^{2}-6x-2x^{2}-10x-12)$$
Collecting terms gives $f(x)=a(x^{4}+4x^{3}-x^{2}-16x-12)$
Hence, the coefficient on $x^{2}$ is $-a$
To make the coefficient on $x^{2}$ $6$, let $a=-6$.
Jean-Claude Arbaut
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the x² coefficient should be 6 so in your answer shouldn't we pick a = -6? – imranfat Mar 22 '13 at 19:38
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Let's start with a polynomial of degree 4 :
$$p(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4$$
Since you know that $a_2=6$ and $p(-3)=p(2)=p(-1)=p(-2)=0$, you have 4 equations and 4 unknown ($a_0$, $a_1$, $a_3$ and $a_4$). This is a linear system that can be solved easily.
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1i think the OP's solution (or at least the direction pointed out) is far more elegant and easier... – long tom Mar 22 '13 at 19:36