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How does one write the polynomial $p(x)=\frac{1}{2}x^3+(-\frac{3}{2})x^2+1$ using the standard basis $\{1,x,x^2,x^3\}$ ?

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$p(x)=\frac12 x^3+\left(-\frac32\right) x^2+0x+1$. – ՃՃՃ Nov 11 '12 at 2:31
That's it? For some reason I imagined I had to do something substantially more complex. – user1132363 Nov 11 '12 at 2:34
Nope, that's it. – ՃՃՃ Nov 11 '12 at 2:35
up vote 8 down vote accepted

It writes itself! Specifically, you want to represent $p(x)$ as a linear combination of the polynomials $1,x,x^2$, and $x^3$, which simply means that you want to write it in the form $$c_0\cdot1+c_1x+c_2x^2+c_3x^3$$ for some scalars $c_0,c_1,c_2$, and $c_3$. And the polynomial $p(x)$ itself does just that:

$$p(x)=1\cdot1+0\cdot x+\left(-\frac32\right)x^2+\frac12x^3\;,$$

with coefficients $1,0,-\frac32$, and $\frac12$.

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Couldn't be more clear. Thank you! (Will accept your answer as soon as stackexchange lets me) – user1132363 Nov 11 '12 at 2:36
@user1132363: You’re welcome! – Brian M. Scott Nov 11 '12 at 2:37

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