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For polynomial multiplication, if $A(x)$ and $B(x)$ are polynomials of degree-bound $n$, we say that their product $C(x)$ is a polynomial of degree-bound $2n-1$ such that $C(x)=A(x)B(x)$ for all $x$ in the underlying field. A way to express the product $C(x)$ is

$$C(x)= \sum_{j=0}^{2n-2} c_j x^j$$

where

$$c_j= \sum_{k=0}^j a_k b_{j-k}$$

Using the above equations, I want to find the product $A(x)B(x)$, where $A(x)=7x^3-x^2+x-10$, $B(x)=8x^3-6x+3$.

I found the following:

$c_0=-30 \\ c_1=63 \\ c_2=-89 \\ c_3=-53$

$c_4= a_0 b_4+a_1 b_3+a_2 b_2+ a_3 b_1+ a_4 b_0=8+21=29$

$c_5=a_0 b_5+ a_1 b_4+ a_2 b_3+ a_3 b_2+ a_4 b_1+ a_5 b_0=-8$

Is it right so far?

Also, is there a typo at this sum: $C(x)= \sum_{j=0}^{2n-2} c_j x^j$ ? Shouldn't the highest power of the product be $2n$?

If so, then $c_6=a_0 b_6+ a_1 b_5+ a_2 b_4+ a_3 b_3+ a_4 b_2+ a_5 b_1+ a_6 b_0=56$

So is it as follows?

$$C(x)=56 x^6-8x^5+29 x^4-53x^3-89x^2+63 x-30$$

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    $\begingroup$ If "degree bound $n$" means degree less than $n$, i.e. $\le n-1$, then $C$ has degree $\le 2n-2$, i.e. degree bound $2n-1$. $\endgroup$ – Robert Israel Jun 9 '15 at 20:41
  • $\begingroup$ @RobertIsrael So you mean that the result should be $C(x)= \sum_{j=0}^{4} c_j x^j$ ? $\endgroup$ – evinda Jun 9 '15 at 21:09
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    $\begingroup$ No, if $A$ and $B$ have degree $3$, as they do in your example, their degree bounds would be $4$, not $3$. The product has degree $6$. $\endgroup$ – Robert Israel Jun 10 '15 at 1:07

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