Using matrix to count the product of two polynomias

Consider two polynomials, without losing generality, $p(x)=1+x+3x^2,q(x)=3+5x+2x^2$$p(x)=1+x+3x^2,q(x)=3+5x+2x^2$

Represent polynomials to Matrix, that is $\begin{array}{c}{M}_p=\left(\begin{array}{c}1\\ 1\\ 3\end{array}\right),M_p=\left(\begin{array}{ccc}3& 5& 2\end{array}\right)\end{array}$$M_p=\begin{pmatrix}1\\1\\3\end{pmatrix},M_p=\begin{pmatrix}3&5&2\end{pmatrix}$

and consider $\begin{array}{c}{M}_p{M}_q=\left(\begin{array}{ccc}3& 5& 2\\ 3& 5& 2\\ 9& 15& 6\end{array}\right)\end{array}$$M_p M_q=\begin{pmatrix}3&5&2\\3&5&2\\9&15&6\end{pmatrix}$

Observe that $a_{ij},i+j=k$$a_{ij},i+j=k$ is a coefficient of $x^k$$x^k$, and consider $a_{i+1j-1}...$$a_{i+1j-1}...$

The coefficient of $x^k$$x^k$ is ${\displaystyle \sum _{i+j=k}{a}_{ij}}$$\displaystyle\sum_{i+j=k}a_{ij}$

It corresponds to, just watch the slanted-up 45 degrees, in this case, 3,(3,5),(9,5,2),(15,2),(6)

So we get $pq(x)=p(x)q(x)=3+8x+16x^2+17x^3+6x^4$$pq(x)=p(x)q(x)=3+8x+16x^2+17x^3+6x^4$

Actually, we can also use this to count products of natural number

Because you can write $abc$$abc$ as $a\times {10}^2+b\times {10}^1+c\times {10}^0$$a\times 10^2+b\times 10^1+c\times 10^0$

And you need to simplify the final result because all the coefficients $<10$$< 10$

For example, consider $124\times 243$$124\times 243$

$\left(\begin{array}{c}1\\ 2\\ 4\end{array}\right)\left(\begin{array}{ccc}2& 4& 3\end{array}\right)=\left(\begin{array}{ccc}2& 4& 3\\ 4& 8& 6\\ 8& 16& 12\end{array}\right)$$\begin{pmatrix}1\\2\\4\end{pmatrix}\begin{pmatrix}2&4&3\end{pmatrix}=\begin{pmatrix}2&4&3\\4&8&6\\8&16&12\end{pmatrix}$

So we get $(2,8,19,24,12)=(2,8,19,25,2)=(2,8,21,5,2)=(3,0,1,3,2)$$(2,8,19,24,12)=(2,8,19,25,2)=(2,8,21,5,2)=(3,0,1,3,2)$

So the answer is 30132