MHB Is the Degree of the Product of Two Polynomials 2n?

[evinda]

Hello! (Wave)

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: $C(x)= \sum_{j=0}^{2n-2} c_j x^j$ ? Should the upper bound be $2n-1$ as at the above proposition is said?

So is the result $\sum_{k=0}^5 c_k x^k$ with the above calulated $c_k$? (Thinking)Thinking about it again, 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?

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

 

[Nono713]

Come now evinda. What is the product of the two polynomials below (you do know how to multiply polynomials out by hand, yes?):
$$(7x^3−x^2+x−10)(8x^3−6x+3)$$
Once you have found the product manually, compare it with the solution you got, and with the solution you would have gotten following the proposition given. Can the proposition possibly be correct? Always check your work! If something doesn't look right, plug in an example to see if it comes out wrong!

Thinking about it again, shouldn't the highest power of the product be 2n?

What is the product of two linear polynomials (of degree 1)? Certainly it is a quadratic, a polynomial of degree 2. And 2 is twice 1, correct? The proposition concludes that the product of two linear polynomials has degree 0 (or 1, with 2n - 1) which is absurd. So it can't be right, and you are correct that it needs to be fixed to 2n.