# Multiplication of polynomials in value representation as done for Fast Fourier Transform

I am trying to understand the discrete Fast Fourier Transform. I get the idea of switching between coefficient and value representations to and then back but I am stuck in figuring out how the multiplication of 2 polynomials using their value representation works.

The screenshot below is from a Reducible video (at time 6:39) that explains the Fast Fourier transform. I see that we need 5 points to get $$C(x)$$, which is of degree $$4$$, but I don't see how multiplying say the first point of $$A(x)$$ and $$B(x)$$, that is, $$(-2,1) \times (-2,9)$$ gives $$(-2,9)$$

There's a similar answer here but the steps aren't provided just the final answer, which still baffles me.

If $$P, Q, R \in \mathbb{R}[X], R = PQ$$, then for any $$x \in \mathbb{R}$$, $$R(x) = P(x)\times Q(x)$$.
It means that if the value representation of $$P$$ is $$[(x_1, P(x_1)); (x_2, P(x_2)); …, (x_n, P(x_n))]$$ and the value representation of $$Q$$ is $$[(x_1, Q(x_1)); (x_2, Q(x_2)); …, (x_n, Q(x_n))]$$, then the value representation of $$R$$ is $$[(x_1, P(x_1)Q(x_1)); (x_2, P(x_2)Q(x_2)); …, (x_n, P(x_n)Q(x_n))]$$.
In your example, $$(-2, 1)$$ represents a couple $$(x_i, P(x_i))$$ and $$(-2, 9)$$ is $$(x_i, Q(x_i))$$, so it's normal that $$(x_i, R(x_i))$$ is $$(-2, 1\times 9) = (-2, 9)$$.
• Thank you. I didn't realize that I needed to just multiply the $y$ values, what is $R(x)$ in your example, and that the $x$ remains the same. Apr 11, 2021 at 16:37