Question

Prove that `deg( frac(f(x))g(x) * frac(F(x))(G(x)) ) = deg( frac(f(x))g(x)) + deg( frac(F(x))(G(x)))` ( `deg(p(x))` means the degree of the polynomial `p(x)` ) ?

Answer 1

See explanation

Explanation:

Let's first pose `q(x) = f(x)/g(x)` and `r(x) = (F(x))/(G(x))`

Now let's rewrite the expression :

`deg(f(x)/g(x) * (F(x))/(G(x))) = deg(q(x) * r(x))`

We know that `AA a, b ; a/b = a*b^-1 implies a * b = a/b^-1`
So :

`deg(q(x) * r(x)) = deg( (q(x))/(r(x)^-1))`

We know that `deg((s(x))/(t(x))) = deg(s(x)) - deg(t(x))`
So :

`deg( (q(x))/(r(x)^-1)) = deg(q(x)) - deg(r(x)^-1)`

We know that `deg(s(x)^-1) = -deg(s(x))`
So :

`deg(q(x)) - deg(r(x)^-1) = deg(q(x)) - (-deg(r(x))) = deg(q(x)) + deg(r(x))`

Substitute in back the values:

`deg(q(x)) + deg(r(x)) = deg(f(x)/g(x)) + deg((F(x))/(G(x)))`

SO :

`deg(f(x)/g(x) * (F(x))/(G(x))) = deg(f(x)/g(x)) + deg((F(x))/(G(x)))`