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)# ) ?

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Answer 1 · 2017-12-15T17:28:35

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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)))#