Let f (x), G (x) and H (x) be polynomials, and the first coefficient of H (x) is 1. It is proved that: (f (x) H (x), G (x) H (x)) = (f (x), G (x)) H (x)

Let f (x), G (x) and H (x) be polynomials, and the first coefficient of H (x) is 1. It is proved that: (f (x) H (x), G (x) H (x)) = (f (x), G (x)) H (x)

Let (f (x), G (x)) = q (x)
Then f = Q * F1, g = Q * G1, and (F1, G1) = 1
Then there exist U (x), V (x) such that:
f1*u+g1*v=1
Simultaneously multiply Q (x) H (x)
Then F1 * Q * h * U + G1 * Q * h * V = Q * H
fh*u+gh*v=q*h
There are also: Q * h | f * h, Q * h | g * H
So: (f (x) H (x), G (x) H (x)) = (f (x), G (x)) H (x)
If you don't understand, please ask

How to prove that f (x) is irreducible if and only if f (AX + b) is irreducible?
Newton's rational root theorem of higher algebra is similar

If f (x) is reducible, then f (AX + b) is reducible. If f (x) is reducible, Let f (x) = g (x) H (x), where g (x) and H (x) are polynomials with rational coefficients of degree not less than 1. Then f (AX + b) = g (AX + b) H (AX + b)

It is known that the coefficient of the penultimate term in the expansion of [(the fourth root of x ^ - 1) + (the third root of x ^ 2)] ^ n is 45
(1) Items with x ^ 3
(2) The term with the largest coefficient

C(n,n-2)=C(n,2)=45
n*(n-1)/2=45
n^2-n-90=0
(n-10)(n+9)=0
n=10
(1) Finding terms with power x3
C(10,m)*(1/X)^(m/4)*(x)^[2(10-m)/3)
=C(10,m)*x^[2(10-m)/3-m/4)
2(10-m)/3-m/4=3
m=4
That is, the term of X3 power is C (10,4) x ^ 3 = 210x ^ 3
(2) Finding the term with the largest coefficient
Because n = 10, the expansion has 11 terms, that is, the sixth term is the largest
C(10,5)*x^[2(10-5)/3-5/4)=C(10,5)X^(25/12)