A question for factoring 5, a question for factoring

Observing the characteristics of this problem, the polynomial in the problem cannot be factored according to the general method, and conditions need to be created. This problem adopts the method of splitting the item grouping, splitting the xyz term into -xyz and -2xyz, participating in the two groups before and after the factorization, and then putting the decomposition results of the two parts together for factorization, so at least three simple factorization needs to be carried out, so the two parts before and after the grouping should not only be factorized, but also consider the decomposition results of the two parts before and after being put together, and further factorization. This part is one of the more difficult parts of factoring and requires a lot of practice to master.

It exercises the ability to synthesize and flexibly apply the basic methods of factorization, such as formula method and group factorization method.

This shouldn't be able to be decomposed.,Landlord.,You see if there's a mistake in the question.,If there's no error.,Add it to the question.,I'll ask my teacher.。

Ah, finally...

Use the meta-method: Let x 5=t

x^4+x^3+x^2+x+1=m

Easy to obtain: x 5-1=(x-1)*(x 4+x 3+x 2+x+1), i.e. t-1=m(x-1).

Original = (t+m) 2-t

t^2+m^2+2mt-t

t(t-1)+m(m+2t)

tm（x-1）+m（m+2t）

m（tx-t+m+2t）

m（tx+t+m）

x^4+x^3+x^2+x+1）*（x^6+x^5+x^4+x^3+x^2+x+1)

This method is applicable to x 6, x 4, x 3, etc.

x 3-x 2+ax+b=(x+1) 2(x+c) x 3+3cx 2+3cx+c

Two-pass control then -1=3c a=3c b=c

Intercepted a=-1 b=c=-1 3

The other factor is (x-1 3).

a+b=3 =>b=a-3

a-3)*a=-10 => a^2-3a+10=0(a-5)(a+2)=0

i.e. a1 = 5 a2 = -2

Because a and b are interchangeable, i.e. a and b have values of 5, -2, or , respectively.