Factorization x 15 1 detailed process, thank you!

Lower-grade requirements for factorization are generally factorization within the range of real numbers, and for your question, it is mainly to examine your application of the following important factors:

x n+1=(x+1)(x (n-1)-x (n-2)+x (n-3) where n is an odd number.

When n=3,5 then there are: x 3+1=(x+1)(x 2-x+1)x 5+1=(x+1)(x 4-x 3+x 2-x+1).

First deform the x 15+1 and you can use the above two formulas to find :

x^15+1

x^5)^3+1

x^5+1)(x^10-x^5+1)

x+1)(x 4-x 3+x 2-x+1)(x 10-x 5+1) If you think my solution is of some help to you, go to the top.

Solution: x 2+14x 11 factoring process is as follows.

x^2+14x＋（7）^2-（7）^2+11（x＋7）^2-38

x＋7）^2-（√38）^2

x＋7＋√38）（x+7-√38）

25x squared - 10 + 1 = parentheses 5x squared of the brackets minus 10x + 1 and finally = squared of the brackets of 5x-1. Answer: The Yuan Zheng case is the square of the chachun bracket orange without the ode in parentheses 5x-1.

If we want to factor $x 2-15$, we can use the formula of difference squared: $a 2-b 2=(a+b)(a-b)$. $x 2-15$ is expressed as the form of the difference between two numbers of flat width hoods, i.e., $x 2-3 2$, and then it can be factored using the formula of the douwu difference and the square of the square as follows:

x^2-15=(x^2-3^2)= x+3)(x-3) $

Thus, $x 2-15$ can be broken down into $(x+3)(x-3)$.

Summary. Hello, the factorization of this question is the original question = (-5x*7) + (3x*1) pro 3 <>

15x 32x 7 factorization.

Hello, the factorization of this question is the original question = (-5x*7) + (3x*1) pro 3 <> are you factoring this?How to have a number.

Sorry, mistyped <>

Dear, it should be (7-5x) (3x+1)<> I hope mine can help you <>

You asked: Factoring the macro solution x one 5 x ten 4?

x I. Sila 5 x X. 4

x - x - 4x ten 4

x（x²-1）-4（x-1）

x（x-1）（x+1）-4（x-1）

x-1）（x²+x-4）

x-1)[x-(1+ Judgment Book 17) 2)][x-(1-17) 2)]

This auspicious opening question is due to the decomposition of the friends as follows:

x^3-5x+4=0

x^3-1-5x+5=0

x^3-1-5(x-1)=0

x-1)(x^2+x+1)-5(x-1)=0x-1)(x^2+x-4)=0

So x-1=0, i.e. x=1 or x=-1 17 2

x a 5x 10-year-old first 4

x³-x)-(4x-4)

x(x+1)(x-1)-4(x-1)

x elimination+x)(x-1)-4(x-1)(x-1)(x-4).

Original = x 3-x-4 (x-1).

x(x-1)(x+1)-4(x-1)

x-1)(x^2+x-4)

x-1)[x+(1+ 17) 2][x+(1- 17) 2].