(x-2)(x-2)(x+1)+3(x-2)(x+1)=(x-2)P

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Solution for (x-2)(x-2)(x+1)+3(x-2)(x+1)=(x-2)P equation: 



(x-2)(x-2)(x+1)+3(x-2)(x+1)=(x-2)

We move all terms to the left:

(x-2)(x-2)(x+1)+3(x-2)(x+1)-((x-2))=0

We multiply parentheses ..

(+x^2-2x-2x+4)(x+1)+3(x-2)(x+1)-((x-2))=0



We calculate terms in parentheses: -((x-2)), so:

(x-2)

We get rid of parentheses

x-2

Back to the equation:

-(x-2)



We get rid of parentheses

(+x^2-2x-2x+4)(x+1)+3(x-2)(x+1)-x+2=0

We multiply parentheses ..

(+x^2-2x-2x+4)(x+1)+3(+x^2+x-2x-2)-x+2=0

We add all the numbers together, and all the variables

(+x^2-2x-2x+4)(x+1)+3(+x^2+x-2x-2)-1x+2=0

We multiply parentheses

(+x^2-2x-2x+4)(x+1)+3x^2+3x-6x-1x-6+2=0

We add all the numbers together, and all the variables

3x^2+(+x^2-2x-2x+4)(x+1)-4x-4=0

We move all terms containing x to the left, all other terms to the right

3x^2+(+x^2-2x-2x+4)(x+1)-4x=4

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