Question

Descompuneti in factori
(x²-x)(x²-x-22)+40=?

Answer 1

   
[tex]\displaystyle\\ (x^2-x)(x^2-x-22)+40=\\ =x^4-x^3-x^3+x^2-22x^2+22x+40=\\ =x^4-2x^3-21x^2+22x+40=\\ =x^4-2x^3-21x^2+22x+40=~~~~~~~\Big|~22x = 42x - 20x\\ =x^4-2x^3-21x^2+42x -20x+40=\\ =x^3(x-2)-21x(x-2) -20(x-2)=\\ =(x-2)( x^3-21x -20)=\\ =(x-2)( x^3+0x^2-21x -20)=~~~~~~~\Big|~0x^2 = -5x^2+5x^2\\ =(x-2)( x^3-5x^2+5x^2-21x -20)=~~~~~~~\Big|~-21x=-25x+4x\\ =(x-2)( x^3-5x^2+5x^2-25x+4x -20)=\\ =(x-2)( x^2(x-5)+5x(x-5)+4(x -5))=\\ =(x-2)(x-5)( x^2+5x+4)=\\ =(x-2)(x-5)( x^2+5x+4)=~~~~~~~\Big|~5x=x+4x [/tex]

[tex]\displaystyle\\ =(x-2)(x-5)( x^2+x+4x+4)=\\ =(x-2)(x-5)( x(x+1)+4(x+1))=\\ =\boxed{\bf (x-2)(x-5)(x+1)( x+4)} [/tex]