Question

Sa se descompuna in factori : a)5x(x+2)+(x+2)²-3(x+2) b)(5x-4)(3x-2)²+2(3x-2)³ +2(x+4)(3x-2)² c)x²-2x+1 d) 9x²-4 e)9x²-64 f)(x-7)²-64 g)(2x+3)²-x² h)(4x-1)²-4x² i)(3x-4)²-9x² j)x²-3x+2x k)x²-6x+8 l)x²-4x+3 m)x²-5x+6 n)x²+2x-35 o)x²+6x-16 p)x²-6x-7 q)x²-8x-20

Answer 1

[tex]a)\:5x(x+2)+(x+2)^2-3(x+2)=(x+2)(5x+x+2-3)=\\ =(x+2)(6x-1)\\ b)\:(5x-4)(3x-2)^2+2(3x-2)^3+2(x+4)(3x-2)^2=\\ =(3x-2)^2(5x-4+2(3x-2)+2(x-4))=\\ =(3x-2)^2(5x-4+6x-4+2x-8)= =(3x-2)^2(13x-16)\\ c)\:x^2-2x+1=(x-1)^2\\ formula:\:a^2-b^2=(a-b)(a+b) \\ d)\:9x^2-4=(3x-2)(3x+2)\\ e)\:9x^2-64=(3x-8)(3x+8)\\ f)\:(x-7)^2-64=(x-7-8)(x-7+8)=(x-15)(x+1)\\ g)\:(2x+3)^2-x^2=(2x+3-x)(2x+3+x)=(x+3)(3x+3)\\ h)\:(4x-1)^2-4x^2=(4x-1-2x)(4x-1+2x)=(2x-1)(6x-1)\\ i)\:(3x-4)^2-9x^2=(3x-4-3x)(3x-4+3x)=-4(6x-4)\\[/tex]
[tex]j)\:x^2-3x+2x=x^2-x=x(x-1)\\ acum\:smecheria\:este\:sa\:gasesti\:doua\:numere\:care\:adunate\:sa\:dea\\ \:coeficientul\:lui\:x,\:iar\:inmultite\:sa\:dea\:termenul\:liber \\ k)\:x^2-6x+8=x^2-2x-4x+8=x(x-2)-4(x-2)=\\ =(x-2)(x-4)\\ l)\:x^2-4x+3=x^2-x-3x+3=x(x-1)-3(x-1)=\\ =(x-1)(x-3)\\ m)x^2-5x+6=x^2-2x-3x+6=x(x-2)-3(x-2)=\\ =(x-2)(x-3)\\ n)\:x^2+2x-35=x^2+7x-5x-35=x(x+7)-5(x+7)=\\ =(x+7)(x-5)\\ o)\:x^2+6x-16=x^2+8x-2x-16=x(x+8)-2(x+8)=\\ =(x+8)(x-2)\\ p)\:x^2-6x-7=x^2+x-7x-7=x(x+1)-7(x+1)=\\ =(x+1)(x-7)\\[/tex]
[tex]q)\: x^2-8x-20=x^2-10x+2x-20=x(x-10)+2(x-10)=\\ =(x-10))(x+2)[/tex]