Question

-2х^3+5х^2+7х; -2х^3+2х^2-1/4х -6х^3+7х^2-2х​

Answer 1

$\displaystyle 1)-2x^3+5x^2+7x=-x(2x^2-5x-7)=-x(2x^2+2x-7x-7)=\\\\ -x(2x(x+1)-7(x+1))=-x(x+1)(2x-7)\\\\ 2)-2x^3+2x^2-\frac{1}{4}x-6x^3+7x^2-2x=-8x^3+9x^2-\frac{1}{4}x-2x=\\\\ -x(8x^2-9x+\frac{1}{4}+2)=-x(8x^2-9x+\frac{9}{4})=-x*\frac{1}{4}(32x^2-36x+9)=\\ \\ -x*\frac{1}{4}*(32x^2-12x-24x+9)=-x*\frac{1}{4}(4x(8x-3)-3(8x-3))=\\ \\ -x*\frac{1}{4}(8x-3)(4x-3)=-\frac{1}{4}x(8x-3)(4x-3)=-0,25x(8x-3)(4x-3)$