Question

f(x)=(x^2-x-2)(x^2-x-4)-8..............va rogggg dau COROANA!

Answer 1

[( x² - x)  - 2 ] · [( x² -x) - 4 ] - 8 = 
= ( x² -x)² -2·( x² -x) -4·( x² -x)  + ( -2) · ( - 4) - 8 
= ( x² -x)² - 6· ( x² -x)  + 8 - 8 
= ( x² - x)² -6·( x² -x) 
= ( x² -x) ·( x² -x - 6 ) 
= x ·( x -1) · ( x -3)·( x + 2)

Answer 2

F(x)=(x²-x-2)(x²-x-4)-8
F(x)=x(x³-2x²-5x-6)
F(x)=(x-3)(x-1)*x*(x+2)
(x²-x-2)(x²-x-4)=(x² *x²)-(x² *x)-4x²-(x² *x)+x²+4x-2x²+2x+8)=
x⁴-x³-4x²-x³+x²+6x-2x²+8=x⁴-2x³-5x²+6x+8
F(x)=(x²-x-2)(x²-x-4)-8 inlocuim
x⁴-2x³-5x²+6x+8-8=x⁴-2x³-5x²+6x
F(x)≥-9
x=-2
x=0
x=1
x=3

x⁴-2x³-5x²+6x=0
x(x-3)(x-1)(x+2)=0
(x-3)=0=>x=3
(x-1)=0=>x=1
(x+2)=0=>x=-2
x=0
Succes!!