Upon a slight rearrangement this problem gets a lot simpler to see.
x^3-x+2x^2-2=0 now factor 1st and 2nd pair of terms...
x(x^2-1)+2(x^2-1)=0
(x+2)(x^2-1)=0 now the second factor is a "difference of square" of the form:
(a^2-b^2) which always factors to (a+b)(a-b), in this case:
(x+2)(x+1)(x-1)=0
So g(x) has three real zero when x={-2, -1, 1}
Answer:
256
Step-by-step explanation:
Answer:
I know you have a problem
Step-by-step explanation:
= (x - 3)(2x² - 5x + 1)
= 2x³ - 5x² + x - 6x² + 15x - 3
= 2x³ + (-5 - 6)x² + (1 + 15)x - 3
= 2x³ - 11x² + 16x - 3
Option → C
