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}
7/12
12/7
1`2/7
7/12
a reciprocal you just flip the fraction
Turn it into transcript, It's upside down and I can't see it properly.
<span> 3(x+2y)+5x−y+1
Use distributive property
3x+6y+5x-y+1
Add 5x to 3x
8x+6y-y+1
Subtract y from 6y
Final Answer: 8x+5y+1</span>
