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:
1) 11
2) 2
3) 
4) 
5) -2
6) 
Step-by-step explanation:
1) 2
+ 3
- 
=(2 × 2 )+ (3 × 4) - 5
= 4 + 12 - 5
= 11
2) 4
-2
+ 
= (4 × 2) - (2 × 7) + 8
= 8 - 14 +8
= 2
3) 5
- 3
= 5× 
- 3×
+ 4×
+ 2×
= 
=
4) 
= 
=
5) 
= 
= -2
6) 
= 
= 2× - 2× - 
= 
=
Hope the working out is clear and will help you. :)
Answer:
bottom left i think
Step-by-step explanation:
Using the slope formula the answer would be 1-(-1)/(-1)-6. The slope would be 2/-7. It would be a negative slope.
