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}
If you ever times a number by 0 then the answer will be 0 no matter what
Which parabola?

Equation of the parabola
y - y1 = 4p(x - x1)
The Vertex of the parabola given is (0, 0) because it does not have the values of x1 and y1.
Then Vertex = (0,0)
-Look for two values of y to the left and two points to the right.
You can choose the points that you which
x y
-3 y = -(-3)^2 = -9
-1 y = -(-1)^2 = -1
0 y = -(0)^2 = 0
1 y = -(1)^2 = -1
3 y = -(3)^2 = -9
Answer:
a=-5
Step-by-step explanation:
subtract 3 from both sides
15=-3a
divide by -3
a=-5
Answer:
expression: (30/t) - 10
evaluation: 30/2 - 10 = 5
Step-by-step explanation: