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:4 in each package
Step-by-step explanation:
Answer: 29
explanation: you add 1 to get 2, you then add 2 to get 4, then 3 to get to 7, then 4 to get to 11, then 5 to get to 16, then 6 to get to 22, then 7 to get to 29
This is a right angle triangle.
So, by Pythagoras theorem,
√(32^2+20^2) = x
or √(1024+400) = x
or √1424 = x
or <em>37.74 = x</em>
Answer:
1. 2n
Step-by-step explanation:
