Answer:
a + 1
to get 1 more than a you add 1 to a because you don't know what a is
Answer:
Believe it or not, we can determine whether a function is linear or nonlinear simply by looking at its graph! Because the rate at which y is changing with respect to x is constant in a linear function, the graph of a linear function is a line, as the name implies. For example, observe the graph of Sophie's function.
Step-by-step explanation:
For this case, to find the roots of the function, we equate to zero.
![x ^ 3 + 2x ^ 2-x-2 = 0](https://tex.z-dn.net/?f=x%20%5E%203%20%2B%202x%20%5E%202-x-2%20%3D%200)
We rewrite how:
![x ^ 3 + 2x ^ 2- (x + 2) = 0](https://tex.z-dn.net/?f=x%20%5E%203%20%2B%202x%20%5E%202-%20%28x%20%2B%202%29%20%3D%200)
We factor the maximum common denominator of each group:
![x ^ 2 (x + 2) -1 (x + 2) = 0](https://tex.z-dn.net/?f=x%20%5E%202%20%28x%20%2B%202%29%20-1%20%28x%20%2B%202%29%20%3D%200)
We factor the polynomial, factoring the maximum common denominator![(x + 2):](https://tex.z-dn.net/?f=%28x%20%2B%202%29%3A)
![(x + 2) (x ^ 2-1) = 0](https://tex.z-dn.net/?f=%28x%20%2B%202%29%20%28x%20%5E%202-1%29%20%3D%200)
By definition of perfect squares we have to:
![a ^ 2-b ^ 2 = (a + b) (a-b)](https://tex.z-dn.net/?f=a%20%5E%202-b%20%5E%202%20%3D%20%28a%20%2B%20b%29%20%28a-b%29)
ON the expression![(x ^ 2-1):](https://tex.z-dn.net/?f=%28x%20%5E%202-1%29%3A)
![a = x\\b = 1](https://tex.z-dn.net/?f=a%20%3D%20x%5C%5Cb%20%3D%201)
So:
![(x ^ 2-1) = (x + 1) (x-1)](https://tex.z-dn.net/?f=%28x%20%5E%202-1%29%20%3D%20%28x%20%2B%201%29%20%28x-1%29)
Thus, the factorization of the polynomial is:
![(x + 2) (x + 1) (x-1) = 0](https://tex.z-dn.net/?f=%28x%20%2B%202%29%20%28x%20%2B%201%29%20%28x-1%29%20%3D%200)
![x_ {1} = - 2\\x_ {2} = - 1\\x_ {3} = 1](https://tex.z-dn.net/?f=x_%20%7B1%7D%20%3D%20-%202%5C%5Cx_%20%7B2%7D%20%3D%20-%201%5C%5Cx_%20%7B3%7D%20%3D%201)
ANswer:![x_ {1} = - 2\\x_ {2} = - 1\\x_ {3} = 1](https://tex.z-dn.net/?f=x_%20%7B1%7D%20%3D%20-%202%5C%5Cx_%20%7B2%7D%20%3D%20-%201%5C%5Cx_%20%7B3%7D%20%3D%201)