SOLUTION: determine how many times -1 is a root of x^3+2x^2-x-2=0 then find the other roots

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Question 47148This question is from textbook advanced mathematical concepts precalculus with applications
: determine how many times -1 is a root of x^3+2x^2-x-2=0 then find the other roots
This question is from textbook advanced mathematical concepts precalculus with applications

Found 2 solutions by Earlsdon, pizza:
Answer by Earlsdon(6287) About Me  (Show Source):
You can put this solution on YOUR website!
If x = -1 is a root ofx%5E3%2B2x%5E2-x-2+=+0, then x+1 is a factor of this polynomial.
You can find the other factors and, hence, the other roots, by dividing the given polynomial by the factor (x+1)
%28x%5E3%2B2x%5E2-x-2%29%2F%28x%2B1%29+=+x%5E2%2Bx-2 Now factor the right side.
x%5E2%2Bx-2+=+%28x%2B2%29%28x-1%29
Now we have:x%5E3%2B2x%5E2-x-2+=+%28x%2B1%29%28x%2B2%29%28x-1%29 and...
%28x%2B1%29%28x%2B2%29%28x-1%29+=+0 Applying the zero products principle, we get:
x%2B1+=+0 and x+=+-1
x%2B2+=+0 and x+=+-2
x-1+=+0 and x+=+1

Answer by pizza(14) About Me  (Show Source):
You can put this solution on YOUR website!
One way to do this is to factorise this into linear factors and count the times -1 appear. However, while that is the ultimate aim, the question tempts you to take out the factors one at a time.

First, to show that -1 is a root, you plug -1 into the equation.
So if f%28x%29+=+x%5E3+%2B+2x%5E2+-+x+-+2+
Then f(-1) = -1 + 2 + 1 - 2 = 0

This means that -1 is a root of f(x) at least once. This is an application of the remainder theorem. Anyway, knowing this, we can divide the polynomial f(x) by (x -(-1))=(x+1), knowing that there is no remainder.
f%28x%29+=+x%5E3+%2B+2x%5E2+-+x+-+2+=+%28x%2B1%29%28x%5E2+%2B+x+-+2%29
At this point, one can directly factorise the quadratic, or apply remainder theorem again. If you apply the remainder theorem to the quadratic with x = -1, you get (-1)^2 -1 - 2 = -2. So, no, -1 is not a root again. In the end, to find the other roots, we have to factorise the quadratic ourselves to get
f(x) = (x+1)(x-1)(x+2)
So the roots are 1, -1, -2