SOLUTION: How would you go about finding the zeroes for {{{f(x)=x^3+2x^2+x-2}}}? I've tried to solve for the zeroes using synthetic division with positive and negative 2 and 1 but all of my

Question 972180: How would you go about finding the zeroes for ? I've tried to solve for the zeroes using synthetic division with positive and negative 2 and 1 but all of my answers have had remainders.
I've tried using the derivative to solve for the zeroes using negative 1 and 1/3, but I know that's not correct.
Found 2 solutions by Boreal, solver91311:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
x^3+2x^2+x-2. I agree with synthetic division. You've done everything right. When this happens, I would graph it and see what the function looks like.
I graphed it and got a root about 0.7, and on the calculator 0.695. That is very close to 2. I find one root. When I put the 0.695 back in, it is very close to a zero, within rounding error.
The derivative is 3x^2 +4x +1
That factors into (3x + 1) (x+1), so there are critical points at x=-1 and -(1/3), exactly as you describe, but they are a maximum (2nd derivative is negative for the first) and a minimum (second derivative positive for second), not zeros. It is a cubic shape, as expected.

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

According to the rational roots theorem, the possible rational zeros of your function are and . If a polynomial function with integer coefficients has a rational root, then that root will be of the form where divides , the constant term, and divides the lead coefficient.

Synthetic division with each of , , , and yields a remainder. Hence, this function has no rational zeros.

The graph of the function:

reveals that there is only one real root and therefore 2 complex roots. Since there are no rational roots, the real root must be irrational.

By the way, the derivative does you no good because the local extrema are nowhere near (relatively speaking) the -axis, and you have nothing that tells you the value of the derivative at the intercept. You can use the first derivative in Newton-Raphson to approximate the real zero as precisely as you like. Wikipedia Newton's Method

In order to find the exact value of the roots you need to use the formula for the solution of the general cubic. If you are interested, look here: Wikipedia Cubic Function. You can root through this mess to find the solution yourself if you are that big a glutton for punishment. If your professor/teacher/instructor insists that you do this, I would drop the course and go learn from someone who is not a direct descendant of the Marquis de Sade.

John

My calculator said it, I believe it, that settles it

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