Question

What are the real zeros of the function g(x) = x^3 + 2x^2 − x − 2?

Answer 1

If there are real roots to be found for this polynomial, the Rational Root Theorem and synthetic division are the best way to find them. I teach from a book that uses c and d for the possible roots of the polynomial.  C is our constant, 2, and d is the leading coefficient, 1.  The factors of 2 are +/- 1 and +/-2.  The factors for 1 are +/-1 only.  Meaning, in all, there are 4 possibilities as roots for this polynomial.  But there are only 3 total (because our polynomial is a third degree), so we have to find the first one, at least, from our possibilities above.  Let's try x = -1, factor form (x + 1).  If there is no remainder when we do the synthetic division, then -1 is a root.  Put -1 outside the "box" and the coefficients from the polynomial inside: -1  (1  2  -1  -2).  Bring down the first coefficient of 1 and multiply it by the -1 outside to get -1.  Put that -1 up under the 2 and add to get 1.  Multiply 1 times the -1 to get -1 and put that -1 up under the -1 and add to get -2.  -1 times -2 is 2, and -2 + 2 = 0.  So we have our first root of (x+1).  The numbers we get when we do the addition along the way are the coefficients of our new polynomial, the depressed polynomial (NOT a sad one cuz it hates math, but a new polynomial that is one degree less than that of which we started!).  The new polynomial is $x^{2} +x-2=0$.  That can also be factored to find the remaining 2 roots.  Use standard factoring to find that the other 2 solutions are (x+2) and (x-1).  Our solutions then are x = -2, -1, 1, choice B from above.

Answer 2

I chose to use the shortest approach I could think of:  to check each given supposed root to verify that it is (or is not) an actual root.

Dividing 1 into the coefficients    1   2   -1   -2, I found there was no remainder, indicating that 1 is a root.  The coefficients of the quotient were 1   3    2.  Using -1 as divisor, I again got no remainder, indicating -1 is a root.

The coeff. of the quotient this time were 1   2, indicating a factor of (x + 2).  Thus, -2 is the third root.

Roots:  {1, -1, -2}