Question

Find the root of (M^3)+(M^2)-2=0. Show steps.

Answer 1

Answer:

x = 1, -1 + i, and -1 - i are the 3 roots

Step-by-step explanation:

The Rational Root Theorem will help us find any real roots in this 3rd degree polynomial.  The c value in that polynomia is -2 and the d value is 1 (the leading coefficient).  

If c is -2, then the factors of -2 are:

1, -1, 2, -2

If d is 1, then the factors of 1 are:

1, -1

And c/d gives us all the possible roots:

1/1, -1/1, 2/1, -2/1 which of course simplify to

1, -1 2, -2

We will now use synthetic division to find the first real root.  I will use x = 1 first.

For synthetic division:

1|   1     1     0     -2

is how it is set up.  Bring down the first number, 1, multiply it by 1 and put that product up under the second "1":

1|    1     1     0     -2

           1

     1

Now add to get

1|     1     1     0     -2

            1

      1     2

Multiply 2 by 1 and put that product up under the 0 and add:

1|     1     1     0     -2

             1     2

      1     2    2

Again multiply the 2 by 1 and put that product up undeer the -2 and add:

1|     1     1     0     -2

            1      2      2

      1     2     2      0

That remainder of 0 tells us that x = 1 is a root of that polynomial and the depressed polynomial, made up from the bold numbers, is one degree less than what it started with:

$x^2+2x+2=0$

This does not factor without imaginary roots, so putting it into the quadratic formula looks like this:

$x=\frac{-2+/-\sqrt{4-4(1)(2)} }{2}$

which simplifies to

$x=\frac{-2+/-\sqrt{-4} }{2}$

The square root of -4 simplifies to 2i, so

$x=\frac{-2+/-2i}{2}$

which reduces to

x = -1 + i and

x = -1 - i