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

1 Answer

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Answer:

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

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=(-2+/-√(4-4(1)(2)) )/(2)

which simplifies to


x=(-2+/-√(-4) )/(2)

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


x=(-2+/-2i)/(2)

which reduces to

x = -1 + i and

x = -1 - i

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