Question

find all the roots of x3+2x2-19x-20

Answer 1

Hey there Manny!

\(\Large\bf \color{#008353}{\text{Welcome to OpenStudy! :)}}\)

Answer 2

Roots of a cubic?
Hmmmmm let's see...

Answer 3

So we'll use the Rational Root Theorem to find possible roots.

Answer 4

To find the roots of the given cubic expression means to find all the roots such that \(x^3 + 2x^2 - 19x - 20 = 0\)

We know that \(x^3 + 2x^2 -19x = 20\) because \(20 - 20 = 0\)

Also if we can express the left side in factored form so
\(x(x^2 + 2x - 19) = 20\)

Now, there are several factors of 20 i.e 1 and 20, 2 and 10, 4 and 5, etc.

What we want to do is find an x value so that we'll have the correct factor of 20.

Notice that \(x^2 + 2x - 19\) is prime which means it could be equal to 1, 2, or 5.

Let's assume it is equal to 5. If we assume this then

\(x^2 + 2x - 19 = 5\)

I chose 5 because if we moved everything to the left side we'd have

\(x^2 + 2x - 24 = 0\) which is factorable

When I factor it I get
\((x + 6)(x - 4) = 0\)

Now that means x = 4 is one of the solutions because 4 is a factor of 20. Obviously -6 doesn't work because it isn't a factor of 20.

If x = 4, then x - 4 is a factor of the cubic polynomial.

Knowing that x - 4 is a factor of the cubic polynomial, factor as follows:

\(x^3+2x^2-19x-20 = 0\)

First split the cubic so that x - 4 is a factor in each pair of terms:
\(x^3 - 4x^2 + 6x^2 - 24x + 5x - 20 = 0\)

Next factor each pair of terms:
\(x^2(x - 4) + 6x(x - 4) + 5(x - 4) = 0\)

Next factor out x - 4:
\((x - 4)(x^2 + 6x + 5) = 0\)

Finally factor \(x^2 + 6x + 5\) to get
\((x - 4)(x + 1)(x + 5) = 0\)

By zero product property:
x - 4 = 0
x + 1 = 0
x + 5 = 0

Solving for x in each equation gives the roots of the given cubic expression.