Hi I know is is probably basic but can anyone assist with the following :

Find all real Roots, expressing as whole # and # under radical sign, not as fraction. and factor completely

f(x)= X^3-2x^2-15x+30

I got roots at 2 (x-2) and +/- √5

If someone solves and gets factored as
(x-2) (x^2-15) can you show me line by line how to factor x^2-15?

Thank you

To find the roots of the equation f(x) = x^3 - 2x^2 - 15x + 30, you have correctly identified one root as x = 2. Now let's find the roots of the quadratic factor x^2 - 15:

1. Set the quadratic factor equal to zero: x^2 - 15 = 0.
2. Add 15 to both sides of the equation: x^2 = 15.
3. To get rid of the exponentiation, take the square root of both sides: √(x^2) = √15.
4. This leads to two possible solutions: x = √15 and x = -√15.

So the complete set of real roots for the equation f(x) = x^3 - 2x^2 - 15x + 30 is {2, √15, -√15}.

Now, let's focus on factoring the quadratic expression x^2 - 15:

We have x^2 - 15 = 0.

To factor the quadratic expression, we look for two binomials that, when multiplied, give us the quadratic expression. The pattern we'll use is the difference of squares:

1. Write the expression as perfect squares:

x^2 - 15 = (x)^2 - (√15)^2.

2. Recognize that this is a difference of squares, so it can be factored as:

(x - √15)(x + √15).

So, the factorization of x^2 - 15 is (x - √15)(x + √15).

Therefore, the complete factorization of f(x) = x^3 - 2x^2 - 15x + 30 is (x - 2)(x - √15)(x + √15).