To find the real and complex solutions of the equation:

$x^3 - x^2 - x - 15 = 0$

we can use a variety of methods. Let's start by checking possible rational roots using the Rational Root Theorem.

Step 1: Applying the Rational Root Theorem

The Rational Root Theorem suggests that any possible rational root must be a factor of the constant term (-15) divided by a factor of the leading coefficient (1).

The constant term is -15, and the leading coefficient is 1. So, the possible rational roots are the factors of -15: $\pm 1, \pm 3, \pm 5, \pm 15$

Step 2: Testing possible roots

Let's test these possible roots by substituting them into the equation.

Testing $x = -3$:

Substitute $x = -3$ into the equation $x^3 - x^2 - x - 15$:

$(-3)^3 - (-3)^2 - (-3) - 15 = -27 - 9 + 3 - 15 = -48 \neq 0$

So, $x = -3$ is not a root.

Testing $x = 3$:

Substitute $x = 3$ into the equation $x^3 - x^2 - x - 15$:

$3^3 - 3^2 - 3 - 15 = 27 - 9 - 3 - 15 = 0$

So, $x = 3$ is a root.

Step 3: Polynomial division

Now that we know $x = 3$ is a root, we can divide the cubic polynomial $x^3 - x^2 - x - 15$ by $x - 3$ using synthetic or long division.

Performing the division:

$x^3 - x^2 - x - 15 \div (x - 3)$

This gives us:

$x^2 + 2x + 5$

So, we have:

$x^3 - x^2 - x - 15 = (x - 3)(x^2 + 2x + 5)$

Step 4: Solving the quadratic $x^2 + 2x + 5 = 0$

Now, let's solve the quadratic equation $x^2 + 2x + 5 = 0$ using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 2$, and $c = 5$.

Substituting these values into the quadratic formula:

$x = \frac{-2 \pm \sqrt{2^2 - 4(1)(5)}}{2(1)} = \frac{-2 \pm \sqrt{4 - 20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2}$

This simplifies to:

$x = -1 \pm 2i$

So, the complex roots are $x = -1 + 2i$ and $x = -1 - 2i$.

Step 5: Summary of solutions

The real and complex solutions to the equation are:

$x = 3, \quad x = -1 + 2i, \quad x = -1 - 2i$

Answer:

The correct answers are:

• B. 3
• E. -1 - 2i
• F. 1 + 2i

Would you like further explanations or details on any of the steps?

Related Questions:

1. How does the Rational Root Theorem help identify possible rational solutions?
2. What is synthetic division, and how is it used to divide polynomials?
3. Can you explain how complex conjugates are always paired in polynomials with real coefficients?
4. What are the steps to solving a cubic equation with multiple roots?
5. How do you handle equations with imaginary solutions?

Tip:

When dealing with cubic equations, always check for possible rational roots first, then factor the equation and solve any resulting quadratics.