Find all real and imaginary solutions to the equation x^3 + 2x^2 + 9x + 18 = 0.

Question image

Understand the Problem

The question is asking us to find both real and imaginary solutions to the given polynomial equation. The approach involves identifying the roots of the cubic equation by either factoring or using the quadratic formula for any resulting quadratic components after finding one root.

Answer

The solutions are $x = -2$, $x = 3i$, $x = -3i$.
Answer for screen readers

The real and imaginary solutions are:

  • $x = -2$
  • $x = 3i$
  • $x = -3i$

Steps to Solve

  1. Identify possible rational roots

Using the Rational Root Theorem, we list the factors of the constant term (18) and the leading coefficient (1). The possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18$.

  1. Test potential roots

We will test these values by substituting them into the polynomial $f(x) = x^3 + 2x^2 + 9x + 18$.

  • Testing $x = -3$:

$$ f(-3) = (-3)^3 + 2(-3)^2 + 9(-3) + 18 = -27 + 18 - 27 + 18 = -18 \neq 0 $$

  • Testing $x = -2$:

$$ f(-2) = (-2)^3 + 2(-2)^2 + 9(-2) + 18 = -8 + 8 - 18 + 18 = 0 $$

Thus, $x = -2$ is a root.

  1. Factor polynomial using the found root

We can perform polynomial long division of $f(x)$ by $(x + 2)$:

$$ f(x) = (x + 2)(x^2 + 9) $$

  1. Solve the quadratic factor

Now we need to solve $x^2 + 9 = 0$ to find the imaginary roots.

$$ x^2 = -9 $$

Taking the square root gives:

$$ x = \pm 3i $$

  1. Summarize all solutions

The solutions to the original equation are $x = -2$, $x = 3i$, and $x = -3i$.

The real and imaginary solutions are:

  • $x = -2$
  • $x = 3i$
  • $x = -3i$

More Information

The equation $x^3 + 2x^2 + 9x + 18 = 0$ is a cubic polynomial. The factored form $(x + 2)(x^2 + 9)$ highlights that one real root exists and that there are coinciding pairs of imaginary roots.

Tips

  • Overlooking imaginary roots: Often, students stop searching after finding one real root. It is essential to check the remaining quadratic factor for additional solutions.
  • Mistaking the sign when taking square roots: The solutions from $x^2 + 9 = 0$ yield both a positive and a negative imaginary component.

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