# How do you solve (- 2y - 3) ^ { 3} + 25( - 2y - 3) = 0?

Jun 8, 2017

$y = - 1.5$

$y = - \frac{3}{2} + \frac{5}{2} i$ and $y = - \frac{3}{2} - \frac{5}{2} i$ also work.

#### Explanation:

The Shorter and Easier Way
We see that there are 2 $\left(- 2 y - 3\right)$ terms in the equation. This might signal something useful. So, let's temporarily replace $\left(- 2 y - 3\right)$ with $x$. Now, let's rewrite the equation:

${x}^{3} + 25 x = 0$

Using the Distributive Property, we can factor out an $x$ from the left side:

$x \cdot \left({x}^{2} + 25\right) = 0$

Looking at our new equation, we can see that the two terms' products is $0$. This means that either $x$ is $0$ or ${x}^{2} + 25$ is $0$. Let's start with the case of $x = 0$ and solve for $y$. However, before we start, remember that $x$ is actually $\left(- 2 y - 3\right)$:

$x = 0$
$- 2 y - 3 = 0$
$- 2 y = 3$
$y = - \frac{3}{2}$

Besides this single answer of $y = - \frac{3}{2}$, there are two more roots (solutions) which involve complex numbers. To obtain these two other solutions, we have to look at the case where ${x}^{2} + 25 = 0$.

${x}^{2} + 25 = 0$
${x}^{2} = - 25$
$x = \sqrt{-} 25$
$x = \sqrt{-} 1 \cdot \sqrt{25}$
$x = \pm 5 i$

The final statement shows that $x$ can be either positive $5 i$ of negative $5 i$. We'll start by working with positive $5 i$. And, because we let $x = \left(- 2 y - 3\right)$, we can substitute it in and find the second value for $y$:

$- 2 y - 3 = 5 i$
$- 2 y = 5 i + 3$
$y = \frac{5 i + 3}{-} 2$
$y = \frac{5 i}{- 2} + \frac{3}{- 2}$
$y = \left(- \frac{5}{2}\right) i - \frac{3}{2}$
$y = - \frac{3}{2} - \left(\frac{5}{2}\right) i$

The third solution for $y$ involves seeing $x$ as equal to $- 5 i$:

$- 2 y - 3 = - 5 i$
$- 2 y = - 5 i + 3$
$y = \frac{- 5 i + 3}{-} 2$
$y = \frac{- 5 i}{-} 2 + \frac{3}{-} 2$
$y = - \frac{3}{2} + \left(\frac{5}{2}\right) i$

The Long and Hard Way

${\left(- 2 y - 3\right)}^{3} + 25 \left(- 2 y - 3\right) = 0$

Let's expand everything out and use the distributive property to simplify things a bit:

$\left(- 2 y - 3\right) \left(- 2 y - 3\right) \left(- 2 y - 3\right) + \left(25\right) \left(- 2 y\right) - \left(25\right) \left(3\right) = 0$

$\left(- 2 y - 3\right) \left(4 {y}^{2} + 6 y + 6 y + 9\right) + \left(- 50 y\right) - 75 = 0$

$\left(4 {y}^{2}\right) \left(- 2 y - 3\right) + \left(6 y\right) \left(- 2 y - 3\right) + \left(6 y\right) \left(- 2 y - 3\right) + \left(9\right) \left(- 2 y - 3\right) - 50 y - 75 = 0$

$\left(- 8 {y}^{3} - 12 {y}^{2}\right) + \left(- 12 {y}^{2} - 18 y\right) + \left(- 12 {y}^{2} - 18 y\right) + \left(- 18 y - 27\right) - 50 y - 75 = 0$

Now let's break open the parentheses:

$- 8 {y}^{3} - 12 {y}^{2} - 12 {y}^{2} - 18 y - 12 {y}^{2} - 18 y - 18 y - 27 - 50 y - 75 = 0$

And now, let's start rearranging like terms together and simplifying them:

$- 8 {y}^{3} - 12 {y}^{2} - 12 {y}^{2} - 12 {y}^{2} - 18 y - 18 y - 18 y - 50 y - 27 - 75 = 0$

$- 8 {y}^{3} - 36 {y}^{2} - 104 y - 102 = 0$

At this point, we can use a cubic calculator to calculate the value of $y$:

$y = - 1.5$

It's important that two other solutions also work perfectly fine, not just $y = - 1.5$. The two other solutions are: $y = - \frac{3}{2} + \frac{5}{2} i$ and $y = - \frac{3}{2} - \frac{5}{2} i$.
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Checking:

If we were to check whether or not $y$ is indeed $- 1.5$, we would substitute $y = - 1.5$ back into the original equation:

${\left[- 2 \left(- 1.5\right) - 3\right]}^{3} + 25 \left[- 2 \left(- 1.5\right) - 3\right] = 0$

${\left(3 - 3\right)}^{3} + 25 \left(3 - 3\right) = 0$

${0}^{3} + 25 \left(0\right) = 0$

$0 + 0 = 0$

$0 = 0$