# Solve 2x^3 + x^2 = -4 - 8x?

Oct 29, 2017

$x = - \frac{1}{2}$

graph{2x^3 + x^2 + 8x + 4 [-11.06, 11.44, -4.63, 7.09]}

#### Explanation:

First thing you always want to do when solving polynomial equations is set them equal to zero. So:

$2 {x}^{3} + {x}^{2} = - 4 - 8 x$
$\implies 2 {x}^{3} + {x}^{2} + 8 x + 4 = 0$

Now, we're going to use a method of solving called grouping. We're going to split the left hand side of our equation into two groups of 2 terms each, and then try to factor out some common term out of each group.

$\implies \left(2 {x}^{3} + {x}^{2}\right) + \left(8 x + 4\right) = 0$

I see that I can factor out a $2 x + 1$ out of each of my groups. This would leave:

$\implies \left(2 x + 1\right) \left({x}^{2}\right) + \left(2 x + 1\right) \left(4\right) = 0$

Since I have a $2 x + 1$ in each of my terms, I can factor it out, and clump what's left together:

$\implies \left(2 x + 1\right) \left({x}^{2} + 4\right) = 0$

Now that I have a product of factors, I can invoke my zero product property, and know that for this equation to be true, one of those factors must equal zero.

$\implies 2 x + 1 = 0$
$x = - \frac{1}{2}$

$\implies {x}^{2} + 4 = 0$
$x = \pm \sqrt{- 2}$

...but wait, how can we have a negative number under our square root? The answer is we cannot! That is, we cannot have a negative number inside a square root and expect a real number as an answer. So your only real solution to this equation would be $x = - \frac{1}{2}$. However, if we were to consider imaginary solutions, we'd also include:

$x = \pm i \sqrt{2}$

However, you should only ever include this in your answer if it imaginary solutions are specifically asked for.

A handy way to check your answer right after is to graph it. Let's see how that turns out:

graph{2x^3 + x^2 + 8x + 4 [-11.06, 11.44, -4.63, 7.09]}

You'll see that our graph does in fact intersect the x-axis at $x = - \frac{1}{2}$, meaning that we are correct.

Here's a great video by patrickJMT if you want to learn more about the process of grouping;

Hope that helps :)