# How do you solve 7x ^ { 3} + 21x ^ { 4} = 0?

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#### Explanation

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#### Explanation:

I want someone to double check my answer

1
Mar 9, 2018

See a solution process below:

#### Explanation:

First, factor out the common term $\left(7 {x}^{3}\right)$ from each term on the left side of the equation:

$\left(7 {x}^{3} \times 1\right) + \left(7 {x}^{3} \times 3 x\right) = 0$

$7 {x}^{3} \left(1 + 3 x\right) = 0$

Now, solve each term on the left for $0$:

Solution 1:

$7 {x}^{3} = 0$

$\frac{7 {x}^{3}}{\textcolor{red}{7}} = \frac{0}{\textcolor{red}{7}}$

${x}^{3} = 0$

$\sqrt[3]{{x}^{3}} = \sqrt[3]{0}$

$x = 0$

Solution 2:

$1 + 3 x = 0$

$1 - \textcolor{red}{1} + 3 x = 0 - \textcolor{red}{1}$

$0 + 3 x = - 1$

$3 x = - 1$

$\frac{3 x}{\textcolor{red}{3}} = - \frac{1}{\textcolor{red}{3}}$

$x = - \frac{1}{3}$

The Solution Set Is:

$x = \left\{- \frac{1}{3} , 0\right\}$

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