# How do you solve by factoring 30x^3 = 21x^2 + 135x?

Jul 8, 2018

${x}_{1} = - \frac{5}{9}$, ${x}_{2} = 0$ and ${x}_{3} = \frac{5}{2}$

#### Explanation:

$30 {x}^{3} = 21 {x}^{2} + 135 x$

$30 {x}^{3} - 21 {x}^{2} - 135 x = 0$

$3 x \cdot \left(10 {x}^{2} - 7 x - 45\right) = 0$

$3 x \cdot \left(10 {x}^{2} - 25 x + 18 x - 45\right) = 0$

$3 x \cdot \left[5 x \cdot \left(2 x - 5\right) + 9 \cdot \left(2 x - 5\right)\right] = 0$

$3 x \cdot \left(5 x + 9\right) \cdot \left(2 x - 5\right) = 0$

So, ${x}_{1} = - \frac{5}{9}$, ${x}_{2} = 0$ and ${x}_{3} = \frac{5}{2}$

Jul 8, 2018

See a solution process below:

#### Explanation:

First, put the equation in standard form:

$30 {x}^{3} - \textcolor{red}{21 {x}^{2}} - \textcolor{b l u e}{135 x} = 21 {x}^{2} - \textcolor{red}{21 {x}^{2}} + 135 x - \textcolor{b l u e}{135 x}$

$30 {x}^{3} - 21 {x}^{2} - 135 x = 0 + 0$

$30 {x}^{3} - 21 {x}^{2} - 135 x = 0$

Next, divide each side of the equation by $\textcolor{red}{3}$ to minimize the coefficients:

$\frac{30 {x}^{3} - 21 {x}^{2} - 135 x}{\textcolor{red}{3}} = \frac{0}{\textcolor{red}{3}}$

$\frac{30 {x}^{3}}{\textcolor{red}{3}} - \frac{21 {x}^{2}}{\textcolor{red}{3}} - \frac{135 x}{\textcolor{red}{3}} = 0$

$10 {x}^{3} - 7 {x}^{2} - 45 x = 0$

Then, factor out the common term:

$\left(x \cdot 10 {x}^{2}\right) - \left(x \cdot 7 x\right) - \left(x \cdot 45\right) = 0$

$x \left(10 {x}^{2} - 7 x - 45\right) = 0$

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

Solution 1:

$x = 0$

Solution 2:

We can use the quadratic equation to solve for this term:

$\textcolor{red}{10} {x}^{2} - \textcolor{b l u e}{7} x - \textcolor{g r e e n}{45} = 0$

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{10}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 7}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 45}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{- 7} \pm \sqrt{{\textcolor{b l u e}{- 7}}^{2} - \left(4 \cdot \textcolor{red}{10} \cdot \textcolor{g r e e n}{- 45}\right)}}{2 \cdot \textcolor{red}{10}}$

$x = \frac{7 \pm \sqrt{49 - \left(- 1800\right)}}{20}$

$x = \frac{7 \pm \sqrt{49 + 1800}}{20}$

$x = \frac{7 \pm \sqrt{1849}}{20}$

$x = \frac{7 \pm 43}{20}$

$x = \frac{7 - 43}{20}$; $x = \frac{7 + 43}{20}$

$x = - \frac{36}{20}$; $x = \frac{50}{20}$

$x = - \frac{9}{5}$; $x = \frac{5}{2}$

The Solution Set Is:

$x = \left\{- \frac{9}{5} , 0 , \frac{5}{2}\right\}$

Jul 8, 2018

Solution: $x = 0 , x = 2.5 , x = - 1.8$

#### Explanation:

$30 {x}^{3} = 21 {x}^{2} + 135 x$ or

$30 {x}^{3} - 21 {x}^{2} - 135 x = 0$ or

$3 x \left(10 {x}^{2} - 7 x - 45\right) = 0$ or

$3 x \left(10 {x}^{2} - 25 x + 18 x - 45\right) = 0$ or

$3 x \left\{5 x \left(2 x - 5\right) + 9 \left(2 x - 5\right)\right\} = 0$ or

$x \left(2 x - 5\right) \left(5 x + 9\right) = 0$

$\therefore x = 0 , 2 x - 5 = 0 \therefore x = \frac{5}{2} = 2.5 \mathmr{and} 5 x + 9 = 0$

$\therefore x = - \frac{9}{5} = - 1.8$

Solution: $x = 0 , x = 2.5 , x = - 1.8$ [Ans]