# Question #cfaee

Sep 11, 2017

The possible values for $x$ are 0, 8 or -3

#### Explanation:

What we have to do is, somehow, come up with an equation that resembles $a {x}^{2} + b x + c = 0$, in our case we have:

${x}^{3} - 24 x = 5 {x}^{2}$

first lets make it so one side equals zero:

${x}^{3} - 5 {x}^{2} - 24 x = 0$

Now, we can factor out one $x$ out of each term, leaving us with:

$x \cdot \left({x}^{2} - 5 x - 24\right) = 0$

So, for any multiplication to be equal to zero, one of the terms is equal to zero. That leaves us with the two following conditions:

Either $x = 0 \text{ }$ or $\text{ } {x}^{2} - 5 x - 24 = 0$

Now we take our $a , b$ and $c$ (1, -5 and -24) from the second condition and plug them into the quadratic formula to find the other values of x:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{5 \pm \sqrt{{\left(- 5\right)}^{2} - 4 \left(1\right) \left(- 24\right)}}{2 \left(1\right)}$

$x = \frac{5 \pm \sqrt{25 - \left(- 96\right)}}{2}$

$x = \frac{5 \pm \sqrt{121}}{2}$

$x = \frac{5 \pm 11}{2}$

$x = \frac{16}{2} = 8 \text{ }$ or $\text{ } x = - \frac{6}{2} = - 3$

Sep 11, 2017

$x = 8 \mathmr{and} x = - 3 \mathmr{and} x = 0$

#### Explanation:

Note : This is not a quadratic equation, a quadratic equation is of degree 2. Since this equation is of degree 3, it is correct to say that it is a cubic equation.

To factor, or factorise means to write an expression as the product of its prime factors.

Factoring $42 : \text{ } 42 = 2 \times 3 \times 7$

There are several ways to factorise:

• divide out a common factor
• divide out a common bracket
• grouping terms
• difference of squares

We have ${x}^{3} - 24 x = 5 {x}^{2}$

Move all the terms to one side and make the other side $0$

${x}^{3} - 5 {x}^{2} - 24 x = 0 \text{ } \leftarrow$ divide out $x$ from each term

$x \left({x}^{2} - 5 x - 24\right) = 0$

Find factors of $24$ which differ by $5$

They are $8 \mathmr{and} 3$

The signs will be different, the bigger is negative.
$- 8 \times + 3 = 24 \text{ } \mathmr{and} - 8 + 3 = - 5$

$\therefore x \left(x - 8\right) \left(x + 3\right) = 0$

There are $3$ factors, so there will be $3$ solutions.

Set each factor equal to zero:

$x = 0$
$x - 8 = 0 \Rightarrow x = 8$
$x + 3 = 0 \Rightarrow x = - 3$

These are the 3 possible solutions.