# How do you solve 8x^5 + 10x^4 = 4x^3 + 5x^2?

Mar 10, 2018

$x = 0 \mathmr{and} x = - \frac{5}{4} \mathmr{and} x = \pm \sqrt{\frac{1}{2}}$

#### Explanation:

Move all the terms to the left.

$8 {x}^{5} + 10 {x}^{4} - 4 {x}^{3} - 5 {x}^{2} = 0$

Group in pairs and factor each pair.

$\left(8 {x}^{5} + 10 {x}^{4}\right) + \left(- 4 {x}^{3} - 5 {x}^{2}\right) = 0$

$2 {x}^{4} \left(4 x + 5\right) - {x}^{2} \left(4 x + 5\right) = 0$

Factor out the common factor and common bracket:

${x}^{2} \left(4 x + 5\right) \left(2 {x}^{2} - 1\right) = 0$

There are now three factors. Set each equal to zero and solve.

${x}^{2} = 0 \text{ } \rightarrow x = 0$

$4 x + 5 = o \text{ } \rightarrow x = - \frac{5}{4}$

$2 {x}^{2} - 1 = 0 \text{ "rarr x^2 = 1/2" } \rightarrow x = \pm \sqrt{\frac{1}{2}}$

Mar 10, 2018

$x = 0 , \pm \frac{1}{\sqrt{2}} , - \frac{5}{4}$

#### Explanation:

We can easily rewrite $8 {x}^{5} + 10 {x}^{4} = 4 {x}^{3} + 5 {x}^{2}$ in the form

$2 {x}^{4} \left(4 x + 5\right) = {x}^{2} \left(4 x + 5\right)$

Subtracting ${x}^{2} \left(4 x + 5\right)$ from both sides, leads to

$2 {x}^{4} \left(4 x + 5\right) - {x}^{2} \left(4 x + 5\right) = 0$

which means that
'
$\left(2 {x}^{4} - {x}^{2}\right) \left(4 x + 5\right) = 0$,

i.e.

${x}^{2} \left(2 {x}^{2} - 1\right) \left(4 x + 5\right) = 0$

Now, for a polynomial to vanish, one of its factors have to be zero. So, either ${x}^{2} = 0$, or $2 {x}^{2} - 1 = 0$ or $4 x + 5 = 0$.

If ${x}^{2} = 0$, we must have $x = 0$

or if $2 {x}^{2} - 1 = 0$, we get ${x}^{2} = \frac{1}{2}$, i.e. $x = \pm \frac{1}{\sqrt{2}}$

or, finally, $4 x + 5 = 0$ which implies that $x = - \frac{5}{4}$