# How do you find the real solutions of the polynomial 2x^5+3x^4=18x+27?

Feb 15, 2017

The solutions are $S = \left\{- \frac{3}{2} , \sqrt{3} , - \sqrt{3}\right\}$

#### Explanation:

Ler's rewrite the polynomial and factorise

$2 {x}^{5} + 3 {x}^{4} = 18 x + 27$

${x}^{4} \left(2 x + 3\right) = 9 \left(2 x + 3\right)$

${x}^{4} \left(2 x + 3\right) - 9 \left(2 x + 3\right) = 0$

$\left(2 x + 3\right) \left({x}^{4} - 9\right) = 0$

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

$\left(2 x + 3\right) \left(x - \sqrt{3}\right) \left(x + \sqrt{3}\right) \left({x}^{2} + 3\right) = 0$

Therefore,

the solutions are

$2 x + 3 = 0$, $\implies$, $x = - \frac{3}{2}$

$x - \sqrt{3} = 0$, $\implies$, $x = \sqrt{3}$

$x + \sqrt{3} = 0$, $\implies$, $x = - \sqrt{3}$

${x}^{2} + 3 = 0$, no real solutions