How do you find all zeros with multiplicities of f(x)=2x^5+3x^4-18x-27?

Nov 21, 2017

(1) The Set of zeroes$= \left\{- \frac{3}{2} , \pm \sqrt{3}\right\} \subset \mathbb{R} .$

(2) The Set of zeroes$= \left\{- \frac{3}{2} , \pm \sqrt{3} , \pm i \sqrt{3}\right\} \subset \mathbb{C} .$

Explanation:

Let $Z \left(f\right)$ be the set of zeroes of $f .$

We have,

$f \left(x\right) = \underline{2 {x}^{5} + 3 {x}^{4}} - \underline{18 x - 27} ,$

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

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

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

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

$\therefore Z \left(f\right) = \left\{- \frac{3}{2} , \pm \sqrt{3}\right\} \subset \mathbb{R} .$

But, in $\mathbb{C} , \because \left({x}^{2} + 3\right) = \left(x + i \sqrt{3}\right) \left(x - i \sqrt{3}\right) ,$

$\therefore Z \left(f\right) = \left\{- \frac{3}{2} , \pm \sqrt{3} , \pm i \sqrt{3}\right\} \subset \mathbb{C} .$