# How do you write a polynomial in standard form given the zeros x=-1/2,0,4?

May 28, 2016

You multiply $x$ minus the solution and you obtain ${x}^{3} - \frac{7}{2} {x}^{2} - 2 x = 0$.

#### Explanation:

The solutions of a polynomial are those numbers that assigned to the $x$ gives zero. So the easiest way to construct the polynomial from the solutions is to multiply together $x$ minus the solution because you are sure that it becomes zero.
In our case we have

$x - \left(- \frac{1}{2}\right)$
$x - 0$
$x - 4$

It is clear that the three of them become zero when x is, respectively $- \frac{1}{2}$, $0$ and $4$. If we multiply the three we have

$\left(x + \frac{1}{2}\right) x \left(x - 4\right)$

we have a polynomial that will become zero for $- \frac{1}{2}$, $0$ and $4$ because these three numbers set as zero one of the three factors and the product for zero is zero, no matter what are the other factors.

Now it is just a multiplication to do to obtain the final result

$\left(x + \frac{1}{2}\right) x \left(x - 4\right)$
$= \left(x + \frac{1}{2}\right) \left({x}^{2} - 4 x\right)$
$= {x}^{3} - 4 {x}^{2} + \frac{1}{2} {x}^{2} - 2 x$
$= {x}^{3} - \frac{7}{2} {x}^{2} - 2 x$

and the corresponding equation is

${x}^{3} - \frac{7}{2} {x}^{2} - 2 x = 0$.