# How do you find the polynomial function with roots -1/2, 1/4, 1, 2?

Nov 23, 2015

$8 {x}^{4} - 22 {x}^{3} + 9 {x}^{2} + 7 x - 2$

#### Explanation:

Consider the equation $\left(x - 2\right) \left(2 x + 3\right) = 0$.

To figure out what $x$ could be, we set each term being multiplied equal to $0$.

$\left\{\begin{matrix}x - 2 = 0 \rightarrow x = 2 \\ 2 x + 3 = 0 \rightarrow x = - \frac{3}{2}\end{matrix}\right.$

If we know one of the roots of the polynomial is $- \frac{1}{2}$, we know $x = - \frac{1}{2}$. Therefore, $x + \frac{1}{2} = 0$, but having it's easier if we don't have fractions. We can multiply everything by $2$ to see that $2 x + 1 = 0$. From this, we know that one of the multiplied terms in our polynomial will be $\left(2 x + 1\right)$.

We can use the same logic for each of the roots:

$\left\{\begin{matrix}- \frac{1}{2} \rightarrow \left(2 x + 1\right) \\ \frac{1}{4} \rightarrow \left(4 x - 1\right) \\ 1 \rightarrow \left(x - 1\right) \\ 2 \rightarrow \left(x - 2\right)\end{matrix}\right.$

Then, we multiply them:

$f \left(x\right) = \left(2 x + 1\right) \left(4 x - 1\right) \left(x - 1\right) \left(x - 2\right)$

If we foil, we find that $f \left(x\right) = 8 {x}^{4} - 22 {x}^{3} + 9 {x}^{2} + 7 x - 2$.