# How do you write a polynomial with zeros: -3,1,5,6?

Feb 4, 2016

For a zero, when $x$ has that value the overall bracket (parenthesis) needs to sum to zero. For example $\left(x + 3\right) = \left(- 3 + 3\right) = 0$. That means the solution is $\left(x + 3\right) \left(x - 1\right) \left(x - 5\right) \left(x - 6\right)$.

#### Explanation:

A 'zero' or 'root' of an equation is a point at which it crosses the $x$-axis: that is, the line $y = 0$ (or $f \left(x\right) = 0$).

For that to happen, one of the factors must go to zero. If we take the example of $x = - 3$, we need a factor that, when $x$ has that value, will equal zero. If it's in the form $\left(x + a\right)$ and we put in the value $x = - 3$, the value of $a$ required is $3$.

For a positive value of $x$ such as $x = 5$, the factor needs to be $\left(x - 5\right)$ so that it becomes zero when $x = 5$.

Using this approach, over all we get:

$\left(x + 3\right) \left(x - 1\right) \left(x - 5\right) \left(x - 6\right)$

We could multiply this through to get an expression in ${x}^{4}$ and all lower values, but it's probably best left in this factorized form.