# How do you find a polynomial of degree 2 that has a zero of -2?

##### 1 Answer
Nov 15, 2016

$P \left(x\right) = {x}^{2} + \left(k - 2\right) x - 2 k$ where $k$ is a real number.

#### Explanation:

A polynomial $P \left(x\right)$ has some number $\alpha$ as a zero if and only if $x - \alpha$ is a factor of $P \left(x\right)$. To generate a polynomial with desired zeros, then, we can multiply any such factors.

As our desired polynomial has $- 2$ as a zero, it must have a factor of $x - \left(- 2\right) = x + 2$. As no other specific zero is given, we can make that choice ourselves. Suppose the other zero (possibly also being $- 2$), is $k$. Then the polynomial would be

$P \left(x\right) = \left(x + 2\right) \left(x - k\right)$

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

Choosing any value for $k$ will give a degree $2$ polynomial with $- 2$ as a zero. For example, $k = 0$ gives ${x}^{2} + 2 x$, or $k = 2$ gives ${x}^{2} - 4$. Multiplying by any nonzero constant also will result in a valid polynomial.