# How do you find all the zeros of y=x^2-11x+30 with its multiplicities?

Apr 4, 2016

$x = 5 , 6$

#### Explanation:

Write out a set of brackets like this

$\left(x + a\right) \left(x + b\right)$

You should find two constants, $a$ and $b$, such that

$a + b = - 11$
$a \cdot b = 30$

Which you can do by trial and error, so

$a = - 5$
$b = - 6$

Therefore the fully factorised form of the equation is

$\left(x - 5\right) \left(x - 6\right) = 0$

Solving for the roots or zeros of this:

For the entire quadratic to equal zero, one or both of the pairs of brackets must equal $0$, so solve for each individual one and you will end up with two answers.

$x - 5 = 0 \to x = 5$
$x - 6 = 0 \to x = 6$

Therefore,

$x = 5 , 6$