# How do you find all the zeros of x^2 + 24 = 11x with its multiplicities?

Aug 16, 2016

This equation has solutions $x = 8$ and $x = 3$ (both with multiplicity $1$).

#### Explanation:

Subtract $11 x$ from both sides to get:

${x}^{2} - 11 x + 24 = 0$

Note that $11 = 8 + 3$ and $24 = 8 \cdot 3$.

So we find:

$0 = {x}^{2} - 11 x + 24 = \left(x - 8\right) \left(x - 3\right)$

Hence zeros $x = 8$ and $x = 3$ (both with multiplicity $1$).