# How do you find all the zeros of f(x) = x^2 - 12x + 20 with its multiplicities?

Mar 27, 2016

$f \left(x\right) = 0$ when $x = 10$ or $x = 2$ each with multiplicity $1$

#### Explanation:

Here are a couple of methods:

Completing the square

$f \left(x\right) = {x}^{2} - 12 x + 20$

$= {\left(x - 6\right)}^{2} - 36 + 20$

$= {\left(x - 6\right)}^{2} - 16$

$= {\left(x - 6\right)}^{2} - {4}^{2}$

$= \left(\left(x - 6\right) - 4\right) \left(\left(x - 6\right) + 4\right)$

$= \left(x - 10\right) \left(x - 2\right)$

So $f \left(x\right) = 0$ when $x = 10$ or $x = 2$ with multiplicity $1$

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Product and sum

Note that $10 \cdot 2 = 20$ and $10 + 2 = 12$

So:

$f \left(x\right) = {x}^{2} - 12 + 20 = {x}^{2} - \left(10 + 2\right) x + \left(10 \cdot 2\right) = \left(x - 10\right) \left(x - 2\right)$

So $f \left(x\right) = 0$ when $x = 10$ or $x = 2$ with multiplicity $1$