# How do you find all the zeros of f(x)= x^2-7x+10?

Feb 25, 2016

The zeroes are $2$ and $5$

#### Explanation:

If we start with ${x}^{2} - 7 x + 10$, we can factor this by considering what multiplys to $10$ ($1 \cdot 10$) and adds to $- 7$. I like to do this via table, like this:

$\textcolor{w h i t e}{.}$x$10$$\textcolor{w h i t e}{\ldots \ldots .}$ add to:
..................................................
$1 \cdot 10 \textcolor{w h i t e}{00. \ldots} = 10$

$- 1 \cdot - 10 \textcolor{w h i t e}{} = - 10$

$1 \cdot - 10 \textcolor{w h i t e}{00} = - 9$

$10 \cdot - 1 \textcolor{w h i t e}{\left(.\right)} = 9$

$2 \cdot 5 \textcolor{w h i t e}{\ldots \ldots .0 .} = 7$

$- 2 \cdot - 5 \textcolor{w h i t e}{c t} = - 7$

$2 \cdot - 5 \textcolor{w h i t e}{\ldots \ldots} = 3$

$5 \cdot - 2 \textcolor{w h i t e}{\ldots \ldots} = 3$

This might look like a mess (and I admit, it kinda is), but we now we know that the only factors of ${x}^{2} - 7 x + 10$ that multiply to 10 and add to $- 7$ are: $- 2$ and $- 5$. That means that we can factor ${x}^{2} - 7 x + 10$ to $\left(x - 2\right) \left(x - 5\right)$. Now, to find the zeroes, we just set each parentheses equal to zero and solve for $x$, like this:

$x - 2 = 0$$\textcolor{w h i t e}{\ldots \ldots \ldots .}$$x - 5 = 0$
$x = 2$$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots .}$$x = 5$

Now we've got it. $x = 2$ and $5$. Nice job!