# How do you factor x²-11x+30?

Mar 18, 2018

$\left(x - 5\right) \left(x - 6\right)$
Refer to explanation.

#### Explanation:

This is a quadratic equation in the form of $a {x}^{2} + b x + c$, so you simply factor b into two parts such that their sum is b and product is ac.

${x}^{2} - 11 x + 30$
${x}^{2} - 5 x - 6 x + 30$
$\left(- 5 \times - 6 = 30 \mathmr{and} - 5 - 6 = - 11\right)$
$x \left(x - 5\right) - 6 \left(x - 5\right)$
$\left(x - 5\right) \left(x - 6\right)$

That's it.
Hope this helps :)

Mar 18, 2018

${x}^{2} - 11 x + 30 = \textcolor{b l u e}{\left(x - 5\right) \left(x - 6\right)}$

#### Explanation:

If we attempt to factor ${x}^{2} - 11 x + 30$
into the product of two binomials: $\left(x + a\right) \left(x + b\right)$
since $\left(x + a\right) \left(x + b\right) = {x}^{2} + \left(a + b\right) x + a b$
we see that
$a$ abd $b$ must have the same sign (since the constant term is $\textcolor{red}{+} 30$)
$a b$ must be equal to $30$
$a + b$ must add up to $- 11$ (which implies they both must be negative).

{: ("negative factor pairs of "30":"," | ",-1xx-30," | ",-3xx-10," | ",color(blue)(-5xx-6)), ("sum of factor pairs:"" | ",,-31," | ",-13," | ",color(blue)(-11)) :}

Mar 18, 2018

$x = 5$ $, 6$

#### Explanation:

On comparing this equation with the equation $a {x}^{2} + b x + c = 0$
$a = 1 , b = - 11 , c = 30$
Now you have to split the middle term i.e. -11 in two parts such that the multiplication of both the parts is equal to $a \cdot c$ while the addition of both the parts is equal to $b$ itself.
We can divide $- 11$ as $- 6$ and $- 5$
So, the equation becomes
$\left({x}^{2} + \left(- 6 - 5\right) x + 30\right) = 0$
$\left({x}^{2} - 6 x - 5 x + 30\right) = 0$
Now taking out common terms
$\left(x \left[x - 6\right] - 5 \left[x - 6\right]\right) = 0$
$\left(\left[x - 5\right] \left[x - 6\right]\right) = 0$
$x = 5 x = 6$