# How do you factor the trinomial x^2+14x+24?

Dec 5, 2016

${x}^{2} + 14 x + 24 = \left(x + 2\right) \left(x + 12\right)$

#### Explanation:

The rule to factorise any quadratic is to find two numbers such that

$\text{product" = x^2 " coefficient "xx" constant coefficient}$
$\text{sum" \ \ \ \ \ \ = x " coefficient}$

So for ${x}^{2} + 14 x + 24$ we seek two numbers such that

$\text{product} = 1 \cdot 24 = 24$
$\text{sum} \setminus \setminus \setminus \setminus \setminus \setminus = 14$

So if we looks at the factors of $24$ and compute their sum we get (as all the terms re positive we only need to consider positive factors);

$\left.\begin{matrix}\text{factor1" & "factor2" & "sum} \\ 24 & 1 & 25 \\ 12 & 2 & 14 \\ 6 & 4 & 10 \\ 3 & 8 & 11\end{matrix}\right.$

So the factors we seek are $12$ and $2$

Therefore we can factorise the quadratic as follows:

${x}^{2} + 14 x + 24 \setminus \setminus \setminus \setminus \setminus = {x}^{2} + 12 x + 2 x + 24$
$\therefore {x}^{2} + 14 x + 24 = x \left(x + 12\right) + 2 x \left(x + 12\right)$
$\therefore {x}^{2} + 14 x + 24 = \left(x + 2\right) \left(x + 12\right)$

This approach works for all quadratics (assuming it does factorise) , The middle step in the last section can usually be skipped with practice.