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

Mar 30, 2016

${x}^{2} + 2 x + 24 = \left(x + 1 - \sqrt{23} i\right) \left(x + 1 + \sqrt{23} i\right)$

#### Explanation:

${x}^{2} + 2 x + 24$ is in the form $a {x}^{2} + b x + c$ with $a = 1$, $b = 2$ and $c = 24$. This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {2}^{2} - \left(4 \cdot 1 \cdot 24\right) = 4 - 96 = - 92$

Since this is negative, our quadratic has no factors with Real coefficients.

It does have zeros given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{- b \pm \sqrt{\Delta}}{2 a}$

$= \frac{- 2 \pm \sqrt{- 92}}{2}$

$= - 1 \pm \sqrt{23} i$

and hence factors:

${x}^{2} + 2 x + 24 = \left(x + 1 - \sqrt{23} i\right) \left(x + 1 + \sqrt{23} i\right)$

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Alternative Method

Use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = x + 1$ and $b = \sqrt{23} i$ as follows:

${x}^{2} + 2 x + 24$

$= {x}^{2} + 2 x + 1 + 23$

$= {\left(x + 1\right)}^{2} - {\left(\sqrt{23} i\right)}^{2}$

$= \left(\left(x + 1\right) - \sqrt{23} i\right) \left(\left(x + 1\right) + \sqrt{23} i\right)$

$= \left(x + 1 - \sqrt{23} i\right) \left(x + 1 + \sqrt{23} i\right)$

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Footnote

If the sign on the last term was $-$ rather than $+$, then the factorisation would be a whole lot simpler:

${x}^{2} + 2 x - 24 = \left(x + 6\right) \left(x - 4\right)$

To find this, you might look for a pair of factors of $24$ that differ by $2$ and find that $6 , 4$ works.