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

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Answer 1 · 2016-03-30T20:34:29

$x^2+2x+24=(x+1-sqrt(23)i)(x+1+sqrt(23)i)$

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

$Delta = b^2-4ac = 2^2-(4*1*24) = 4 - 96 = -92$

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

It does have zeros given by the quadratic formula:

$x = (-b+-sqrt(b^2-4ac))/(2a)$

$= (-b+-sqrt(Delta))/(2a)$

$= (-2+-sqrt(-92))/2$

$=-1+-sqrt(23)i$

and hence factors:

$x^2+2x+24 = (x+1-sqrt(23)i)(x+1+sqrt(23)i)$

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Alternative Method

Use the difference of squares identity:

$a^2-b^2 = (a-b)(a+b)$

with $a=x+1$ and $b=sqrt(23)i$ as follows:

$x^2+2x+24$

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

$=(x+1)^2-(sqrt(23)i)^2$

$=((x+1)-sqrt(23)i)((x+1)+sqrt(23)i)$

$=(x+1-sqrt(23)i)(x+1+sqrt(23)i)$

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Footnote

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

$x^2+2x-24 = (x+6)(x-4)$

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

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