How do you factor $x^2-4x+24$ ?

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Answer 1 · 2015-05-16T22:13:29

$x^2-4x+24$ is of the form $ax^2+bx+c$ with $a=1$ , $b=-4$ and $c=24$ .

This has discriminant

$Delta = b^2-4ac$
$= (-4)^2-(4xx1xx24) = 16-96 = -80$

Since $Delta < 0$ , $x^2-4x+24=0$ has 2 complex, non-real roots and $x^2-4x+24$ has no linear factors with real coefficients.

In other words $x^2-4x+24$ is already in simplest terms for real coefficients.

How do you factor x^2-4x+24x^2-4x+24?