How do you factor $6x^2 + 13x + 8$ ?

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Answer 1 · 2016-08-26T18:13:42

$6x^2+13x+8 = 6(x+13/12-sqrt(23)/12i)(x+13/12+sqrt(23)/12i)$

$6x^2+13x+8$ is in the form $ax^2+bx+c$ with $a=6$ , $b=13$ and $c=8$

This has discriminant $Delta$ given by the formula:

$Delta = b^2-4ac = 13^2-(4*6*8) = 169 - 192 = -23$

Since $Delta < 0$ this quadratic has no simpler factors with Real coefficients, but it does have Complex zeros given by the quadratic formula:

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

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

$=(-13+-sqrt(-23)/12)$

$=-13/12+-sqrt(23)/12i$

Hence factorisation:

$6x^2+13x+8 = 6(x+13/12-sqrt(23)/12i)(x+13/12+sqrt(23)/12i)$

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Footnote

I suspect a sign error in the question since the following is much simpler:

$6x^2+13x-8 = (3x+8)(2x-1)$

How do you factor 6x^2 + 13x + 86x^2 + 13x + 8?