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)#