How do you factor #4w^2 -13w -27#?

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Answer 1 · 2015-05-23T12:02:39

#4w^2-13w-27# is of the form #aw^2+bw+c#, with #a=4#, #b=-13# and #c=-27#.

This has discriminant given by the formula:

#Delta = b^2-4ac = (-13)^2-(4xx4xx-27)#

#= 169 + 432 = 601#

Though positive, this is not a perfect square. So the values of #w# for which #4w^2-13w-27 = 0# are real, but irrational. These roots are:

#w = (-b +- sqrt(Delta))/(2a) = (13+-sqrt(601))/8#

Hence:

#4w^2-13w-27#

#=4(w-(13+sqrt(601))/8))(w-(13-sqrt(601))/8))#

#=(2w-(13+sqrt(601))/4)(2w-(13-sqrt(601))/4))#

If the original problem were to factor #4w^2-12w-27# the result would have been the much simpler:

#4w^2-12w-27 = (2w-9)(2w+3)#

which you could spot using the rational root theorem and a little trial and error.