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.

How do you factor 4w^2 -13w -274w^2 -13w -27?