How do you factor #10w² + 11w - 8#?

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Answer 1 · 2015-05-17T11:59:35

#10w^2+11w-8# is of the form #aw^2+bw+c# with #a=10#, #b=11# and #c=-8#.

The discriminant of this quadratic is given by the formula

#Delta = b^2-4ac = 11^2-(4xx10xx-8) = 121+320 = 441 = 21^2#

Being a positive square integer the equation #10w^2+11w-8 = 0# has two distinct rational roots, given by the formula:

#w = (-b+-sqrt(Delta))/(2a) = (-11+-21)/20#

That is

#w = (-11+21)/20 = 10/20 = 1/2#

and

#w = (-11-21)/20 = -32/20 = -8/5#

These give us factors #(2w-1)# and #(5w+8)#

So #10w^2+11w-8 = (2w-1)(5w+8)#