How do you factor #5x^2 + 9x - 81#?

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Answer 1 · 2015-05-31T22:52:32

#5x^2+9x-81# is of the form #ax^2+bx+c#
with #a=5#, #b=9#, #c=-81#

The discriminant #Delta# is given by the formula:

#Delta = b^2-4ac = 9^2-(4xx5xx-81) = 9^2 * 21#

#Delta > 0# so #5x^2+9x-81 = 0# has two distinct real roots. Unfortunately #Delta# is not a perfect square, so those roots are irrational.

The roots of #5x^2+9x-81 = 0# are given by the formula:

#x = (-b+-sqrt(Delta))/(2a)#

#= (-9+-9sqrt(21))/10#

#= 9/10(-1+-sqrt(21))#

Hence

#5x^2+9x-81#

#= 5(x - 9/10(-1+sqrt(21)))(x - 9/10(-1-sqrt(21)))#

#= 5(x + 9/10(1-sqrt(21)))(x + 9/10(1+sqrt(21)))#