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How do you factor #11x^2-54x-5#?

1 Answer

Aug 9, 2015

#(11x+1)(x-5)#

Explanation:

Noting that 11 is a prime number we need a factorisation of the form:
#(11x-a)(x-b)=11x^2-(11b+a)x+ab#

So we need two numbers #a,b# where:
#11b+a=54#
#ab=-5#

By inspection we have:
#b=5,a=-1#

So we have:
#11x^2-54x-5=(11x+1)(x-5)#

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