How do you factor #14k^(2)-9k-18#?

1 Answer
George C.
Apr 1, 2016

Use an AC method to find:

#14k^2-9k-18=(7k+6)(2k-3)#

Explanation:

Look for a pair of factors of #AC = 14*18 = 252# which differ by #B=9#.

Since #252# is close to #256 = 16^2#, look for a pair close to #16+-9/2#. We find the pair #21, 12# works:

#21 xx 12 = 252#

#21 - 12 = 9#

Use this pair to split the middle term, then factor by grouping:

#14k^2-9k-18#

#=14k^2-21k+12k-18#

#=7k(2k-3)+6(2k-3)#

#=(7k+6)(2k-3)#

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