How do you factor #14n^2+23n-15#?

1 Answer
George C.
May 7, 2016

#14n^2+23n-15=(2n-1)(7n+15)#

Explanation:

Given: #14n^2+23n-15#

Use an AC method.

Find a pair of factors of #AC = 14*15=210# which differ by #B=23#

The pair #30, 7# works.

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

#14n^2+23n-15#

#=14n^2+30n-7n-15#

#=(14n^2+30n)-(7n+15)#

#=2n(7n+15)-1(7n+15)#

#=(2n-1)(7n+15)#

#color(white)()#
Alternatively, multiply by #4*14 = 56#, make into a difference of squares, factor using the difference of squares identity:

#a^2-b^2=(a-b)(a+b)#

with #a=(28n+23)# and #b=37#, then divide by #56#:

#56(14n^2+23n-15)#

#=784n^2+1288n-840#

#=(28n)^2+2(28n)(23)-840#

#=(28n+23)^2-23^2-840#

#=(28n+23)^2-529-840#

#=(28n+23)^2-1369#

#=(28n+23)^2-37^2#

#=((28n+23)-37)((28n+23)+37)#

#=(28n-14)(28n+60)#

#=(14(2n-1))(4(7n+15))#

#=56(2n-1)(7n+15)#

So:

#14n^2+23n-15 = (2n-1)(7n+15)#

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