How do you factor the expression # 6x^2 - 23x + 15#?

1 Answer
George C.
May 8, 2016

#6x^2-23x+15=(6x-5)(x-3)#

Explanation:

Use an AC method:

Find a pair of factors of #AC=6*15=90# with sum #B=23#

The pair #18, 5# works in that #18xx5 = 90# and #18+5=23#.

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

#6x^2-23x+15#

#=6x^2-18x-5x+15#

#=(6x^2-18x)-(5x-15)#

#=6x(x-3)-5(x-3)#

#=(6x-5)(x-3)#

#color(white)()#
Alternatively, you can complete the square and use the difference of squares identity:

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

with #a=(12x-23)# and #b=13# as follows:

I will multiply by #4*6 = 24# first to avoid some fractions:

#24(6x^2-23x+15)#

#=144x^2-552x+360#

#=(12x)^2-2(12x)(23)+360#

#=(12x-23)^2-23^2+360#

#=(12x-23)^2-529+360#

#=(12x-23)^2-169#

#=(12x-23)^2-13^2#

#=((12x-23)-13)((12x-23)+13)#

#=(12x-36)(12x-10)#

#=(12(x-3))(2(6x-5))#

#=24(x-3)(6x-5)#

Dividing both ends by #24# we find:

#6x^2-23x+15 = (x-3)(6x-5)#

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