How do you factor completely #9x^2 + 53x +40#?

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Answer 1 · 2016-05-06T21:31:12

#9x^2+53x+40=(9x+8)(x+5)#

Use an AC method:

Find a pair of factors of #AC=9*40 = 360# with sum #B=53#

The pair #45, 8# works.

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

#9x^2+53x+40#

#=9x^2+45x+8x+40#

#=(9x^2+45x)+(8x+40)#

#=9x(x+5)+8(x+5)#

#=(9x+8)(x+5)#

#color(white)()#
Footnote

How did I find the pair #45, 8#?

Note that #53# is odd, so as a sum of two numbers one must be odd and the other even.

The prime factorisation of #360# is:

#360 = 2 * 2 * 2 * 3 * 3 * 5#

So the only possible splits into two factors which put all the #2#'s on the right hand side are:

#1 * 360#

#3 * 120#

#5 * 72#

#9 * 40#

#15 * 24#

#45 * 8#

The last of these works, in that #45+8=53#