How do you factor by grouping #30x^2 + 76x + 48 #?

2 Answers
George C.
Jun 7, 2015

#30x^2+76x+48#

#= 2(15x^2+38x+24)#

#= 2(15x^2+20x+18x+24)#

#= 2((15x^2+20x) + (18x+24))#

#= 2(5x(3x+4)+6(3x+4))#

#= 2(5x+6)(3x+4)#

How did I find a splitting of the middle #38x# term into #20x+18x# that would work?

The coefficients of #15x^2+38x+24# are:

#A=15#, #B=38#, #C=24#

We need to find a factorization of #AC=15*24=360# into a pair of factors whose sum is #B=38#.

#15+24 = 39# is quite close to the #38# we want, so the pair of numbers we're looking for will be similarly close to one another. In fact, notice that #361 = 19^2# and #19+19 = 38# so
#(19-1)(19+1) = 19^2-1^2 = 361 - 1 = 360# as required.

Nghi N.
Jun 7, 2015

#f(x) = 2(15x^2 + 38x + 24).#
If you like to avoid guessing, or avoid the lengthy factoring by grouping, use the systematic New AC Method.
#f(x) = 2a(x - p)(x - q)#
Convert f(x) to f'(x) = x^2 + 38x + 360. = (x - p')(x - q'). Compose factor pairs of a.c = 360. a and c have same sign. Proceed:...(15, 24)(18, 20). OK. This sum is 18 + 20 = 38 = b. Then, p' = 18 and q' = 20
Then #p = (p')/a = 18/15 = 6/5, and q = 20/15 = 4/3.#

Factored form:# f(x) = 30(x + 4/3)(x + 6/5) = 2(3x + 4)(5x + 6)#

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