How do you factor #16x^2 + 80x + 84#?

1 Answer
George C.
May 2, 2016

#16x^2+80x+84=4(2x+3)(2x+7)#

Explanation:

Separate out the common scalar factor #4#, complete the square, then use the difference of squares identity:

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

with #a=(2x+5)# and #b=2# as follows:

#16x^2+80x+84#

#=4(4x^2+20x+21)#

#=4((2x)^2+2(2x)(5)+21)#

#=4((2x+5)^2-25+21)#

#=4((2x+5)^2-4)#

#=4((2x+5)^2-2^2)#

#=4((2x+5)-2)((2x+5)+2)#

#=4(2x+3)(2x+7)#

#color(white)()#
Alternatively, we can use an AC method to factor #4x^2+20x+21#:

Look for a pair of factors of #AC=4*21 = 84# with sum #B=20#.

The pair #14, 6# works.

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

#4x^2+20x+21#

#=4x^2+14x+6x+21#

#=(4x^2+14x)+(6x+21)#

#=2x(2x+7)+3(2x+7)#

#=(2x+3)(2x+7)#

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