How do you factor #72x^3-228x^2+140x#?

1 Answer
George C.
Feb 18, 2017

#72x^3-228x^2+140x = 4x(3x-7)(6x-5)#

Explanation:

Complete the square then use the difference of squares identity:

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

with #a=(12x-19)# and #b=9# as follows:

#72x^3-228x^2+140x = 1/2x(144x^2-456x+280)#

#color(white)(72x^3-228x^2+140x) = 1/2x(144x^2-456x+361-81)#

#color(white)(72x^3-228x^2+140x) = 1/2x((12x)^2-2(12x)(19)+19^2-81)#

#color(white)(72x^3-228x^2+140x) = 1/2x((12x-19)^2-9^2)#

#color(white)(72x^3-228x^2+140x) = 1/2x((12x-19)-9)((12x-19)+9)#

#color(white)(72x^3-228x^2+140x) = 1/2x(12x-28)(12x-10)#

#color(white)(72x^3-228x^2+140x) = 1/2x(4(3x-7))(2(6x-5))#

#color(white)(72x^3-228x^2+140x) = 4x(3x-7)(6x-5)#

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