How do you factor completely #16y^3 - 14y^2 - 12y#?

1 Answer
Feb 18, 2017

Answer:

#16y^3-14y^2-12y = 16y(y-7/16-sqrt(241)/16)(y-7/16+sqrt(241)/16)#

Explanation:

Complete the square then use the difference of squares identity:

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

with #a=(16y-7)# and #b=sqrt(241)# as follows:

#16y^3-14y^2-12y = 1/16y(256y^2-224y-192)#

#color(white)(16y^3-14y^2-12y) = 1/16y((16y)^2-2(16y)(7)+(7)^2-241)#

#color(white)(16y^3-14y^2-12y) = 1/16y((16y-7)^2-(sqrt(241))^2)#

#color(white)(16y^3-14y^2-12y) = 1/16y(16y-7-sqrt(241))(16y-7+sqrt(241))#

#color(white)(16y^3-14y^2-12y) = 16y(y-7/16-sqrt(241)/16)(y-7/16+sqrt(241)/16)#