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

Question

Eva Abramson · Accepted Answer

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

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)#

Genesis Allen · Answer

undefined To factor completely $16y^3 - 14y^2 - 12y$, first, factor out the common factor of $2y$:

$2y(8y^2 - 7y - 6)$

Next, factor the quadratic expression $8y^2 - 7y - 6$ by splitting the middle term:

$2y(8y^2 - 9y + 2y - 6)$

Now, factor by grouping:

$2y((8y^2 - 9y) + (2y - 6))$

$2y( y(8y - 9) + 2( y - 3))$

Therefore, the completely factored form is $2y(y - 3)(8y - 9)$.