How do you divide #(y^3 + 216)/ (y+6)#?

2 Answers
Jun 1, 2016

quotient #y^2-6y+36# and remainder #0#

Explanation:

In a division of #p_n(x)# by #(x-x_0)# we have
#p_n(x) = (x - x_0)q_{n-1}(x) + r_0# where #q_{n-1}(x)# is the quotient and #r_0# is the remainder. In our case we have

#p_3(y) = y^3+216# so the quotient will have the structure
#q_2(y)=y^2+a y+b# and #r_0 = c# then equating

# y^3+216 = (y+6)(y^2+a y+b)+c#

grouping the coefficients

#{(216 - 6 b + c=0), (6 a + b=0), (6 + a=0):}#

solving for #a,b,c#

#(a = -6, b = 36, c = 0)#

Hint. #c=0# warns us that #(y+6)# is a factor of #y^3+216#

Finally #q_2(y) = y^2-6y+36#

Jun 2, 2016

#y^2-6y+36#

Explanation:

#y^3+216# is expressible as a sum of cubes, which takes the form #a^3+b^3=(a+b)(a^2-ab+b^2)#.

Thus, #y^3+216=y^3+6^3=(y+6)(y^2-y*6+6^2)# which simplifies to be #(y+6)(y^2-6y+36)#.

Therefore, #(y^3+216)/(y+6)=((y+6)(y^2-6y+36))/(y+6)#. The #(y+6)# terms cancel:

#(color(red)(cancel(color(black)((y+6))))(y^2-6y+36))/color(red)(cancel(color(black)((y+6))))=y^2-6y+36#