Question

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

Answer 1

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`

Answer 2

`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`