Polynomials: (x^3 - 27) divided by (x^3 + 3x^2 + 9x)

[chelser13]

I am a tutor and need help solving two problems. I have the answer but I keep getting it wrong and I can't instruct my student correctly on these two problems. I have tried every possible way I can think of but am not getting them right. Here are the problems:

1. (x^3 - 27) divided by (x^3 + 3x^2 + 9x)

2. (x^3+3x^2 - 2x - 6) divided by (x^3 + 27)

 

[galactus]

Re: Polynomials

The first one. Notice the difference of two cubes?.

\(\displaystyle \frac{x^{3}-27}{x^{3}+3x^{2}+9x}\)

Factor the numerator:

\(\displaystyle \frac{(x-3)(x^{2}+3x+9)}{x(x^{2}+3x+9)}\)

See the huge cancellation?.

You end up with:

\(\displaystyle \frac{x-3}{x}=1-\frac{3}{x}\)

However you wish to write it.

Do the same in the second. Note the sum of two cubes.

 

[stapel]

I have tried every possible way I can think of but am not getting them right.

Unfortunately, it is not possible to find errors we cannot see.

As a tutor, you of course understand the importance of clear communication for solving problems. So kindly reply showing your work and reasoning so far. For instance, for the second exercise, you applied to the denominator the sum-of-cubes formula that you've told your student to memorize, you factored the numerator "by grouping", and... then what?

Thank you!

Eliz.

 

[chelser13]

thank you both for your help. for the second problem I just want to check to make sure I am correct.

2. (x^3 + 3x^2 - 2x - 6)/ (x^3 + 27)

I factored out by: (x^2 -2)(x+3)/(x^2 - 3x +9)(x + 3)
so I canceled the (x + 3) and got:
(x^2 - 2)/(x^2 - 3x + 9). Am I missing anything else? Or is this simplified enough?

 

[galactus]

Nope. You're not missing anything. That's it.