1. ## Simplifying expressions

Hi everyone! I would like you all to check these problems and if there's an easier way please do give the suggestion. The directions are to simplify.

#1 $\displaystyle 8 + \frac{25}{\sqrt{5}}$

I though I should get one fraction: $\displaystyle \frac{8\sqrt{5}\,+ 25}{\sqrt{5}}$

Rationalize: $\displaystyle \frac{40\,+\,25\sqrt{5}}{5} = \boxed{8 + 5\sqrt{5}}$

# 2 $\displaystyle 9x^3 - 35x^2 - 4x$

$\displaystyle \boxed{x(9x^2 - 35x - 4)}$

I don't see how that can simplify any more... do you agree?

# 3 $\displaystyle 27x^6 + 125y^6 = (3x^2)^3 + (5y^2)^3$

I factored this as: $\displaystyle \boxed{(3x^2 + 5y^2)(9x^4 - 3x^2 5y^2 + 25y^4)}$

# 4 $\displaystyle 4x^3 - 4x^2 - 9x + 9$

Factor by grouping: $\displaystyle 4x^2(x - 1) -9(x - 1) \Rightarrow (4x^2 - 9)(x - 1)$

Which goes to: $\displaystyle \boxed{(2x - 3)(2x + 3)(x - 1)}$

Look good guys/gals?

2. Originally Posted by Jonboy
Hi everyone! I would like you all to check these problems and if there's an easier way please do give the suggestion. The directions are to simplify.

#1 $\displaystyle 8 + \frac{25}{\sqrt{5}}$

I though I should get one fraction: $\displaystyle \frac{8\sqrt{5}\,+ 25}{\sqrt{5}}$

Rationalize: $\displaystyle \frac{40\,+\,25\sqrt{5}}{5} = \boxed{8 + 5\sqrt{5}}$

# 2 $\displaystyle 9x^3 - 35x^2 - 4x$

$\displaystyle \boxed{x(9x^2 - 35x - 4)}$

I don't see how that can simplify any more... do you agree?

# 3 $\displaystyle 27x^6 + 125y^6 = (3x^2)^3 + (5y^2)^3$

I factored this as: $\displaystyle \boxed{(3x^2 + 5y^2)(9x^4 - 3x^2 5y^2 + 25y^4)}$

# 4 $\displaystyle 4x^3 - 4x^2 - 9x + 9$

Factor by grouping: $\displaystyle 4x^2(x - 1) -9(x - 1) \Rightarrow (4x^2 - 9)(x - 1)$

Which goes to: $\displaystyle \boxed{(2x - 3)(2x + 3)(x - 1)}$

Look good guys/gals?
Good job on #1. For the second one, now factor $\displaystyle 9x^2-35x-4$. The third one looks good, although multiply out the 3 and the 5 to get $\displaystyle 15x^2y^2$. And, the fourth one looks good. Nice job.

3. Why not just notice that $\displaystyle \frac{{25}}{{\sqrt 5 }} = 5\sqrt 5$?

4. Originally Posted by topsquark
I'll add a bit here to Plato's answer: An expression is typically considered to be unsimplified if there is a radical in the denominator.

-Dan
There are many ways to do this problem. Although he did unnecessary steps, his final answer does not have a radical in the denominator and is in simplified form.

5. Originally Posted by AfterShock
There are many ways to do this problem. Although he did unnecessary steps, his final answer does not have a radical in the denominator and is in simplified form.
Oops! I didn't look down the page far enough.

-Dan