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 $8 + \frac{25}{\sqrt{5}}$

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

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

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

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

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

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

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

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

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

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

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 $8 + \frac{25}{\sqrt{5}}$

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

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

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

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

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

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

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

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

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

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

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

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

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