# How do you simplify the following expressions: (x)/(sqrt(x+5)-sqrt(5))? (1/x)/(x-1/2)? x^4+11x^2-80? e^x(2x+1)^3+e^2x(2x+1)^2? x^3-xy^2+x^2y-y^3?

Aug 4, 2016

You have posted too many questions on the one page. Please allocate one solution request page to each question.

$\sqrt{x + 5} + \sqrt{5}$

#### Explanation:

Solution to the first one.

Given:$\text{ } \frac{x}{\sqrt{x + 5} - \sqrt{5}}$

'.......................................................................................................................
Known that if you have ${a}^{2} - {b}^{2}$ then it is the same as $\left(a - b\right) \left(a + b\right)$
We have the equivalent of $a - b$
,.........................................................................................................................

Multiply by 1 but in the form of

$1 = \frac{\sqrt{x + 5} + \sqrt{5}}{\sqrt{x + 5} + \sqrt{5}} \leftarrow \text{ equivalent to the examples } \left(a + b\right)$ but
$\text{ }$ as $\frac{a + b}{a + b}$

This will result in squaring the square roots in the denominator

$\frac{x}{\sqrt{x + 5} - \sqrt{5}} \times \frac{\sqrt{x + 5} + \sqrt{5}}{\sqrt{x + 5} + \sqrt{5}}$

$\frac{x \left(\sqrt{x + 5} + \sqrt{5}\right)}{\left(x + 5\right) - \left(5\right)} = \sqrt{x + 5} + \sqrt{5}$

Aug 9, 2016

I will factorize ${x}^{4} + 11 {x}^{2} - 80$ for you.

#### Explanation:

Problems like this are often meant to be factored as $\left({x}^{2} + a\right) \left({x}^{2} + b\right)$, $a \mathmr{and} b$ being integers.

We use the same strategy to factor trinomials of this type as we do to factor those of the form $y = a {x}^{2} + b x + c$, namely finding two numbers that multiply to $a c$ and that add to $b$. Two numbers that do this are $16$ and $- 5$.

Hence, ${x}^{4} + 11 {x}^{2} - 80 = \left({x}^{2} + 16\right) \left({x}^{2} - 5\right)$

This is as far as we can go, because nothing can be factored further.

Hopefully this helps!