# Advanced Algebra, factor and simplify

1. Jan 12, 2009

### pooker

I am having problems understanding this. Could someone explain how to arrive at the answer for this problem .

(-5/2)(x)(x+3)^(-3/2) + (5)(x+3)^(-1/2)

Thank you

It says to factor and simplify. Express the answer as a fraction without negative exponents.

This is where I get to

(-5x/2)(1/cubed root of (x+3)^3) + 5 / (cubed root of x+3 )

edit here is more i got

-5x/2(x+3)^2) + 5/(x+3)

get common denominator times the second by x+3 then by 2

so

-5x/2(x+3)^2 + 10x + 30 / 2(x+3)^2

I assume I have to get a common denominator but I am unsure of how to do that with negative fraction exponents that are not already equal.

Last edited: Jan 12, 2009
2. Jan 12, 2009

### Дьявол

(-5/2)(x)(x+3)^(-3/2) + (5)(x+3)^(-1/2)

$$\frac{-5x}{2\sqrt{(x+3)^3}}+\frac{5}{\sqrt{x+3}}$$

$$\frac{-5x}{2(x+3)\sqrt{x+3}}+\frac{5}{\sqrt{x+3}}$$

$$\frac{-5x+5*2(x+3)}{2(x+3)\sqrt{x+3}}$$

Do you feel comfortable to continue now?

3. Jan 12, 2009

### pooker

let me try now

attain common denominator

so multiply other side by 2(x+3)

so we get

-5x/2(x+3)(square root x+3) + 10(x+3)/2(x+3)(square root x+3)

simplify

10x+30

since they are know common denominators do the addition

5x + 30 / 2x+6(square root x+3)

Is this correct or am I wrong ?

4. Jan 12, 2009

### Дьявол

You're right. Also, it is even better when you rationalize (multiply the whole equation with sqrt(x+3)/sqrt(x+3)) so that you eliminate the square root in the denominator.

Regards.

5. Jan 12, 2009

### pooker

thank you very much aaron you really helped me out with future problems. Our teacher said only 25% of students pass this class out of 24.

Here is my next one I tried to work

((-4/x+h) + (-4/x) ) / h

I remember something called ltw, where if you had addition in the denominator you multipled each side by the others denominator.

So we would get

-4x/(x^2 + h) + 4x + h / x^2 + h

of course all divided by h

now that the top has a common denominator we simplay add the two together leaving us

h / x^2 + h the H should cancel out leaving

1/ x^2

now for the remainder the total problem is now

(1/x^2) / h we can reverse the bottom and multiply fractions so

1/(x^2) * 1/h

we get

1/ (x^2H)

is this correct Aaron?

btw you are really smart. :)

6. Jan 12, 2009

### NoMoreExams

When you multiply the denominator x + h by x how do you end up with x^2 + h?

7. Jan 13, 2009

### Дьявол

((-4/x+h) + (-4/x) ) / h

$$\frac{\frac{-4x}{x(x+h)}+\frac{-4(x+h)}{x(x+h)}}{h}$$

$$\frac{-4x-4x-4h}{xh(x+h)}$$

The final result:

$$-\frac{4(2x+h)}{xh(x+h)}$$