Advanced Algebra, factor and simplify

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

[tex]\frac{-5x}{2\sqrt{(x+3)^3}}+\frac{5}{\sqrt{x+3}}[/tex]

[tex]\frac{-5x}{2(x+3)\sqrt{x+3}}+\frac{5}{\sqrt{x+3}}[/tex]

[tex]\frac{-5x+5*2(x+3)}{2(x+3)\sqrt{x+3}}[/tex]

Do you feel comfortable to continue now?
 
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 ?
 
pooker said:
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 ?

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.
 
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. :)
 
When you multiply the denominator x + h by x how do you end up with x^2 + h?
 
((-4/x+h) + (-4/x) ) / h

[tex]\frac{\frac{-4x}{x(x+h)}+\frac{-4(x+h)}{x(x+h)}}{h}[/tex]

[tex]\frac{-4x-4x-4h}{xh(x+h)}[/tex]

The final result:

[tex]-\frac{4(2x+h)}{xh(x+h)}[/tex]