Polynomials

[tertie]

Can you please advise if I did this problem correctly?

2 - 3
5w+10 2w-4


LCD of 10(w-2)(w+2)

4w-8 - 15w-10
10(w+2)(w-2) 10(w+2)(w-2)


-11w-18
10(w+2)(w-2) Is this right? If so, it cannot be factored anyore, right?[/u]

 

[Denis]

Can you please advise if I did this problem correctly?
2 - 3
5w+10 2w-4

is that: 2 / (5w + 10) - 3 / (2w - 4) ?

and: what are you asked to do with it ? simplify?

 

[tertie]

Yes to both questions, I am sorry it didnt come out the way I hoped it would.

 

[Denis]

Well, all I can see is multiply by the LCD, IF that's simplifying :?:
Are you sure the expression is not equal to some number, then
the job would be solving for w ?

 

[tertie]

just need to simplify.

I just wanted to make sure I did the problem correctly.

 

[Denis]

Can you please advise if I did this problem correctly?
2 - 3
5w+10 2w-4
LCD of 10(w-2)(w+2)
4w-8 - 15w-10
10(w+2)(w-2) 10(w+2)(w-2)
-11w-18
10(w+2)(w-2) Is this right? If so, it cannot be factored anyore, right?[/u]

Sorry, but I have no idea what you're doing...
What does this mean : LCD of 10(w-2)(w+2) ?

Your first step should be:
[2(2w - 4) - 3(5w + 10)] / [(5w + 10)(2w - 4)]

Leads to:
(-11w - 38) / (10w^2 - 40)

If you want to make sure that's correct, substitute any value
for w in the original equation and in the equation I ended up with;
you'll get same result for both.[/code]

 

[tertie]

As per the text we need to find the least common denomiator in order to subtract. So the first step is to factor out both denominators. After doing this I got 5(w+2) for the first denominator and 2(w-2) for the second.

Then since both denominators need common factors I get 10(w+2)(w-2)

Then, of course, multiply the numerator accordingly.

I didn't do it wrong, did I? :-( That would mean I have about 15 problems to do over.

 

[Denis]

4w-8 - 15w-10
10(w+2)(w-2) 10(w+2)(w-2)

-11w-18
10(w+2)(w-2) [/u]

You were close; you made a slight error in the numerator:
-3(5w+10) = -15w - 30 ; you have -15w - 10

So I'll grade you 85% :wink: