Math Junkies - Algebra!!!

[calvinbiss]

I am slowly preparing for the PE exam in Texas, starting by brushing up on calculus, but I am stuck on the algebra.

The problem is looking for the integral of x^2/(x^2-x-6). The book solution says first to simply the expression to this: 1-[(x-6)/(x^2-x-6)]

I cannot figure how the expression is simplified in that manner!! Is there an error? Am I missing some addition of terms? Anyone wanna give it a go?

 

[RESmonkey]

partial fraction decomposition?

 

[RESmonkey]

o wait, i can do this. hold on.

 

[nonameo]

what is the PE exam? I'm lost.

 

[RESmonkey]

x^2 / (denominator) = (x^2 + x + 6 -x -6) / (denominator)

now separate it to

(x^2 - x-6)/ (denominator) + (-x -6)/(denominator)

=

1 + (-x -6)/(denominator)

 

[edro]

I use that allll the time in my engineering job... :rolleyes;

That's the kind of crap that scares me from taking the FE and PE.

 

[TecHNooB]

Try actually dividing it. It's cuz the degree of the numerator and denomator are the same. I'm guessing the next step is partial fractions.

 

[TridenT]

x^2 / (denominator) = (x^2 + x + 6 -x -6) / (denominator)

now separate it to

(x^2 - x-6)/ (denominator) + (-x -6)/(denominator)

=

1 + (-x -6)/(denominator)

Wut... How can you go from X^2/(x^2-x-6) to (x^2 + x + 6 - x - 6)/(x^2-x-6) ???

 

[calvinbiss]

what is the PE exam? I'm lost.

Professional Engineering Exam. Its this scary test where they make you do all the crazy math you learned in school but have rarely used in the past 5 years of your career due to computers. But you get to bring books.

And you become liable for anything you stamp.

YAY!

 

[OCGuy]

Thank the lawdy that I found a way to make money without having to take tests like that.

 

[calvinbiss]

partial fraction decomposition?

Yes, Yes, Yes!

I forgot about partial fractions!!!!

 

[TecHNooB]

Btw, the simplified expression you posted is wrong :O

 

[syrillus]

Btw, the simplified expression you posted is wrong :O

was just thinking that

 

[calvinbiss]

The simplified expression is directly out of the book, which may have an error.

RESmonkey - TecHNooB is correct: partial fractions need to have the numerator a degree lower than the denom, have to do polynomial division first.

TecHNooB - do you have the correct simplification?

 

[calvinbiss]

Now I need help on the polynomial division!! What is x^2/(x^2-x-6)?!

 

[TecHNooB]

Now I need help on the polynomial division!! What is x^2/(x^2-x-6)?!

Remember when you learned long division in 3rd grade? Write the fraction in that form and you should immediately see how you would evaluate it.

 

[shortylickens]

looks to me like you need to factor out first.

I.E. a^2 + 2ab + b^2 is equal to (a+b)(a+b).

Then you can divide easily.

amirite, or amiwrong?

In the OP problem I believe its (x-3)(x+2), yes?

 

[JTsyo]

Checking to see if the OP question has a problem or not is pretty simple, just plug in 1 for X.

-1/6 != 1/6

It should be 1-[(-x-6)/(x^2-x-6)]

looks to me like you need to factor out first.

I.E. a^2 + 2ab + b^2 is equal to (a+b)(a+b).

Then you can divide easily.

amirite, or amiwrong?

In the OP problem I believe its (x-3)(x+2), yes?

So break it up to (x/(x-3))*(x/(x+2))? then integrate them separately? Man I don't remember the rules for integration. Are you allowed to do that?

x^2 / (denominator) = (x^2 + x + 6 -x -6) / (denominator)

now separate it to

(x^2 - x-6)/ (denominator) + (-x -6)/(denominator)

=

1 + (x+6)/(denominator)

Fix'd