Factor and simplify (PREcalculus)

[Conzen]

Hello,

First time poster here. After lurking about the forums for a couple days I feel as if this question may be overly rudimentary; regardless, I'm stuck. This is not a homework assignment but preparation for an upcoming precalculus class.

(x^2-4)((x^2+3)^1/2) - ((x^2-4)^2) ((x^2+3)^3/2) - Factor and simplify

I understand that (x^2-4) by itself can be broken down to (x-2)(x+2). However I don't think that is getting me anywhere. Beyond that, I know that obviously (x^2-4) is a factor of ((x^2-4)^2) as well as ((x^2+3)^1/2) being a factor of ((x^2+3)^3/2).

With that said, I would rearrange to (x^2-4)(1+(x^2-4)) and ((x^2+3)^1/2)(1+(x^2-4)). Outside of that, I am lost. I realize that I'm seemingly ignoring that subtraction sign there, but in all honesty, I just don't know how to make it fit.

Any help would be appreciated.

 

[HallsofIvy]

You appear to be making a major error. You seem to think that if you factor a term out of an expression, it leaves "1+ " the rest of the expression. That is not true. If you factor "a" out of "ab", all that is left is "b", not "1+ b".

Yes, you have $x^2- 4$ in one term and $(x^2- 4)^2$ in the other so there is a "$x^2- 4$ in both and you can take that out leaving $(x^2+ 3)^{1/2}- (x^2- 4)(x^2+ 3)^{3/2}$/ And [itex](x^2+ 3)^{3/2}= (x^2+ 3)(x^2+ 3)^{1/2}[/tex] so each term has a "$(x^2+ 3)^{1/2}$ that can be factored out, leaving $(1- (x^2- 4)(x^2+ 3)^{1/2}$ <br /> <br /> That is, $(x^2- 4)(x^2+ 3)^{1/2}- (x^2- 4)^2(x^2+ 3)^{3/2}= (x^2- 4)(x^2- 3)^{1/2}(1- x^2- 4(x^2+ 3)^{1/2}$<br /> And now you can factor $x^2- 4$.[/itex]

 

[Conzen]

Thank you for the swift reply.

I definitely understand the mistake I was making. For some reason the subtraction in the middle threw me off. You put me in the right direction. I eventually came to the following:

$[(x^2-4)(x^2+3)^{1/2}]-[(x^2-4)^2(x^2+3)^{3/2}]$

$((x^2-4)(x^2+3)^{1/2})(1-(x^2-4)(x^2+3))$

$((x^2-4)(x^2+3)^{1/2})(1-x^4+x^2+12)$

With a final answer of

$((x+2)(x-2)(x^2+3)^{1/2})(-x^4+x^2+13)$ which is what the book agrees with.

This problem was in the review section (prerequisite) of my precalculus book. Unfortunately, it being a review, there isn't much more of it's kind. Everything problem before and after it I seem to have a very good grounding in. Does anyone have any suggestion of a website or program where I can practice multiple problems of this type? I've checked Khan Academy already. He doesn't seem to have anything quite like the above in his practice exercises; they are either too easy or too advanced (for me at this time).

 

[verty]

Something like http://www.coolmath.com/crunchers/algebra-problems-factoring-by-grouping.htm perhaps?

 

[Conzen]

Thanks for the suggestion; I didn't know about that site. Unfortunately, unless I'm missing something, I run into the same issue as with Khan Academy. These sample problems are extremely simple. I'm looking for something where you actually factor out whole binomials (as well as positive/negative exponents). I understand the basic concept is the same in either case, but I still seem to be having issues here and there...

 

[Conzen]

For instance:

Problems like $x^{-3/2} - 2x^{-1/2} + x^{1/2}$ are extremely simple for me, whereas problems such as $(x^2+1)^{1/2} - 10(x^2+1)^{-1/2}$ are hit or miss and always confuse me somewhere in the process. I know I'm missing something very simple somewhere in the process, but it eludes me.

 

[SammyS]

For instance:

Problems like $(x^-3/2) - (2x^-1/2) + (x^1/2)$ are extremely simple for me, whereas problems such as $((x^2+1)^1/2) - 10((x^2+1)^-1/2)$ are hit or miss and always confuse me somewhere in the process. I know I'm missing something very simple somewhere in the process, but it eludes me.

When doing LaTeX: To include more than one character in an exponent, enclose it in braces, { } .

As in x^{-3/2} which LaTeX renders as $x^{-3/2} \ .$

 

[Conzen]

Thanks for that.