Faster way of factoring?

  • Thread starter OrbitalPower
  • Start date
  • #1
OrbitalPower

Homework Statement



[tex]t[\frac{1}{2} (4- t)^\frac{-1}{2}(-1)] + (4 - t)^\frac{1}{2}[/tex]


The way I'd do it is

[tex]\frac{-t}{2(4-t)^\frac{1}{2}}[/tex] + [tex]\frac{\sqrt{4-t}}{1}}[/tex]

Get it over a common denominator

[tex]\frac{-t+\sqrt{4-t}*2\sqrt{4-t}}{2(4-t)^\frac{1}{2}}[/tex]


simplify:

[tex]\frac{-t + 2(4-t)}{2(4-t)^\frac{1}{2}}[/tex]

=

[tex]\frac{8-3t}{2(4-t)^\frac{1}{2}}[/tex]


The way I see books do it:

[tex]t[\frac{1}{2} (4- t)^\frac{-1}{2}(-1)] + (4 - t)^\frac{1}{2}[/tex]

=

[tex]\frac{1}{2}(4-t)^\frac{-1}{2}[-t+ 2(4-t)][/tex]

=

[tex]\frac{8-3t}{2(4-t)^\frac{1}{2}}[/tex]


What exactly is he sending through here? How does he know what to pull out in the second step? Is he mostly just skipping steps or is there a maneuver I'm missing here? I think I see the relationship between my third step (common denominator) and his second step, but I wouldn't know how to get that quickly without doing the algebra.

Here's another example of what I'm talking about:


x^2 * [ (1/2) * (1-x^2)^(-1/2) * (-2x)] + (1 - x^2)^(1/2) * (2x) 1

=

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

=

x(1 - x^2)^(-1/2) * [-x^2(1) + 2(1 - x^2)] 3

=

x * (2 - 3x^2)/ (sqrt(1 - x^2)) 4


Again, same questions. This actually looks confusing to me. What's going on at steps 2 and 3?

How did he know how to separate them like that without writing over a common factor. I've seen it done like this in other places, and it looks to me as if steps are being skipped.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Hi OrbitalPower! :smile:

I'm sorry, but that is the correct way of factoring, and you're just going to have to get used to it, both reading it and doing it.

It is not necessary to put things into quotients first, and examiners will probably deduct marks if you do.

I expect you can see immediately that [tex]x^2 y^3 x^3\,=\,x^5 y^3[/tex], because you can just move the x through the y, and combine all the x's?

Well, it's the same if it's an inverse power: [tex]x^2 y^3 x^{-3}\,=\,x^{-1} y^3[/tex], or [tex]x^{1/2} y^3 x^{-3/2}\,=\,x^{-1} y^3[/tex].

You just move the x through the y, and writing over a common factor is totally unnecessary. :smile:
 
  • #3
OrbitalPower
Thanks for the help. I definitely see what you're saying and I do that all the time during derivatives and so on.

I still can't figure out what's going on, though.

For example, the first one:

[tex]t[\frac{1}{2} (4- t)^\frac{-1}{2}(-1)] + (4 - t)^\frac{1}{2}
[/tex]


Assuming I get (4-t)^(-1/2) + (4-t)^(1/2) together somehow by pulling out the 1/2, and negative one, that's adding the two terms, and the exponents are different. So I don't see the process at all. Maybe I'm just not used to it.

What exactly are you taking out here to set it up like that.
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
251
[tex]t[\frac{1}{2} (4- t)^\frac{-1}{2}(-1)] + (4 - t)^\frac{1}{2}[/tex]

Assuming I get (4-t)^(-1/2) + (4-t)^(1/2) together somehow by pulling out the 1/2, and negative one, that's adding the two terms, and the exponents are different. So I don't see the process at all.
Yes, you're right, of course - the exponents are different - but all that means is that you can't add the two terms.

However, you can factor the second one in your head, and then add them! :smile:

It's the same as, for example, [tex]e^x\,+\,e^{-x}\,=\,e^{-x}(e^{2x}\,+\,1)\,.[/tex]

You need to be happy with doing this! :smile:
 
  • #5
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0
You want to factorise it as much as possible.. taking out the term with a negative power is normally since it means the other term can then have a positive power. 1/2(4-t)^-1/2 is a common factor in that you can multiply by something 'simple' to get the second term. You just need to work out what that is, to get positive power of a half you need to add on 1.. so the second term ends up as 2(4-t). Hope that helps.
 

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