# Thread: rational exponents for factoring

1. ## rational exponents for factoring

i am having pre cal woes

today was the first day of school and i cant remember how to do this home work

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

step by step would be awesome

2. Originally Posted by rohizzle121
i am having pre cal woes

today was the first day of school and i cant remember how to do this home work

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

step by step would be awesome
the common term is $x^2 + 1$, start by factoring out the lowest power of it (you can also factor out a 1/4 to make things look nice, but lets keep things simple.

so you get $(x^2 + 1)^{-1/4}( \cdots$

can you finish?

3. Originally Posted by Jhevon
the common term is $x^2 + 1$, start by factoring out the lowest power of it (you can also factor out a 4 to make things look nice, but lets keep things simple.

so you get $(x^2 + 1)^{-1/4}( \cdots$

can you finish?
when i try that i get
$(x^2 + 1)^{-1/4}({3/4}-{1/2}^{1/2})$

i am like 100% i did somthing wrong

after i take the common i am not sure what to do with the rest

4. Originally Posted by rohizzle121
when i try that i get
$(x^2 + 1)^{-1/4}({3/4}-{1/2}^{1/2})$

i am like 100% i did somthing wrong

after i take the common i am not sure what to do with the rest
indeed, you did do something wrong. if you factor out $(x^2 + 1)^{-1/4}$ from $(x^2 + 1)^{3/4}$ you are left with $(x^2 + 1)$ since $(x^2 + 1)^{3/4} = (x^2 + 1)(x^2 + 1)^{-1/4}$

you have to make sure that when you factor something out, that the powers of the base add up to the original power

5. Originally Posted by Jhevon
indeed, you did do something wrong. if you factor out $(x^2 + 1)^{-1/4}$ from $(x^2 + 1)^{3/4}$ you are left with $(x^2 + 1)$ since $(x^2 + 1)^{3/4} = (x^2 + 1)(x^2 + 1)^{-1/4}$

you have to make sure that when you factor something out, that the powers of the base add up to the original power

i am still quite confused
${3/4}(x^2+1)^{-1/4}-{1/2}(x^2+1)^{3/4}$

$(x^2+1)^{-1/4}-{1/2}(x^2+1)$
then
i am completely lost
i plan on going tomorrow for tutorials after school i just want to try right now

6. Originally Posted by rohizzle121
i am still quite confused
${3/4}(x^2+1)^{-1/4}-{1/2}(x^2+1)^{3/4}$

$(x^2+1)^{-1/4}-{1/2}(x^2+1)$
then
i am completely lost
i plan on going tomorrow for tutorials after school i just want to try right now
$\frac 34(x^2 + 1)^{-1/4} - \frac 12 (x^2 + 1)^{3/4} = (x^2 + 1)^{-1/4} \bigg( \frac 34 - \frac 12 (x^2 + 1) \bigg)$ ........note that if you multiply this out, you get back the original. this was not true with what you wrote.

to make it look nicer, we can factor out a 1/4, to get:

$\frac 14 (x^2 + 1)^{-1/4} [3 - 2(x^2 + 1)] = \frac 14 (x^2 + 1)^{-1/4}(1 - 2x^2) = \frac 14 (1 - 2x^2)(x^2 + 1)^{-1/4}$

look carefully at what i did. now's your chance to ask questions

7. Originally Posted by Jhevon
indeed, you did do something wrong. if you factor out $(x^2 + 1)^{-1/4}$ from $(x^2 + 1)^{3/4}$ you are left with $(x^2 + 1)$ since $(x^2 + 1)^{3/4} = (x^2 + 1)(x^2 + 1)^{-1/4}$

you have to make sure that when you factor something out, that the powers of the base add up to the original power
i understand up to that part why is this true $(x^2 + 1)^{3/4} = (x^2 + 1)(x^2 + 1)^{-1/4}$

and on the second part after you took out the 1/4 i am totally confused

8. Originally Posted by rohizzle121
i understand up to that part why is this true $(x^2 + 1)^{3/4} = (x^2 + 1)(x^2 + 1)^{-1/4}$

and on the second part after you took out the 1/4 i am totally confused
forget factoring out the 1/4. did you get up to the step before that?