I'd appreciate any help that I get in solving these problems.
I'll upload an image because it'll be easier for you all to read the problems.
http://img102.imageshack.us/img102/5...lifyingzo8.png
I'd appreciate any help that I get in solving these problems.
I'll upload an image because it'll be easier for you all to read the problems.
http://img102.imageshack.us/img102/5...lifyingzo8.png
Rules you need to know:
$\displaystyle x^a \cdot x^b = x^{a + b}$
$\displaystyle \frac {x^a}{x^b} = x^{a - b} $
$\displaystyle \left( x^a \right)^b = x^{ab}$
to learn how to convert roots to powers so you can apply the rules above, see here
so, that should take care of question 3, try it
for question 4:
(a)
notice that $\displaystyle 1 + 8x^{-1} + 15x^{-2} = 1 + 8 \left(x^{-1} \right) + 15 \left( x^{-1} \right)^2$ ......this is a quadratic
(b)
notice that $\displaystyle x^{\frac 12} - x^{\frac 52} = \left( x^{\frac 12} \right) - \left( x^{\frac 12} \right)^5$ .........factor out the common term
(c) try a similar maneuver as in (a)
(d)
factor out the lowest power of $\displaystyle x$, that is, $\displaystyle x^{- \frac 12}$
3.a) $\displaystyle a^{2p + 2q - (2p - 2q)}$
$\displaystyle = a^{4q}$
3.b) $\displaystyle ( \frac{ x^{\frac{1}{3}} x^{\frac{1}{2}} }{ x^{\frac{2}{3}} } )^{\frac{1}{3}}
$
$\displaystyle = ( \frac{ x^{\frac{5}{6}} }{ x^{\frac{2}{3}} } )^{\frac{1}{3}}$
$\displaystyle = (x^{\frac{1}{6} })^{\frac{1}{3}}$
$\displaystyle = (x^{\frac{1}{18} })$
3.c) $\displaystyle \frac{ x^{( a-b ) \times ( a+b )} }{ x^{a^2} }$
$\displaystyle = \frac{ x^{ ( a^2 - b^2 ) } }{ x^{a^2} }$
$\displaystyle = x^{( - {b^2} )}$
$\displaystyle = \frac{1}{ x^{b^2} }$
---------------------------------
I'm working on the rest...
I did make a mistake. I'll change my previous post in a moment. But there is a problem with how you are looking at all this (I think.)
Try looking at it this way. I'm going to change the variable momentarily to $\displaystyle y = x^{1/2}$.
Thus
$\displaystyle x^{1/2} - \left ( x^{1/2} \right )^5 = y- y^5$
$\displaystyle = y(1 - y^4) = y(1 - y)(y^3 + y^2 + y + 1)$
As it happens, this last factor factors again:
$\displaystyle y^3 + y^2 + y + 1 = (y^3 + y^2) + (y + 1) = y^2(y + 1) + (y + 1) = (y^2 + 1)(y + 1)$
So we have
$\displaystyle = y(1 - y)(y + 1)(y^2 + 1) = x^{1/2}(1 - x^{1/2})(x^{1/2} + 1)(x + 1)$
which is what I should have posted before.
-Dan
Edit: I did the job here better anyway. I'll just leave this one as my answer to your question.
what about treating the second factor in your third line as the difference of two squares?
$\displaystyle y - y^5 = y \left( 1 - y^4 \right) $
$\displaystyle = y \left( 1 - \left( y^2 \right)^2 \right) $
$\displaystyle = y \left( 1 - y^2 \right) \left( 1 + y^2 \right) $ ...the middle factor is the difference of two squares as well
$\displaystyle = y(1 + y)(1 - y) \left( 1 + y^2 \right)$
now replace $\displaystyle y$ with $\displaystyle x^{ \frac 12}$