# Need help on simplifying each of the following.

• Sep 23rd 2007, 09:52 AM
Jeavus
Need help on simplifying each of the following.
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

http://img102.imageshack.us/img102/5...lifyingzo8.png
• Sep 23rd 2007, 09:59 AM
Jhevon
Quote:

Originally Posted by Jeavus
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

http://img102.imageshack.us/img102/5...lifyingzo8.png

Rules you need to know:

$x^a \cdot x^b = x^{a + b}$

$\frac {x^a}{x^b} = x^{a - b}$

$\left( x^a \right)^b = x^{ab}$

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 $1 + 8x^{-1} + 15x^{-2} = 1 + 8 \left(x^{-1} \right) + 15 \left( x^{-1} \right)^2$ ......this is a quadratic

(b)
notice that $x^{\frac 12} - x^{\frac 52} = \left( x^{\frac 12} \right) - \left( x^{\frac 12} \right)^5$ .........factor out the common term

(c) try a similar maneuver as in (a)

(d)
factor out the lowest power of $x$, that is, $x^{- \frac 12}$
• Sep 23rd 2007, 10:08 AM
Jeavus
I have no idea as to where to start on 3a.

>_>
• Sep 23rd 2007, 10:10 AM
Jhevon
Quote:

Originally Posted by Jeavus
I have no idea as to where to start on 3a.

>_>

you have a raised to some power divided by a raised to some power, use the second rule i showed you
• Sep 23rd 2007, 10:12 AM
janvdl
Quote:

Originally Posted by Jeavus
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

http://img102.imageshack.us/img102/5...lifyingzo8.png

3.a) $a^{2p + 2q - (2p - 2q)}$

$= a^{4q}$

3.b) $( \frac{ x^{\frac{1}{3}} x^{\frac{1}{2}} }{ x^{\frac{2}{3}} } )^{\frac{1}{3}}
$

$= ( \frac{ x^{\frac{5}{6}} }{ x^{\frac{2}{3}} } )^{\frac{1}{3}}$

$= (x^{\frac{1}{6} })^{\frac{1}{3}}$

$= (x^{\frac{1}{18} })$

3.c) $\frac{ x^{( a-b ) \times ( a+b )} }{ x^{a^2} }$

$= \frac{ x^{ ( a^2 - b^2 ) } }{ x^{a^2} }$

$= x^{( - {b^2} )}$

$= \frac{1}{ x^{b^2} }$

---------------------------------

I'm working on the rest...
• Sep 23rd 2007, 10:14 AM
Jeavus
Well I figured out 3a now.

Onto b.

EDIT: I just figured out b and I didn't even see janvdl's post. :P
• Sep 23rd 2007, 10:27 AM
janvdl
3.d) $( 16^{p+q} ) ( 8^{p-q} )$

$= 2 \times 8^{p^2 - q^2}$
• Sep 23rd 2007, 10:32 AM
Jhevon
Quote:

Originally Posted by janvdl
3.d) $( 16^{p+q} ) ( 8^{p-q} )$

$= 2 \times 8^{p^2 - q^2}$

4.a) $1 + 8x^{-1} + 15x^{-2}$

Multiply straight through with $x^2$

$= x^2 + 8x + 15$

$= (x + 5)(x + 3)$

4.b) $\sqrt{x} - \sqrt{x^5}$

= $\sqrt{x} ( 1 - \sqrt{x^4} )$

4.c) Multiply straight through with $x^3$

$= x^2 + x - 12$

$= (x + 4)(x - 3)$

you can't multiply through like that. you are changing the value of the expression. if you expand your final form, you do not get back the original expression
• Sep 23rd 2007, 10:34 AM
janvdl
Quote:

Originally Posted by Jhevon
you can't multiply through like that. you are changing the value of the expression. if you expand your final form, you do not get back the original expression

Sorry Jhevon, my mistake.

Mistook it for an equation.
• Sep 23rd 2007, 11:11 AM
Jeavus
I've figured out 4a and 4c, but I'm still having troubles with 4c and 4d.

Can anyone show me how?
• Sep 23rd 2007, 11:17 AM
Jhevon
Quote:

Originally Posted by Jeavus
I've figured out 4a and 4c, but I'm still having troubles with 4c and 4d.

Can anyone show me how?

as i said, factor out the smallest power of $x$

so, $x^{\frac 32} - 25x^{- \frac 12} = x^{- \frac 12} \left( x^2 - 25 \right)$

now continue

was it 4c or 4b you are having problems with?
• Sep 23rd 2007, 11:19 AM
Jeavus
4b. I understand 4a, b, and d now.

x^(1/2) - x^(5/2)
= (x^1/2) - (x^1/2)^5

I don't know what the factored equation looks like with the 5 as the exponent. Can't recall doing that.
• Sep 23rd 2007, 11:26 AM
Jeavus
Wouldn't making it (1 - x^5) not equal x^5/2?

Multiplying x^1/2 and x^5 doesn't give you x^5/2, I thought you had to add the exponents when multiplying?
• Sep 23rd 2007, 11:34 AM
topsquark
Quote:

Originally Posted by Jeavus
Wouldn't making it (1 - x^5) not equal x^5/2?

Multiplying x^1/2 and x^5 doesn't give you x^5/2, I thought you had to add the exponents when multiplying?

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 $y = x^{1/2}$.

Thus
$x^{1/2} - \left ( x^{1/2} \right )^5 = y- y^5$

$= y(1 - y^4) = y(1 - y)(y^3 + y^2 + y + 1)$

As it happens, this last factor factors again:
$y^3 + y^2 + y + 1 = (y^3 + y^2) + (y + 1) = y^2(y + 1) + (y + 1) = (y^2 + 1)(y + 1)$

So we have
$= y(1 - y)(y + 1)(y^2 + 1) = x^{1/2}(1 - x^{1/2})(x^{1/2} + 1)(x + 1)$

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.
• Sep 23rd 2007, 11:40 AM
Jhevon
Quote:

Originally Posted by topsquark
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 $y = x^{1/2}$.

Thus
$x^{1/2} - \left ( x^{1/2} \right )^5 = y- y^5$

$= y(1 - y^4) = y(1 - y)(y^3 + y^2 + y + 1)$

As it happens, this last factor factors again:
$y^3 + y^2 + y + 1 = (y^3 + y^2) + (y + 1) = y^2(y + 1) + (y + 1) = (y^2 + 1)(y + 1)$

So we have
$= y(1 - y)(y + 1)(y^2 + 1) = x^{1/2}(1 - x^{1/2})(x^{1/2} + 1)(x + 1)$

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?

$y - y^5 = y \left( 1 - y^4 \right)$

$= y \left( 1 - \left( y^2 \right)^2 \right)$

$= y \left( 1 - y^2 \right) \left( 1 + y^2 \right)$ ...the middle factor is the difference of two squares as well

$= y(1 + y)(1 - y) \left( 1 + y^2 \right)$

now replace $y$ with $x^{ \frac 12}$