Need help on simplifying each of the following.

**Jeavus** · September 23rd 2007, 09:52 AM

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

**Jhevon** · September 23rd 2007, 09:59 AM

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

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, $x^{- \frac 12}$

**Jeavus** · September 23rd 2007, 10:08 AM

I have no idea as to where to start on 3a.

>_>

**Jhevon** · September 23rd 2007, 10:10 AM

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

**janvdl** · September 23rd 2007, 10:12 AM

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

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

$= a^{4q}$

3.b) $( \frac{ x^{\frac{1}{3}} x^{\frac{1}{2}} }{ x^{\frac{2}{3}} } )^{\frac{1}{3}}<br />$

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

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

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

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...

**Jeavus** · September 23rd 2007, 10:14 AM

Well I figured out 3a now.

Onto b.

EDIT: I just figured out b and I didn't even see janvdl's post. :P

**janvdl** · September 23rd 2007, 10:27 AM

3.d) $( 16^{p+q} ) ( 8^{p-q} )$

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

**Jhevon** · September 23rd 2007, 10:32 AM

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

**janvdl** · September 23rd 2007, 10:34 AM

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.

**Jeavus** · September 23rd 2007, 11:11 AM

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

Can anyone show me how?

**Jhevon** · September 23rd 2007, 11:17 AM

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?

**Jeavus** · September 23rd 2007, 11:19 AM

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.

**Jeavus** · September 23rd 2007, 11:26 AM

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?

**topsquark** · September 23rd 2007, 11:34 AM

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.

**Jhevon** · September 23rd 2007, 11:40 AM

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 $x^{ \frac 12}$

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