Just a simple order of operations question

**Reckoner** · May 30th 2012, 07:49 AM

Originally Posted by SplashDamage

In this example (which is actually from a previous test that i failed hard)
...
I am not sure what to do.

Is that your work there? Because that looks good to me (so long as we agree that x is positive and that y and z are nonzero).

You are correct when you say that raising a product to an nth power is the same as multiplying that product by itself n times. The result of this is that the power gets "distributed" to each factor. That is, in general,

$(ab)^n=\overbrace{(ab)(ab)(ab)\cdots(ab)}^{n\ \mathrm{factors}}=a^nb^n$ .

Another useful property is

$(a^m)^n = a^{m\cdot n} = a^{mn}$ .

So,

$\frac{27z^9\left(y^2x^{-1/2}\right)^2}{\left(3z^3x^{-1/3}\right)^3y^4}$

$=\frac{27z^9\left(y^2\right)^2\left(x^{-1/2}\right)^2} {\left(3\right)^3\left(z^3\right)^3\left(x^{-1/3}\right)^3y^4}$

$=\frac{27z^9y^4x^{-1}}{27z^9x^{-1}y^4} = 1$ ,

again with $x>0$ and $y,z\neq0$ .

Edit: But be sure you see that we cannot distribute powers over addition or subtraction. That is, $(a+b)^n\neq a^n+b^n$ for arbitrary $a$ and $b$ (though there are specific values for which this will be true). In this case, we would have to multiply $(a+b)$ by itself $n$ times.

**Reckoner** · May 30th 2012, 07:54 AM

Originally Posted by SplashDamage

I mean, if you had something like (x + 3) (y + 4) (x + y)... How do you multiply them all together? Like you would with just two.

Multiply any two factors together. This will give you a new expression with one fewer factor than the previous one. Then repeat the process until everything is multiplied out and you are left with only one factor.

To use your example,

$(x+3)(y+4)(x+y)$

$=\left[(x+3)(y+4)\right](x+y)$

$=(xy+4x+3y+12)(x+y)$

Now we distribute,

$=xy(x+y)+4x(x+y)+3y(x+y)+12(x+y)$

$=x^2y + xy^2 + 4x^2 + 4xy + 3xy + 3y^2 + 12x + 12y$

and we could combine like terms. Using this procedure, you can multiply any finite number of polynomials together.

**SplashDamage** · May 30th 2012, 08:00 AM

Wow that is amazingly helpful. You're epic. What would happen if I ended up doing the ( ) ( ) ( ) thing depending on ^n? And would you recommend the way you did it over that method?

Also another thing. This concerns getting the lowest common denominator in order to -/+ fractions together.

For example I am confronted with this. (1/2) - (2/x+4)

The way I was taught to find the LCD was split the lower halves into their prime factors, which I thought was 2 and 2^2. Although I think I got that wrong. Can you explain to me how to properly find the LCD as my lecturer clearly fails at explaining these things.

**Reckoner** · May 30th 2012, 08:19 AM

Originally Posted by SplashDamage

Wow that is amazingly helpful. You're epic. What would happen if I ended up doing the ( ) ( ) ( ) thing depending on ^n? And would you recommend the way you did it over that method?

Yes, if you have an expression with more than one term, you don't have much choice other than multiplying the expression by itself over and over again.

$(a+b)^n = \overbrace{(a+b)(a+b)(a+b)\cdots(a+b)}^{n\ \mathrm factors}$ ,

and then multiply everything out.

Originally Posted by SplashDamage

For example I am confronted with this. (1/2) - (2/x+4)

The way I was taught to find the LCD was split the lower halves into their prime factors, which I thought was 2 and 2^2. Although I think I got that wrong. Can you explain to me how to properly find the LCD as my lecturer clearly fails at explaining these things.

A simple procedure for finding the LCD is to completely factor each denominator (so that you have prime factors), and then take the highest power of each factor from either denominator. For example, to find the LCD of

$\frac1{3a^2+6ab+3b^2},\quad\frac1{4a^2+4ab},\quad \mathrm{and}\quad\frac1{2a^3}$

Factor each denominator completely:

$3a^2+6ab+3b^2 = 3\left(a^2+2ab+b^2\right) = 3(a+b)^2$

$4a^2+4ab=4a(a+b)=2^2a(a+b)$

$2a^3 = 2a^3$

To find the LCD, multiply the highest power of each factor:

$\mathrm{LCD} = 2^2\cdot3^1\cdot a^3\cdot(a+b)^2 = 12a^3(a+b)^2$ .

For your example, I assume you mean $\frac12 - \frac2{x+4}$ and not $\frac12 - \left(\frac2x + 4\right)$ ?

In this case the denominators are already factored completely. And since there are no common factors, our LCD would simply be the product: $\mathrm{LCD}=2(x+4)$ .

**SplashDamage** · May 30th 2012, 08:28 AM

Originally Posted by Reckoner

To find the LCD, multiply the highest power of each factor:

$\mathrm{LCD} = 2^2\cdot3^1\cdot a^3\cdot(a+b)^2 = 12a^3(a+b)^2$ .

Not sure what you did/meant here

and yeah I meant 2 / (x + 4)

**SplashDamage** · May 30th 2012, 08:40 AM

Also I figured at if we had 2 and (x + 4)...

wouldnt the factors be 2, x, 4

and then wouldnt it be 2 * x * 4, making 8x?

**Reckoner** · May 30th 2012, 08:40 AM

Originally Posted by SplashDamage

Not sure what you did/meant here

The first denominator is: $3\cdot(a+b)^2$

The second: $2^2\cdot a\cdot(a+b)$

And the third: $2\cdot a^3$

In all of these we have four different factors: 2, 3, $a$ , and $(a+b)$ .

So we first look for the highest power of 2. The first denominator has zero factors of 2 ( $2^0$ ), the second one has two factors ( $2^2$ ), and the third denominator has one factor ( $2^1$ ). In our LCD, we use the highest power of 2 among these three, so we take $2^2$ . Similarly, $3^1$ is the highest power of 3 (the other denominators have zero factors of three), $a^3$ is the highest power of $a$ , and $(a+b)^2$ is the highest power of $(a+b)$ . We take each of these and multiply them together.

**SplashDamage** · May 30th 2012, 08:43 AM

OMG i get it. That makes the other question I just made clear.

**Reckoner** · May 30th 2012, 08:43 AM

Originally Posted by SplashDamage

Also I figured at if we had 2 and (x + 4)...

wouldnt the factors be 2, x, 4

No, factors are the things that are being multiplied to form the final product. $x$ and 4 are being added, not multiplied. So $(x+4)$ , in its entirety, is a single factor, and cannot be split further.

**SplashDamage** · May 30th 2012, 09:11 AM

So something like 5x can be split into 5 and x, or something like 25x could be split into 5^2 and x. And two factors can be grouped as one factor if there is parenthesis around them for example (x + 4)?

and if that parenthesis was not there, would the LCD indeed be 4x?

actually no it would still be 2 (x + 4) wouldnt it.

**Reckoner** · May 30th 2012, 09:15 AM

Originally Posted by SplashDamage

So something like 5x can be split into 5 and x, or something like 25x could be split into 5^2 and x.

Yes

Originally Posted by SplashDamage

And two factors can be grouped as one factor if there is parenthesis around them for example (x + 4)?

I'm not sure I understand what you mean. We can group factors as we please, for multiplication is associative: $abc=(ab)c=a(bc)$ . $x+4$ is a single factor, however (though it consists of two terms).

**SplashDamage** · May 30th 2012, 09:17 AM

So with or without the brackets, it would still be 2(x + 4)?

**Reckoner** · May 30th 2012, 09:29 AM

Originally Posted by SplashDamage

So with or without the brackets, it would still be 2(x + 4)?

Right. When $x+4$ is by itself, we don't need to write the parentheses. But it's still a single factor, and we can't factor it further.

**SplashDamage** · May 30th 2012, 09:54 AM

What would happen if we had (x + 8) instead? since 8 can be factored.

also if we had x + 8

**Reckoner** · May 30th 2012, 09:59 AM

Originally Posted by SplashDamage

What would happen if we had (x + 8) instead? since 8 can be factored.

8 can be factored as $2^3$ , yes. But even though 2 is a factor of 8, it is not a factor of $(x+8)$ (because it is not a factor of $x$ ). Since $x$ and 8 have no common factors, the overall expression $(x+8)$ cannot be factored any further.

If, on the other hand, we had something like $(2x+8)$ , we could factor out a 2 from each term:

$2x+8 = 2\cdot x+2\cdot4 = 2(x+4)$ .

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