Multiplying Rational Expressions
Objective
Reviewing over LCD
Well, to be honest, we still need to review before we actually get into it.
But what do we need to review?
Well, a little more 3rd grade math, which is adding fractions!
So, how do we add fractions?
Well, if we have the same denominator, we just add the numerators together.
Like so:
But what if it isn’t easy?
What if we don’t have the same denominators?
Well, that’s when we need to find the least common multiple!
Finding the Least Common Multiple
So here’s how we add fractions.
Let’s start off with an example:
We can’t just add these right?
We have no idea how to add them.
However, what if we changed them to something we can add?
But, we can’t change the equation right?
So, how do we change the numbers, without changing the amount?
Well, we can try to multiply the fractions by 1, but what kind of one?
This is where the Least Common Multiple comes in.
So, to start, let’s see which multiple the two denominators share.
To find it, let’s list out the multiples of 2:
2 4 6 8 10 12 14 16 18 20
Now, let’s list out the multiples of 3:
3 6 9 12 15 18 21 24 27 30
Now, which number do they both share?
6
6
Now we have to find out how to make 3 into 6.
We multiply it by 2!
Now we have to find out how to make 2 into 6.
We multiply it by 3!
THE CHEATING WAY
So, there is actually another way of creating a common denominator instead of finding the least common multiple
But, it’s not the best way.
However, it will work every single time.
We multiply by opposite denominators.
Sounds weird, but let’s try it:
So, what we do, is look at:
( )
= 35
Now, we multiply each side by the opposite number to get 35:
Again, this works every time, but it can take more work.
SO…….WHAT DOES THIS HAVE TO DO WITH ALGEBRA 2?
Again, this is not to insult your intelligence.
The truth is, you probably haven’t needed to find the least common denominator for years.
So, a little refresher on how to do it isn’t that bad of a thing.
But why do we need it?
Well, mainly because we are going to be adding fractions…….
With variables.
So, for example:
EXAMPLE 1
Add the following:
Alright, so this one seems a little weirder than what we’ve done before.
However, we can still do this.
First things first, we can’t add them together the way they are.
Mainly because we need a least common denominator.
Well, what we have so far is just x and 2.
Is there a possible way for us to maybe use the cheat way to find the LCD?
Well sure!
So, to use the cheat way, let’s first multiply the left side by 2/2:
And the right side by x/x:
Now, if we multiply the left side, we get:
And if we multiply the right side, we get:
Now we can add them together!
So, adding them together we get:
THAT WASN’T SO BAD, WAS IT?
Pretty easy right?
Well, it does get a little more complicated.
However, we’re usually using the cheat method to find the LCD.
So, let’s try a few more to make sure you got it.
EXAMPLE 2:
Add the following:
Again, we can’t seem to add these together the way they are.
Mainly because we need a least common denominator.
Well, what we have so far is x + 1 and x.
Is there a possible way for us to maybe use the cheat way to find the LCD?
Well sure!
However, it gets a little complicated.
First, we need to multiply the left side by x/x:
And the right side by x+1/x+1:
Now, if we multiply the left side, we get:
And if we multiply the right side, we get:
Now we can add them together!
So, adding them together we get:
And that’s our answer!
Now let’s try a little harder one:
Example 3:
Add the following:
Again, we can’t seem to add these together the way they are.
Mainly because we need a least common denominator.
Well, what we have so far is x + 1 and x - 2.
Is there a possible way for us to maybe use the cheat way to find the LCD?
Well sure!
However, it gets a little complicated.
First, we need to multiply the left side by x - 2/x - 2:
And the right side by x + 1/x + 1:
Now, if we multiply the left side, we get:
And if we multiply the right side, we get:
Now we can add them together!
So, adding them together we get:
And that’s our answer!
So, now, what if there is something harder?
But we don’t really want to multiply by a binomial?
Something like:
Example 4:
Add the following:
Again, we can’t seem to add these together the way they are.
Mainly because we need a least common denominator.
However, do we really want to multiply this monster by anything to make it even bigger?
Or, maybe there’s a way to make it smaller?
Well, can we factor this?
In fact we can!
When we factor the left side, we see that:
And when we factor the right side, we see:
Now that we see we have some common factors, we can see how easy we can get a LCD.
All we need to do is multiply the right side by -4/-4, so we can get:�12(x + 13) on both sides.
So:
And now we add, and we get:
Example 5:
Add the following:
Again, we can’t seem to add these together the way they are.
Mainly because we need a least common denominator.
However, do we really want to multiply this monster by anything to make it even bigger?
Or, maybe there’s a way to make it smaller?
Well, can we factor this?
In fact we can!
When we factor the left side, we see that:
And when we factor the right side, we see:
Now that we see we have some common factors, we can simplify some.
And what we’re left with is:
Adding what’s left, we get:
So now, how do we multiply rational functions?
Well, honestly there are two ways:
We can just multiply them together and hope for the best, then simplify it later.
Orrrrr……
We can simplify what we have and the multiply.
To do this, however, we need to review over
FACTORING
Factoring a quadratic
Example 1
Let’s say we are given:
We can see that it’s just 1, so there’s nothing to add to our factored equation.
However, we can see that we have 12 as our number.
So we break 12 into its multiples, and see if we can add two of those multiples together to get 7.
Well, we can see that
12 = 6 * 2
12 = 3 * 4
We know 6 + 2 = 8
But 3 + 4 = 7!
So, now we have our pieces!
Factoring the quadratic
So now we just plug in what we have:
So:
(x + )(x + )
And we know that 12 = 3 * 4, and 3 + 4 = 7, so:
(x + 3)(x + 4)
Now we solve!
(x + 3)(x + 4) = 0
x + 3 = 0
-3 -3
x = -3
And
x + 4 = 0
-4 -4
x = -4
So x = -3, -4!
So let’s try another one
Let’s say we’re given:
(x + )(x + )
And we know 2 = 2 * 1
And 2 + 1 = 3
So:
(x + 2)(x + 1) = 0
Now we solve!
x + 2 = 0
- 2 -2
x = -2
And
x + 1 = 0
-1 -1
x = -1
What about if there are negatives?
So the thing about factoring is, the quadratic that you are trying to factor actually gives you all of the information you need.
Whether you need to make a certain number negative or positive, as well as whether or not you need to subtract versus add, you’ll find it in the original quadratic.
You just have to know where to look:
WHAT A QUADRATIC TELLS US:
Let’s take a look at a sample quadratic:
Looking at this quadratic, we can see that the last number is negative:
So, this means that we are going to have alternating signs.
Or, in other words, we’re going to be subtracting.
The reason this works is because the only way to get a negative sign for the last number in our quadratic, is if the two numbers in parenthesis also have different signs.
We can also see that 3 is negative:
Which means the two numbers that are being subtracted, need to end up negative.
So, we’re going to find the factors of 4:
/\
4 1
2 2
And we’re going to subtract them to find what we need.
So:
- = 3
- = 0
As we can see, 4 – 1 = 3, so our factors are going to be:
(x – 4)(x + 1) = 0
Why is 4 negative?
Because we need to make a -3x
x – 4 = 0
x + 1 = 0
+ 4 + 4 -1 -1
x = 4 x = -1
WHAT A QUADRATIC ALSO TELLS US:
Let’s take a look at a sample quadratic:
Looking at this quadratic, we can see that the last number is positive:
So, this means that we are going to have the same sign.
Or, in other words, we’re going to be adding.
This is a little different than the last one.
Because the last number is a positive, this means we will be adding, and both numbers will need to be the same sign.
We can also see that 3 is negative:
Which means the two numbers that are being added, need to end up negative.
So, we’re going to find the factors of 2:
/\
2 1
And we’re going to add them to find what we need.
So:
+ = 3
As we can see, 2 + 1 = 3, so our factors are going to be:
(x – 2)(x - 1) = 0
Why are they both negative?
Because we need to make a -3x
x – 2 = 0
x - 1 = 0
+ 2 + 2 +1 +1
x = 2 x = 1
SO NOW THAT WE HAVE REVIEWED
Let’s look at some examples
Because like a lot of Algebra 2, this is much easier to show than explain:
EXAMPLE 1:
Multiply the following:
Now, again, we could just multiply and then simplify
But dividing a 4th exponential polynomial by another 4th exponential polynomial may be a little bit of a nightmare.
So instead, let’s see if we can factor this and make it more simple.
So, now substituting in our factored expressions, we have:
Now we can cancel!
Lastly, we need to determine the excluded values of the expression.
Going back to the original expression, we saw that we had:
Setting up the denominators to 0, we see that:
x + 2 = 0 x – 4 = 0 x – 5 = 0
-2 -2 +4 +4 +5 +5
x = -2 x = 4 x = 5
Example 2:
Multiply the following:
Again, we need to factor the expression that we have to make it easier to multiply it.
So:
So, now substituting in our factored expressions, we have:
Now we can cancel!
And what we have left over is:
Simplifying our fraction, we finally get:
Lastly, we need to determine the excluded values of the expression.
Going back to the original expression, we saw that we had:
x + 3 = 0 x + 5 = 0 x + 8 = 0
+3 +3 -5 -5 -8 -8
x = 3 x = -5 x = -8