I see that both denominators are factorable. The first denominator is a case of difference of two squares. The second denominator is easy because I can pull out a factor of x.
This is how it looks.
Factor the denominators
Now, I can multiply across the numerators and across the denominators by placing them side by side.
At this point, I compare the top and bottom factors and decide which ones can be crossed-out.
By color coding the common factors, it is clear which ones to eliminate.
The factors (2x+1) and (x+1) should both be cancelled out as shown.
What remains on top is just the number 1.
And so we have this as our final answer.
Try not to distribute it back and keep it in factored form. However, if your teacher wants the final answer to be distributed, then do so. Either case should be correct. It's just a matter of preference.
Example 2: Multiply the rational expressions .
We need to factor out all the trinomials. The good news is that this type of trinomial, where the coefficient of the squared term is 1, is very easy to handle. Starting with the first numerator, find two numbers where its product gives the last term (10) and its sum gives the middle coefficient (7). I'm thinking of +5 and +2. They should work, agree? For the second numerator, the two numbers must be −7 and +1 since their product is the last term (−7) while its sum is the middle coefficient (−6).
To factor the first denominator, find two numbers with a product of the last term (14) and a sum of the middle coefficient (-9). By trial and error, the numbers are −2 and −7. However, you should always verify it. Now for the second denominator, think of two numbers when multiplied gives the last term (5) and when added gives (6). Obviously they are +5 and +1.
The correct factors for each trinomial should appear like this.
Factor the numerators and denominators completely.
Multiply them together - numerator against numerator and denominator against denominator.
In other words, place side by side in a single fractional symbol.
The color schemes should aid in identifying common factors that we can get rid of.
Yep! We cleaned it out beautifully. This is our final answer.
Example 3:Multiply the rational expressions.
In this problem, I have six terms that need factoring. However, most of them are easy to handle and I will provide suggestions on how to factor each. I will also show a quick side calculation how to factor 4x2+x-3 because it can be challenging to some.
Factoring out all the terms.
Multiply all of them at once by placing them side by side. All numerators stay on top and denominators at the bottom.
I decide to cancel common factors one or two at a time so that I can keep track of them accordingly. Otherwise, I may commit "careless" errors.
I will first get rid of the terms 4x−3 and x−4.
Next, I will eliminate the factors x+4 and x+1.
Simplify. Now I am thinking to cancel the 2's.
Let's do that.
At this point, there's really nothing to cancel. However there's something I can simplify by division.
Notice that (−5) ÷ (−1) = 5.
There you go! This is our final answer.
Example 4:Multiply the rational expressions.
As you can see, there are so many things going on in this problem. However, don't be intimidated by how it looks. Start by factoring each term completely. The problem will become easier as you go along.
In fact, once we have factored the terms correctly, the rest of the steps becomes manageable.
Factor out each term completely.
Multiply by placing them in a single fractional symbol. All numerators are written side by side on top while the denominators at the bottom.
I will first cancel all the (x+5) terms.
Next, cross out the (x+2) and (4x−3) terms.
At this point, I can also simplify the monomials with variable x.
There are five x's on top and two x's at the bottom.
Cancelling the x's with one for one correspondence should leave us three x's in the numerator.
This is the final answer.
Caution: Don't do this!
Note that the x in the denominator is not by itself. It is part of the entire term x−7. This is a common error of many students. Don't fall into this common mistake.
Example 5:Multiply the rational expressions .
I am sure that by now, you are getting better on how to factor. The only thing I need to point out is the denominator of the first rational expression, x3−1. This is a special case called difference of two cubes.
In this problem, I will use Case 2 because of the "minus"; hence the word difference. Case 1 is known as the sum of two cubes because of the "plus" symbol.
These are the factors...
Factorize all the terms as much as possible.
Multiply the numerators together and do the same with the denominators.
I will first cross out the term x2+x+1.
Next, I will eliminate the terms (x−1) and (x−3).
Here it goes. I hope the color coding help you keep track which terms are being cancelled out.
I see a single x term on both the top and bottom.
Cross out that x as well.
At this point, I will multiply the constants on the numerator.
Since (−3) (7) = −21.
Cancel the 21's but leaving −1 on top.
I can keep this as the final answer. However, it will look better if I distribute −1 into (x+3).