Multiplying polynomials involves using the distributive property. A lot. We'll start with the easiest case, and gradually work our way up to multiplying multiple polynomials together multiple times. Wow, that was a mouthful.
The easiest case of polynomial multiplication is multiplying a monomial and a polynomial. In this case, we "distribute" the monomial to each term in the polynomial. We really do distribute it, though. We weren't using the quotation marks to be sarcastic or ironic.
What is (2x)(3x2 + 4x + 9)?
We use the distributive property to distribute (2x) over the longer polynomial, then simplify the resulting terms.
|(2x)(3x2 + 4x + 9)||=||(2x)(3x2) + (2x)(4x) + (2x)(9)|
|=||6x3 + 8x2 + 18x|
What is (5x2)(4x3 + 7x)?
We distribute (5x2) over both terms in the second set of parentheses, then simplify.
(5x2)(4x3) + (5x2)(7x) =
20x5 + 35x3
Multiplying two binomials is still an application of the distributive property. In fact, we can use the distributive property even more than we did in the first example. It's like the stuffing and mac 'n cheese that you go back for over and over again at Thanksgiving dinner, while the poor broccoli casserole languishes in the corner.
What is (4x2 + 3)(12x2 + 5x)?
We use the distributive property to distribute the expression (4x2 + 3) over the polynomial (12x2 + 5x), like this:
(4x2 + 3)(12x2) + (4x2 + 3)(5x)
Now we have a new expression with two terms. We use the distributive property again on the first term to distribute (12x2):
(4x2)(12x2) + (3)(12x2) + (4x2 + 3)(5x)
Then use the distributive property a whopping third time on the last term to distribute (5x):
(4x2)(12x2) + (3)(12x2) + (4x2)(5x) + (3)(5x)
Now simplify everything:
48x4 + 36x2 + 20x3 + 15x
And finally write it all down in order of descending exponents:
48x4 + 20x3 + 36x2 + 15x
By this point, you're probably stuffed. However, those au gratin potatoes certainly do seem to be calling your name...
What is (3x – 1)(4x + 2)?
Distribute (3x – 1) over (4x + 2):
(3x – 1)(4x) + (3x – 1)(2)
Now distribute (4x) over (3x – 1) and distribute (2) over (3x – 1):
(3x)(4x) + (-1)(4x) + (3x)(2) + (-1)(2)
Simplify the terms:
12x2 – 4x + 6x – 2
And combine like terms to get:
12x2 + 2x – 2
The terms are already written in descending order by exponent, so we're done. Delicious...time for dessert!
Clearly, there's a whole lotta distributin' going on. With a bit of practice and some elbow grease, you'll get quicker and hopefully cut down on the amount of pencil scribbling when you're rocking two binomials. Even if "elbow grease" does sound pretty disgusting now that we think about it.
Let's run through one more of these guys before we call it quits.
Find (4x2 + 3x)(5x – 6).
First we distribute (4x2 + 3x) over all the stuff in the second term:
(4x2 + 3x)(5x) + (4x2 + 3x)(-6)
Then distribute (5x) and (-6), remembering to distribute the minus sign as well. He'll feel so left out if we forget about him.
(4x2)(5x) + (3x)(5x) + (4x2)(-6) + (3x)(-6)
Now we clean things up a bit with one more round of simplifyin':
20x3 + 15x2 – 24x2 – 18x
Hey, look at that: we've got two x2 terms, so we can combine them to finish up:
20x3 – 9x2 – 18x
It's a thing of beauty, isn't it?
There are some special cases of binomial multiplication that every algebra student should know. Some are useful because they can save you from doing more work than you absolutely need to. Others are good to know because if you don't know them, you're likely to be tripped up and give a completely wrong answer somewhere, which could be devastating. Little-known fact: that's how the War of 1812 started. (Don't check with your history teacher.)
The first special case is multiplication of two binomials of the form
(▲ + ■) and (▲ – ■).
In English, the binomials have the same first term while their second terms are additive inverses of each other. That second square may look the same as the first, but don't forget about the negative sign. It certainly hasn't forgotten about you.
Find (2x + 4)(2x – 4).
Using the distributive property, we calculate that this is:
(2x + 4)(2x) + (2x + 4)(-4) =
4x2 + 8x – 8x – 16
And we can simplify it a bit:
4x2 – 16
Hmm, sneaky. The first and last terms stuck around, but the middle terms went away. Let's try another example and see if this happens again. (Spoiler alert: it will.)
Find (x + y)(x – y).
Using our mighty powers of distribution, we get:
(x + y)(x) + (x + y)(-y) =
x2 + xy – xy – y2
Those middle terms are additive inverses, meaning that when you add them, you have zero. It's almost like they've been erased. From existence. Sorry, we were having another Back to the Future moment.
The inner terms cancel each other out, and we're left with x2 – y2.
To summarize, whenever we perform a multiplication of the form (▲ + ■)(▲ – ■), here's what we get:
(▲ + ■)(▲ – ■) = ▲2 – ■2
Or, if you're more of a letter fan:
(a + b)(a – b) = a2 – b2
This is called a difference of two squares. Because there are...and we're taking the...oh, you get it.
Our next special case is squaring a binomial. Or, in other words, multiplying a binomial by itself. We should be able to accomplish this feat without resorting to the cloning process; the technology is still too new and untested anyway.
To square a binomial, we can, of course, write out the binomial two times and use distribution as normal.
Find (7x + 4)2.
First, we recognize that (7x + 4)2 means (7x + 4)(7x + 4). Now, we distribute to find:
(7x + 4)(7x) + (7x + 4)(4) =
49x2 + 28x + 28x + 16
We can simplify this a little to get:
29x2 + (2 × 28)x + 16 =
29x2 + 56x + 16
In the above example, look at the middle terms we got: they were totally identical. Let's do another example and see if this happens again. We have a feeling it might. Are you getting the same feeling?
Find (x + y)2.
This means we want to find (x + y)(x + y). Let's do this thing.
(x + y)(x) + (x + y)(y) =
x2 + xy + xy + y2 =
x2 + 2xy + y2
In general, whenever we're squaring a binomial of the form (▲ + ■), here's what we'll get:
(▲ + ■)2 = ▲2 + 2▲■ + ■2
Or in letter form:
(a + b)2 = a2 + 2ab + b2
Be careful: (a + b)2 is not the same as a2 + b2. There's a big, glaring "2ab" in there. Remember the Alamo, and remember the 2ab.
Find a binomial whose square is 4x2 + 4x + 1.
From the wording of the problem, we can assume there is such a binomial. If not, this example is sending us on a wild goose chase, and we really don't appreciate it. It could have at least sent us on a domesticated goose chase. They're much easier to catch.
The binomial (if it does indeed exist) will look like (▲ + ■) and we need to figure out what ▲ and ■ are. We know ▲2 = 4x2, so, taking the positive square root to be tidy, ▲ = 2x.
We also know that ■2 = 1 so, once again taking the positive square root, ■ = 1.
At this point we can probably guess that the binomial we're looking for is (2x + 1).
We have a fairly good track record with our guesses, but to make absolutely sure we're right, we can square (2x + 1) and see if we actually do find 4x2 + 1. Thankfully, we do, and our streak of 237 consecutive correct guesses is extended. We should probably try out for Jeopardy!
|(2x + 1)2||=||(2x)2 + 2(2x)(1) + (1)2|
|=||4x2 + 4x + 1|
Now try squaring a binomial that has a minus sign instead of a plus sign. Stand back...there's no telling what might happen.
What is (6x – 2)2?
Use the distributive property on (6x – 2)(6x – 2) to find 36x2 – 12x – 12x + 4.
This simplifies to 36x2 – 24x + 4.
This is very similar to the answer we got in the previous example. However, the second term in the answer is now negative. Negative signs tend to have that effect.
The general pattern is that when we square a binomial of the form (▲ – ■), we get:
(▲ – ■)2 = ▲2 – 2▲■ + ■2
With variables instead of shapes, that's:
(a – b)2 = a2 – 2ab + b2
Alas, there are no quick and easy patterns to use when we're multiplying any two polynomials that don't fit the description of one of our special cases. We just need to apply the distributive property and apply the distributive property again. And maybe once more, being careful to write down all the intermediate work. As the polynomials become more complex, there's more work to write down, but the problems don't become harder...just longer. You may want to free up some time in your afternoon schedule.
Find (5x2 + 3x + 7)(-2x3 – 4x2 + 3x + 4).
Hang on, here we go. First application of the distributive property:
(5x2 + 3x + 7)(-2x3) + (5x2 + 3x + 7)(-4x2) + (5x2 + 3x + 7)(3x) + (5x2 + 3x + 7)(4)
In English you can have a run-on sentence; we're pretty sure this is a run-on expression. It's too monstrous for one line, so we'll break it into several:
(5x2 + 3x + 7)(-2x3)
+ (5x2 + 3x + 7)(-4x2)
+ (5x2 + 3x + 7)(3x)
+ (5x2 + 3x + 7)(4)
Now apply the distributive property to each line and simplify. Quickly, before this expression thinks of something else it wants to say.
The first line will go like this:
|(5x2 + 3x + 7)(-2x3)||=||(5x2)(-2x3) + (3x)(-2x3) + 7(-2x3)|
|=||-10x5 – 6x4 – 14x3|
After we're done following the same steps for each of the other lines, we'll find this sum:
|-10x5||– 6x4||– 14x3|
|– 20x4||– 12x3||– 28x2|
|+ 15x3||+ 9x2||+ 21x|
|+ 20x2||+ 12x||+ 28|
To make things as un-monstrous as possible, we've lined everything up so like terms are stacked on top of each other. It's like Bejeweled, but with coefficients, variables, and exponents!
To find our final answer, we add down the columns:
-10x5 – 26x4 – 11x3 + x2 + 33x + 28
Hopefully you won't be asked to do too many problems like this, but if you are, you'll know what to do.
Another thing we can do that's more messy than it is difficult is multiplying three or more polynomials together. We promise no run-on expressions this time.
Find (x + 1)(2x + 3)(x2 + x).
To do this problem, break it into pieces so that we only need to multiply two polynomials together at a time. Any time you can break something down into simpler parts, we encourage you to do so. Unless it's your car. That should probably stay in one piece.
First we use the distributive property to find the product of the first two polynomials:
(x + 1)(2x + 3) = 2x2 + 5x + 3
Then we find (2x2 + 5x + 3)(x2 + x).
Apply the distributive property once to get:
(2x2 + 5x + 3)(x2)
+ (2x2 + 5x + 3)(x)
Then apply the distributive property to each line and simplify:
|2x4||+ 5x3||+ 3x2|
|+ 2x3||+ 5x2||+ 3x|
Add down the columns to find the final answer:
|2x4||+ 7x3||+ 8x2||+ 3x|
Whenever we're multiplying together more than three polynomials, it doesn't matter which ones we multiply first. We could choose the last two, or the first and last, or whichever two we want. This may be one of the only times in algebra that you're able to exercise some degree of freedom or creativity, so live it up.
However, there's often a rhyme and reason to it. Sometimes we choose which two polynomials to multiply first because we know their product will be nice and attractive. Basically, we choose our polynomials like we choose our prom dates.
Polynomials and integers have many similarities, but when it comes to dividing, they each go their own way. You might even say they "divide." Ooh, see what we did there?
If we add any two integers, subtract one integer from another, or multiply two integers, we always find an answer that's also an integer. When we divide one integer by another, however, we don't necessarily have an integer. We might find a rational number. It would be nice if division followed the example set by addition, subtraction, and multiplication instead of being a pain in the hiney, but there's always one...
It's the same deal with polynomials. If we add, subtract, or multiply two polynomials, we get an answer that's also a polynomial. When we try to divide one polynomial by another, however, the answer won't necessarily be a polynomial. For example, "1" is a polynomial, and "x" is a polynomial, but 1/x is not a polynomial. This is an example of a rational function, which we'll talk more about in the next unit. Sorry, we don't mean to be such a tease.