# Algebra: Multiplying Polynomials

## Multiplying Polynomials

Unlike addition and subtraction, you don't need like terms in order to multiply polynomials (nor do you need like terms to divide polynomials, but I'll discuss that in the next section). In fact, multiplying polynomials is actually pretty easy. All you have to do is apply exponential rules and the distributive property, both of which you learned in Encountering Expressions.

### Products of Monomials

Here's what you should do to multiply two monomials together:

**Multiply their coefficients**. The result is the coefficient of the answer.**List all the variables that appear in either term**. These should follow the coefficient you got in step 1, preferably in alphabetical order.**Add up the powers**. Determine the sums of matching variables' exponents and write them above the corresponding variable in the answer.

##### How'd You Do That?

Step 3 tells you to add the powers of matching variables because of the exponential rule from Encountering Expressions stipulating that *x*^{a} *x*^{b} = x^{a + b}. (The product of exponential expressions with matching bases equals the base raised to the sum of the powers.)

Even if the steps seem weird at first, don't worry. Multiplying monomials is a skill you'll understand very quickly.

**Example 3**: Calculate the products.

- (a) (-3
*x*^{2}*y*^{3}*z*^{5})(7*xz*^{3}) **Solution**: First multiply the coefficients: -3 7 = -21; then, list all the variables that appear in the problem in alphabetical order. (It doesn't matter that the second monomial doesn't contain a*y*. As long as a variable appears anywhere in the problem, you should list it next to the coefficient you just found.)- -21
*xyz* - Add up the exponents for each variable you listed. The first term has
*x*to the 2 power, and the second term has*x*to the 1 power, so the answer will have*x*to the 2 + 1 = 3 power. Similarly, the*z*power of the answer should be 8, since there's a z to the 5 power in the first monomial and a z to the 3 in the second. Since there's only one y term, you just copy its power to the final answer; there's nothing to add. - -21
*x*^{3}*y*^{3}*z*^{8} - (b) 3
*w*^{2}*x*(2*wxy*-*x*^{2}*y*^{2}) **Solution**: Apply the distributive property, multiplying both terms by 3*w*^{2}*x*.- 3
*w*^{2}*x*(2*wxy*) + 3*w*^{2}*x*(-*x*^{2}*y*^{2}) - Find each product separately.
- 6
*w*^{3}*x*^{2}*y*- 3*w*^{2}*x*^{3}*y*^{2}

##### You've Got Problems

Problem 3: Calculate the product.

3*x*^{2}*y* (5*x*^{3} + 4*x*^{2}*y* - 2*y*^{5})

### Binomials, Trinomials, and Beyond

##### Critical Point

Some algebra teachers focus on the FOIL method, a technique for multiplying two binomials. Each letter stands for a pair of terms in the binomials, the first, outside, inside, and last terms.

If you've never heard of FOIL, that's fine, because it only works for the special case of multiplying two binomials, whereas my multiple distribution technique works for all polynomial products. Besides, if you use my method, you actually end up doing FOIL anyway, even though it's unintentional.

##### Kelley's Cautions

Once you multiply, always make sure to see if you can simplify the result. Just about every algebra teacher in the world demands simplified answers, and if you don't comply, they've been known to do things like mark answers wrong, take points off, or (in extreme cases) get so angry that they send a cybernetic organism back in time to kill you before you sign up for their class.

Calculating polynomial products is kind of freeing. As I've said, two terms need not have anything in common to be multiplied together. (Based on couples I've met, I think the same is true for people, but I digress.) However, so far you can only multiply polynomial expressions if one of them is a monomial. In Example 3(a), you had two monomials, and in Example 3(b) and Problem 3, you were distributing a monomial. It turns out that multiplying polynomials with more than one term can be accomplished through a slightly modified version of the distributive property.

##### You've Got Problems

Problem 4: Find the product and simplify. (2*x* + *y*)(*x* - 3*y*)

Thanks to the distributive property, you already know that the expression *a*(*b* + *c*) can be rewritten as *ab* + *ac*; just multiply the *a* by each thing in the parentheses. In a similar fashion, you can calculate the product of the expression (*a* + *b*)(*c* + *d*), even though in this case, you're multiplying binomials. Instead of just distributing *a*, like you did moments ago, you'll distribute each term in the first binomial through the second binomial, one at a time.

In other words, you'll multiply everything in the second binomial by *a* and then go through and do it again, this time multiplying everything by *b*.

*ac*+*ad*+*bc*+*bd*

So, you're still distributing, you're just doing it twice, that's all. What if you were multiplying a trinomial by a trinomial? Follow the same procedure; distribute each term in the first polynomial through the second, one at a time.

- (
*a*+*b*+*c*)(*d*+*e*+*f*) =*ad*+*ae*+*af*+*bd*+*be*+*bf*+*cd*+*ce*+*cf*

In case you're wondering, the numbers of terms in the polynomials don't have to match. You could multiply a binomial times a trinomial just as easily, as you'll see in Example 4.

**Example 4**: Find the product and simplify.

- (
*x*- 2*y*)(*x*^{2}+ 2*xy*-*y*^{2})

**Solution**: Each term of the left polynomial, *x* and -2*y*, should be distributed through the second polynomial, one at a time.

- (
*x*)(*x*^{2}) + (*x*)(2*xy*) + (*x*)(-*y*^{2}) + (-2*y*)(*x*^{2}) + (-2*y*)(2*xy*) + (-2*y*)(-*y*^{2})

If you place all of the terms in parentheses, you don't have to worry about signs right away. It doesn't matter if some terms are positive and some are negative; just write them all in parentheses and add all the products together.

Now all you have to do is multiply pairs of monomials together.

*x*^{3}+ 2*x*^{2}*y*-*xy*^{2}- 2*x*^{2}*y*- 4*xy*^{2}+ 2*y*^{3}

The directions for the problem tell you to simplify, which means you should now look for like terms which can be combined. If you look closely, you'll see that the terms 2*x*^{2}*y* and -2*x*^{2}*y* have the same variable, so they can be combined to get 0 (they're opposites of one another, so they'll cancel each other out). In addition, you can combine the terms -*xy*^{2} and -4*xy*^{2} to get -5*xy*^{2}.

*x*^{3}- 5*xy*^{2}+ 2*y*^{3}

Excerpted from The Complete Idiot's Guide to Algebra 2004 by W. Michael Kelley. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with **Alpha Books**, a member of Penguin Group (USA) Inc.

You can purchase this book at Amazon.com and Barnes & Noble.

**See also:**