## Idiot's Guides: Algebra I (2015)

### Part IV. Polynomials

### Chapter 12. Multiplying and Dividing Polynomials

**In This Chapter**

· Multiplying a monomial by a polynomial using the distributive property

· Multiplying two binomials using the FOIL rule

· Multiplying larger polynomials in vertical format

· Dividing by a monomial

· Polynomial long division

In the last chapter, we looked at how polynomials could fit into our understanding of addition and subtraction. Now it’s time to revisit multiplication and division. We need to see what ideas we’ve already learned that are still useful, and what modifications we might need to make. In this chapter, we’ll review methods of multiplication we learned earlier that will still be useful, examine some special cases, and explore what division is possible when working with polynomials.

**Multiplying Polynomials**

To multiply two polynomials, you’ll employ some methods you’ve already learned. Because each monomial involves a power of the variable, the rule for multiplying powers of the same base will be key. You’ll also use the distributive property and the FOIL rule.

To multiply two monomials, like (5*x*^{3})·(−8*x*^{2}), you multiply coefficient by coefficient and variable by variable. Because each of the variables is raised to a power, you need to use the rule for exponents, and add the exponents.

(5*x*^{3})·(−8*x*^{2}) = 5 · (−8)·(*x*^{3} · *x*^{2}) = −40*x*^{3 + 2} = −40*x*^{5}

If one of the monomial factors is a constant, multiply the coefficients, and the variable factor is unchanged.

(−7)·(4*t*^{7}) = (−7)·(4)·(*t*^{7}) = − 28*t*^{7}

If one (or both) of the factors is a first degree monomial, remember that you can write in the exponent of 1 that usually isn’t shown, and then apply the exponent rule.

Multiplying one monomial by one monomial is the basis of all polynomial multiplication. Each of the methods that we’ll explore breaks problems down to a collection of small multiplications. Let’s take what we know about multiplying monomials and build to larger problems.

**Review of Distributive Property**

To multiply a single monomial by a larger polynomial, we’re going to turn to our old friend, the distributive property. You’ve used the distributive property before. It states that multiplying a single term by a sum of terms is equivalent to multiplying that one term by each of the terms in the sum. Applying that to polynomials results in a problem like this one.

3*x*^{2} (5*x*^{3} − 7*x*^{2} + 2*x* − 6) = (3*x*^{2})(5*x*^{3}) + (3*x*^{2})(−7*x*^{2}) + (3*x*^{2}) (2*x*) + (3*x*^{2}) (−6)

= 15*x*^{5} − 21*x*^{4} + 6*x*^{3} − 18*x*^{2}

The distributive property lets you change the problem into a collection of multiplications, all monomial multiplied by monomial. Follow the same rules for those as you have before. Multiply the coefficients and multiply the powers of the variable by keeping the variable and adding the exponents. The distributive property says you can add like terms after multiplying, but if the polynomials were properly simplified before you started, there shouldn’t be any like terms to combine.

**THINK ABOUT IT**

When you multiply a polynomial by a monomial using the distributive property, each term of the polynomial is multiplied by the monomial. Each of the terms in the polynomial was a different power of the variable, and each was multiplied by the same term. The power of each term was increased by the same amount so the terms of the product will all be distinct powers of the variable, and no combining will be necessary.

You might encounter a larger problem, involving a multiplication as well as other operations, or more than one multiplication, in which you will have simplifying to do at the end. The following example has two different multiplications.

4*t* (8*t*^{2} + 3*t* + 4) − 7*t*^{2} (*t*^{2} − 5*t* + 2)

The first multiplication, 4*t*(8*t*^{2} + 3*t* + 4), is a monomial multiplied by a trinomial. The trinomial is in simplest form, so after that multiplication, there won’t be any like terms.

4*t* (8*t*^{2} + 3*t* + 4) = 32*t*^{3} + 12*t*^{2} + 16*t*

The second multiplication, −7*t*^{2} (*t*^{2} − 5*t* + 2), is also a monomial times a trinomial, with the trinomial in simplest form, so that product has no like terms.

−7*t*^{2} (*t*^{2} − 5*t* + 2) = −7*t*^{4} + 35*t*^{3} − 14*t*^{2}

When the whole problem gets put together, however, there are terms from the first multiplication that you can combine with terms from the second.

4*t* (8*t*^{2} + 3*t* + 4) − 7*t*^{2} (*t*^{2} − 5*t* + 2)

32*t*^{3} + 12*t*^{2} + 16*t* − 7*t*^{4} + 35*t*^{3} − 14*t*^{2}

− 7*t*^{4} + (32*t*^{3} + 35*t*^{3}) + (12*t*^{2} − 14*t*^{2}) + 16*t*

− 7*t*^{4} + 67*t*^{3} − 2*t*^{2} + 16*t*

**CHECK POINT**

Complete each multiplication using the distributive property.

1. 5*a* (2*a*^{2} + 3*a*)

2. −3*b*^{2} (2*b*^{2} − 3*b* + 5)

3. 2*x*^{2} (11*x*^{2} − 3*x* − 7)

4. −3*x*^{3} (2 + 4*x* − 5*x*^{2})

5. 3*y*^{4} (2*y*^{6} − y^{2})

6. 8*x* (*x*^{2} − 3)

7. −5*a*^{3} (2*a*^{4} − 4*a*^{2} + 6)

8. 7*x*^{5} (9 − 3*x*^{3})

9. −6*x*^{4} (*x*^{2} + *x* + 3)

10. −*x*^{3} (*x*^{5} − *x*^{4} + *x*^{3} − *x*^{2} + *x* − 1)

The other technique that’s already part of your toolbox and is a crucial skill for multiplying polynomials is the FOIL rule. It’s useful in only one situation, the multiplication of two binomials, but that’s a very common event. Multiplying two terms by two terms requires four separate multiplications that are designated as First, Outer, Inner, and Last.

**Review of FOIL**

The FOIL rule states that (*a* + *b*)(*c* + *d*) = *ac* + *ad* + *bc* + *bd* but writing it in that form doesn’t really communicate it very well. Let’s try this instead. Every binomial is made up of two monomials. Call them the First and the Last, so in the binomial 5*x*^{2} + 3*x*, 5*x*^{2} is the First and 3*x* is the Last. In the binomial 2*x*^{3} − 7*x*^{2}, 2*x*^{3} is the First and −7*x*^{2} is the Last.

When you multiply (5*x*^{2} + 3*x*)(2*x*^{3} − 7*x*^{2}), you have (First + Last)(First + Last). You want to multiply First × First and Last × Last, but you also want to multiply Firsts and Lasts together. There’s a First × Last pair on the outer ends and a Last × First pair in the middle.

When all the multiplications are done, there may (or may not) be like terms that can be combined. Our earlier example will have like terms.

(5*x*^{2} + 3*x*)(2*x*^{3} − 7*x*^{2}) = 5*x*^{2} · 2*x*^{3} + 5*x*^{2} (– 7*x*^{2}) + 3*x* · 2*x*^{3} + 3*x* (– 7*x*^{2})

= 10*x*^{5} − 35*x*^{4} + 6*x*^{4} − 21*x*^{3}

= 10*x*^{5} − 29*x*^{4} − 21*x*^{3}

Here’s an example of the multiplication of two binomials that doesn’t have any like terms after the multiplication is done.

(*x*^{2} + 3) (2*x* + 4) = *x*^{2} · 2*x* + *x*^{2} · 4 + 3 · 2*x* + 2 · 4

= 2*x*^{3} + 4*x*^{2} + 6*x* + 8

The FOIL rule is most often used for multiplying two first-degree binomials to produce a second-degree trinomial. When you multiply (*x* + 5)(*x* + 2), you get a trinomial beginning with *x*^{2}.

(*x* + 5) (*x* + 2) = *x*·*x* + 2 · *x* + 5 ·*x* + 5 · 2

= *x*^{2} + 2*x* + 5*x* + 10

= *x*^{2} + 7*x* + 10

The product of 3*x* − 7 and −4*x* + 1 is also a second degree trinomial.

(3*x* − 7) (−4*x* + 1) = 3*x* (− 4*x*) + 3*x* · 1 + (−7) (−4*x*) + (−7) · 1

= −12 *x*^{2} + 3*x* + 28*x* − 7

= −12 *x*^{2} + 31*x* − 7

**CHECK POINT**

Complete each multiplication using the FOIL rule.

11. (*y* + 6)(*y* − 1)

12. (*x* + 5)(*x* + 2)

13. (*t* − 3)(*t* − 2)

14. (*b* − 4)(*b* + 6)

15. (*y* + 3)(3*y* − 1)

16. (3*x* − 8) (4*x* + 5)

17. (3*x* + 1)(2*x* − 3)

18. (2*x* − 7)(4*x* + 9)

19. (10*a* − 3) (8*a* − 5)

20. (*x*^{2} + 9)(*x*^{2} − 4)

**Special Products**

The FOIL rule can handle any product of two binomials, but two common patterns are worth pointing out. One occurs when you multiply a binomial by itself, squaring it, and the other when you multiply the difference of two monomials by their sum.

When you square a binomial, the product is a trinomial. Two of the terms of that trinomial are the squares of each term of the binomial, and the third is twice the product of the two terms. Let’s start with a simple example.

**ALGEBRA TRAP**

When squaring a binomial, many people will forget the FOIL rule and just square each term. Don’t be one of them. One of the reasons you learn the pattern of the perfect square is so that you won’t make that mistake. Each term of the first binomial must be multiplied by both terms of the second. (*a* + *b*)^{2} ≠ *a*^{2} + *b*^{2}

Recognizing this pattern lets you square binomials quickly. (*t* + 7)^{2} = *t*^{2} + 2(7*t*) + 7^{2} = *t*^{2} + 14*t* + 49 or (*y* − 3)^{2} = *y*^{2} + 2(−3*y*) + (−3)^{2} = *y*^{2} − 6*y* + 9. The same rule applies for more complicated binomials.

When you multiply a difference of two monomials by their sum, like (*x* + 3)(*x* − 3) or (5*x* − 2)(5*x* + 2), the inner and outer products turn out to be opposites. Adding opposites gives you 0, so the middle term you would usually get disappears.

(*x* + 3)(*x* − 3) = *x*^{2} − 3*x* + 3*x* − 3^{2} = *x*^{2} − 9

(5*x* − 2) (5*x* + 2) = (5*x*)^{2} + 2·5*x* − 2·5*x* − 2^{2} = 25*x*^{2} − 4

Multiplying a sum of two monomials by their difference gives you the difference of the square of the first term and the square of the last term.

(*a* + *b*)^{2} = *a*^{2} + 2*ab* + *b*^{2}

(*a* − *b*)^{2} = *a*^{2} − 2*ab* + *b*^{2}

(*a* + *b*)(*a* − *b*) = *a*^{2} − *b*^{2}

**ALGEBRA TRAP**

Don’t forget to square the coefficient of the variable term when you square a binomial. (3*x* + 5)^{2} = (3*x* + 5)(3*x* + 5), and when you FOIL, 3*x*·3*x* is 9*x*^{2}, not 3*x*^{2}.

**CHECK POINT**

Use special product shortcuts to complete each multiplication.

21. (*t* − 4)^{2}

22. (*y* − 7)(*y* + 7)

23. (2*x* − 1)^{2}

24.

25. (2*t* − 5)(2*t* + 5)

26. (3*x* + 5)^{2}

27. (5*a* − 3)^{2}

28. (*x*^{2} − 5)^{2}

29. (*x*^{2} + 1)(*x*^{2} − 1)

30. (5*t*^{2} − 2*t*)(5*t*^{2} + 2*t*)

You already know some strategies for polynomial multiplication. The distributive property is your method for a monomial multiplied by a larger polynomial, and the FOIL rule applies to the specific but common situation of a binomial multiplied by a binomial. Now we need to have a look at how to deal with larger problems.

**Larger Products**

The general rule for multiplying two polynomials, of any size, is that you multiply each term of the first polynomial by each term of the second polynomial. For small problems, that’s not too difficult to keep track of, but when the problems are larger, we need a way of keeping things organized.

Being asked to multiply (2*x*^{2} + 7*x* − 3)(*x*^{3} − 4*x*^{2} + 8*x* + 3) can be challenging if you look at it all at once. The trick is to break it into lots of little multiplications and then put the like terms together at the end. You could just multiply 2*x*^{2} times each term of the second polynomial, then multiply 7*x* times each term, and then multiply −3 times each, but it’s easy to lose your place or miss a term. So a vertical arrangement will make it easier to stay organized. Place the polynomials one under the other, usually with the longer one on top.

Start with the rightmost term of the bottom polynomial, the −3, and multiply each term of the upper polynomial by −3.

Place a 0 under the rightmost term of that *partial product*. Then multiply each term of the upper polynomial by the 7*x*.

**DEFINITION**

A **partial product** is the polynomial produced by multiplying by just one digit or term in a multiplication of polynomials or in the multiplication of multi-digit numbers.

For the third line, start with two zero terms, then multiply each term of the upper polynomial by 2*x*^{2}.

Once you’ve multiplied by each term of the bottom polynomial, the last step is to combine like terms, and they should be nicely lined up in columns.

If one or both polynomials have missing powers of the variable, you can insert zero terms to hold a place and help keep like terms aligned. The product (*t*^{2} − 3*t* + 2)(*t*^{5} − 7*t*^{3} + 5) can be rewritten as (*t*^{2} − 3*t* + 2)(*t*^{5} + 0*t*^{4} − 7*t*^{3} + 0*t*^{2} + 0*t* + 5). When you set up the vertical format, it looks like this.

**CHECK POINT**

Complete each multiplication. Use the format that feels most comfortable.

31. (4*x* − 3)(3*x*^{2} + *x* − 6)

32. (*x* + 5)(*x*^{2} + 2*x* + 3)

33. (2*y* − 3)(*y*^{2} − 3*y* + 2)

34. (2*a* − 5)(*a*^{3} − 5*a*^{2} + *a* − 3)

35. (3*x* − 2)(*x*^{2} − 3*x* − 6)

36. (2*y* − 3) (4*y*^{3} − *y*^{2} + 2*y* + 5)

37. (*x* − 2)(*x*^{3} + 5*x* − 7)

38. (3*a* + 1)(*a*^{2} − a)

39. (2*x*^{3} − 1)(6*x*^{4} + 3*x*^{2} + 1)

40. (*t* − 4)(*t*^{2} + 4*t* + 16)

**Dividing Polynomials**

Once you know how to multiply polynomials, it’s natural to think about the inverse, or opposite operation. You won’t often need to divide polynomials, but there are two techniques you can use for this operation.

**Single Term Divisor**

Just as the basis of all polynomial multiplication is the multiplication of one monomial by another monomial, the root of all division of polynomials is division by a single term. Dividing a monomial by a monomial means that you divide the coefficients, and then deal with dividing the variables by again following the rules for dividing powers of the same base.

To divide −8*x*^{3} by 4*x*, think of the quotient as . The numerical coefficient will be −2, and , so .

**TIP**

If you’re dividing a polynomial by a monomial and you encounter one or more terms for which the coefficients don’t divide evenly, you’ll find it more convenient to leave the quotients as improper fractions rather than decimals or mixed numbers. Simplify the fractions but don’t fuss with decimals that might not terminate.

To divide a larger polynomial by a monomial, for example, , you’ll break the quotient into a sum of divisions.

Then do each division of monomial by monomial.

Not every division problem involving polynomials will produce a polynomial. Remember that the monomials that make up a polynomial must involve positive integer powers of the variable. The exponents can’t be negative in a polynomial, but dividing powers could certainly produce negative exponents.

You can give that quotient using negative exponents, with the understanding that the quotient is not a polynomial. You can use the language of your early experience with division and talk about a remainder. The first three terms of the quotient form a polynomial. You could take the division that far, and say that 6*y*^{5} + 8*y*^{3} − 4*y*^{2} + 2*y* − 3 divided by 2*y*^{2} is 3*y*^{3} + 4*y* − 2 with a remainder of 2*y* − 3. Or you could write the remainder over the divisor as a fraction and say .

**CHECK POINT**

Perform each division, putting your answer in simplest form.

41. (8*x* + 24) ÷ 8

42. (28*x*^{2} + 84*x*) ÷ 7*x*

43. (6*x*^{5} − 12*x*^{4} + 9*x*^{3}) ÷ 3*x*

44. (15*y*^{3} − 20*y*^{2} − 10*y*) ÷ −5*y*

45. (8*x*^{4} − 12*x*^{3}) ÷ (4*x*^{2})

46. (15*t*^{6} − 21*t*^{3}) ÷ (− 3*t*^{2})

47. (16*y*^{5} − 24*y*^{3}) ÷ (−8*y*^{3})

48. (14*x*^{6} − 42*x*^{4} + 56*x*^{2}) ÷ (−14*x*^{2})

49. (−9*a*^{4} + 27*a*^{3} − 81*a*^{2}) ÷ (−9*a*2)

50. (24*x*^{7} − 16*x*^{5} − 48*x*^{4} + 36*x*^{3}) ÷ (−4*x*^{2})

**Long Division**

When the divisor is a polynomial with more than one term, you can’t use the term-by-term division strategy that you used with a monomial divisor. Instead, you have to fall back on the *algorithm* for long division that we learned in arithmetic.

**DEFINITION**

An **algorithm** is a step-by-step procedure for performing a task.

To divide 12*x*^{3} + 8*x*^{2} − 10*x* + 14 by *x* + 2, set up a long division problem with 12*x*^{3} + 8*x*^{2} − 10*x* + 14 as the dividend and *x* + 2 as the divisor.

In arithmetic, you often used estimates to get you through long division. In polynomial long division, you can estimate using the first term of the divisor. In this case, that’s *x*. Start by asking what 12*x*^{3} ÷ *x* will give you. The answer to that is 12*x*^{2}, so put 12*x*^{2} up in the quotient.

Of course, 12*x*^{2} isn’t the whole quotient. It’s just the first term. Multiply 12*x*^{2} (*x* + 2), to get 12*x*^{3} + 24*x*^{2}, and put that underneath the first two terms of the dividend.

Subtract 12*x*^{3} + 24*x*^{2} from the terms above, remembering that both terms are subtracted, not just the first one.

The 0*x*^{3} can be dropped (the first term of your difference will always go away). Now, bring down the last two terms of the dividend.

At this point, you have to decide if you’re done. Will continuing produce negative exponents in the quotient? The quick way to decide is to compare −16*x*^{2} − 10*x* + 14 to the divisor *x* + 2. Because the degree of −16*x*^{2} − 10*x* + 14 is higher than the degree of *x* + 2, you can continue.

Continuing the division process means dividing −16*x*^{2} − 10*x* + 14 by *x* + 2. Follow the same steps. Estimate by dividing −16*x*^{2} by *x*, then multiply that estimate times *x* + 2, place that product underneath, subtract, and bring down any terms from the dividend not yet used. Here’s how it looks.

Compare 22*x* + 14 to *x* + 2. If the degree of the remainder polynomial is lower than the degree of the divisor, it’s time to stop, but you’re not there yet. 22*x* + 14 and *x* + 2 have the same degree, so you can go one more round.

Once the remainder is an expression with a lower degree than the divisor, you can stop dividing. You can say that 12*x*^{3} + 8*x*^{2} − 10*x* + 14 divided by *x* + 2 is 12*x*^{2} − 16*x* + 22 with a remainder of −30, or you can put that remainder over the divisor as a fraction and say the quotient is . You can check your work by multiplying the quotient times the divisor, then adding on the remainder. If the result is the original dividend, our work is correct.

(*x* + 2) (12*x*^{2} − 16*x* + 22) − 30 = (12*x*^{3} − 16*x*^{2} + 22*x* + 24*x*^{2} − 32*x* + 44) − 30

= (12*x*^{3} + 8*x*^{2} − 10*x* + 44)– 30

= 12*x*^{3} + 8*x*^{2} − 10*x* + 14

If either the dividend or the divisor is missing powers of the variable, you’ll probably find it easier to manage the division if you insert terms with zero coefficients to fill the spaces of the missing terms. If you need to divide *x*^{3} − 8 by *x* − 2, set it up as (*x*^{3} + 0*x*^{2} + 0*x* − 8) ÷ (*x* − 2).

**CHECK POINT**

Complete each long division and express the result as quotient plus remainder over divisor.

51. (*x*^{2} + 11*x* + 30) ÷ (*x* + 5)

52. (*x*^{2} + 10*x* + 24) ÷ (*x* + 4)

53. (*z*^{2} − 15*z* + 56) ÷ (*z* − 8)

54. (*b*^{2} − 3*b* − 28) ÷ (*b* + 4)

55. (6*x*^{2} + 13*x* + 6) ÷ (3*x* + 2)

56. (7*y*^{2} − 3*y* − 4) ÷ (*y* − 1)

57. (9*x*^{2} − 42*x* + 45) ÷ (3*x* − 8)

58. (2*b*^{2} − 7*b* + 3) ÷ (*b* − 3)

59. (*x*^{4} − 2*x*^{2} + 1) ÷ (*x*^{2} − 2*x* + 1)

60. (3*x*^{4} − 17*x*^{2} + 10) ÷ (*x*^{2} − 5*x*)

**The Least You Need to Know**

· To multiply a monomial times a polynomial, use the distributive property.

· To multiply two binomials, use the FOIL rule.

· For larger polynomials, arrange them vertically and multiply each term of the bottom polynomial by each term of the top one, and then combine like terms.

· To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.

· To divide by a larger polynomial, use the long division algorithm.