Q. Evaluate when y = –4 and x = –3.
A. 6
Q. Evaluate when n = 8 and r = 3.
A. 56. What’s with this exclamation (the n!)? The exclamation indicates an operation called a factorial. This operation has you multiply the number in front of the ! by every positive whole number smaller than it. You see a lot of factorials in statistics and higher mathematics. The order of operations is important here, too.
15. Evaluate 3x2 if x = –2.
Solve It
16. Evaluate 9y – y2 if y = –1.
Solve It
17. Evaluate –(3x – 2y) if x = 4 and y = 3.
Solve It
18. Evaluate 6x2 – xy if x = 2 and y = –3.
Solve It
19. Evaluate if x = 4 and y = 1.
Solve It
20. Evaluate if x = 3 and y = –1.
Solve It
21. Evaluate if a = 3, b = –2, and c = –1.
Solve It
22. Evaluate if n = 5 and r = 2.
Solve It
Answers to Problems on Algebraic Expressions
This section provides the answers (in bold) to the practice problems in this chapter.
1. Combine the like terms in . The answer is .
2. Combine the like terms in . The answer is .
3. Combine the like terms in . The answer is .
4. Combine the like terms in . The answer is .
5. Multiply . The answer is .
6. Multiply . The answer is .
7. Multiply . The answer is .
8. Divide (write all exponents as positive numbers) . The answer is –2xy.
9. Divide (write all exponents as positive numbers) . The answer is .
10. Divide (write all exponents as positive numbers) . The answer is .
11.
12.
13.
14.
15. Evaluate . The answer is 12.
16. Evaluate 9y – y2 if y = –1. The answer is –10.
17. Evaluate –(3x – 2y) if x = 4 and y = 3. The answer is –6.
18. Evaluate 6x2 – xy if x = 2 and y = –3. The answer is 30.
19. Evaluate if x = 4 and y = 1. The answer is 3.
20. Evaluate if x = 3 and y = –1. The answer is –3.
21. Evaluate if a = 3, b = –2 and c = –1. The answer is 1.
22. Evaluate if n = 5 and r = 2. The answer is 70.
Part II
Changing the Format of Expressions
In this part . . .
A lot of algebra requires changing an expression involving terms and operations into something more user-friendly. You perform the operations of addition, subtraction, multiplication, division, power, and root to help simplify an expression and cut down on terms. But you have to know when to do what, where it goes, and how it gets there. Another change of format involves factoring, which in algebra means something very specific. The factored form is very useful — you just have to know how to get there. Conquer these skills, and you’re on your way to the challenges of equations and other terrific algebraic activities. Go for it!
Chapter 7
Specializing in Multiplication Matters
In This Chapter
Distributing over algebraic expressions and incorporating FOIL
Squaring and cubing binomials
Calling on Pascal to raise binomials to many powers
Incorporating special rules when multiplying
In Chapter 6, I cover the basics of multiplying algebraic expressions. In this chapter, I expand those processes to multiplying by more than one term. When performing multiplications involving many terms, it’s helpful to understand how to take advantage of special situations, like multiplying the sum and difference of the same two values, such as (xy + 16)(xy – 16), and raising a binomial to a power, such as (a + 3)4, which seem to pop up frequently. Fortunately, various procedures have been developed to handle such situations with ease.
You also see the nitty-gritty of multiplying factor by factor and term by term (the long way) compared to the neat little trick called FOIL. With this information, you’ll be able to efficiently and correctly multiply expressions in many different types of situations. You’ll also be able to expand expressions using multiplication so that you can later go backward and factor expressions back into their original multiplication forms.
Distributing One Factor over Many
When you distribute Halloween candy, you give one piece to each costumed person. When you distribute some factor over several terms, you take each factor (multiplier) on the outside of a grouping symbol and multiply it by each term (all separated by + or –) inside the grouping symbol.
The distributive rule is
a(b + c) = ab + ac and a(b – c) = ab – ac
In other words, if you multiply a value times a sum or difference within a grouping symbol such as 2(x + 3), you multiply every term inside by the factor outside. In this case, multiply the x and 3 each by 2 to get 2x + 6.
Q. Distributing: 6x(3x2 + 5x – 2) =
A. 18x3 + 30x2 – 12x. Now for the details: 6x(3x2 + 5x – 2) = 6x(3x2) + 6x(5x) – 6x(2) = 18x3 + 30x2 – 12x
Q. Distributing: 3abc(2a2bc + 5abc2 – 2abd) =
A. 6a3b2c2 + 15a2b2c3 – 6a2b2cd. The details:
1. Distribute x(8x3 – 3x2 + 2x – 5).
Solve It
2. Distribute x2y(2xy2 + 3xyz + y2z3).
Solve It
3. Distribute –4y(3y4 – 2y2 + 5y – 5).
Solve It
4. Distribute .
Solve It
Curses, FOILed Again — Or Not
A common process found in algebra is that of multiplying two binomials together. A binomial is an expression with two terms such as x + 7. One possible way to multiply the binomials together is to distribute the two terms in the first binomial over the two terms in the second.
But some math whiz came up with a great acronym, FOIL, which translates to F for First, O for Outer, I for Inner, and L for Last. This acronym helps you save time and makes multiplying binomials easier. These letters refer to the terms’ position in the product of two binomials.
When you FOIL the product (a + b)(c + d)
The product of the First terms is ac.
The product of the Outer terms is ad.
The product of the Inner terms is bc.
The product of the Last terms is bd.
The result is then ac + ad + bc + bd. Usually ad and bc are like terms and can be combined.
Q. Use FOIL to multiply (x – 8)(x – 9).
A. x2 – 17x + 72. Using FOIL, multiply the First, the Outer, the Inner, and the Last and then combine the like terms: (x – 8)(x – 9)
= x2 + x(– 9) + (– 8)x + 72
= x2 – 9x – 8x + 72= x2 – 17x + 72
Q. Use FOIL to multiply (2y2 + 3)(y2 – 4).
A. 2y4 – 5y2 – 12. Using FOIL, multiply the First, the Outer, the Inner, and the Last and then combine the like terms: (2y2 + 3)(y2 – 4)
5. Use FOIL to multiply (2x + 1)(3x – 2).
Solve It
6. Use FOIL to multiply (x – 7)(3x + 5).
Solve It
7. Use FOIL to multiply (x2 – 2)(x2 – 4).
Solve It
8. Use FOIL to multiply (3x + 4y)(4x – 3y).
Solve It
Squaring Binomials
You can always use FOIL or the distributive law to square a binomial, but there’s a helpful pattern that makes the work quicker and easier. You square the first and last terms, and then you put twice the product of the terms between the two squares.
The squares of binomials are
(a + b)2 = a2 + 2ab + b2 and (a – b)2 = a2 – 2ab + b2
Q. (x + 5)2 =
A. x2 + 10x +25. The 10x is twice the product of the x and 5: (x + 5)2 = (x)2 + 2(x)(5) + (5)2 = x2 + 10x +25
Q. (3y – 7)2 =
A. 9y2 – 42y + 49. The 42y is twice the product of 3y and 7. And because the 7 is negative, the Inner and O
uter products are, too: (3y – 7)2 = (3y)2 – 2(3y)(7) + (7)2 = 9y2 – 42y + 49
9. (x + 3)2 =
Solve It
10. (2y – 1)2 =
Solve It
11. (3a – 2b)2 =
Solve It
12. (5xy + z)2 =
Solve It
Multiplying the Sum and Difference of the Same Two Terms
When you multiply two binomials together, you can always just FOIL them. You save yourself some work, though, if you recognize when the terms in the two binomials are the same — except for the sign between them. If they’re the sum and difference of the same two numbers, then their product is just the difference between the squares of the two terms.
The product of (a + b)(a – b) is a2 – b2.
This special product occurs because applying the FOIL method results in two opposite terms that cancel one another out: (a + b)(a – b) = a2 – ab + ab – b2 = a2 – b2.
Q. (x + 5)(x – 5) =
A. x2 – 25
Q. (3ab2 – 4)(3ab2 + 4) =
A. 9a2b4 – 16
13. (x + 3)(x – 3) =
Solve It
14. (2x – 7)(2x +7) =
Solve It
15. (a3 – 3)(a3 + 3) =
Solve It
16. (2x2h + 9)(2x2h – 9) =
Solve It
Cubing Binomials
To cube something in algebra is to multiply it by itself and then multiply the result by itself again. When cubing a binomial, you have a couple of options. With the first option, you square the binomial and then multiply the original binomial times the square. This process involves distributing and then combining the like terms. Not a bad idea, but I have a better one.
When two binomials are cubed, two patterns occur.
In the first pattern, the coefficients (numbers in front of and multiplying each term) in the answer start out as 1-3-3-1. The first coefficient is 1, the second is 3, the third is 3, and the last is 1.
The other pattern is that the powers on the variables decrease and increase by ones. The powers of the first term in the binomial go down by one with each step, and the powers of the second term go up by one each time.
To cube a binomial, follow this rule:
The cube of (a + b), (a + b)3, is a3 + 3a2b + 3ab2 + b3.
When one or more of the variables has a coefficient, then the powers of the coefficient get incorporated into the pattern, and the 1-3-3-1 seems to disappear.
Q. (y + 4)3 =
A. y3 + 12y2 + 48y + 64. The answer is built by incorporating the two patterns — the 1-3-3-1 of the coefficients and powers of the two terms. (y + 4)3 = y3 + 3y2(41) + 3y(42) + 43 = y3 + 12y2 + 48y + 64. Notice how the powers of y go down one step each term. Also, the powers of 4, starting with the second term, go up one step each time. The last part of using this method is to simplify each term. The 1-3-3-1 pattern gets lost when you do the simplification, but it’s still part of the answer — just hidden. Note: In binomials with a subtraction in them, the terms in the answer will have alternating signs: +, –, +, –.
Q. (2x – 3)3
A. Starting with the 1-3-3-1 pattern and adding in the multipliers, (2x – 3)3 = (2x)3 + 3(2x)2(–3)1 + 3(2x)( –3)2 + (–3)3 = 8x3 – 36x2 + 54x – 27.
17. (x + 1)3 =
Solve It
18. (y – 2)3 =
Solve It
19. (3z + 1)3 =
Solve It
20. (5 – 2y)3 =
Solve It
Creating the Sum and Difference of Cubes
A lot of what you do in algebra is to take advantage of patterns, rules, and quick tricks. Multiply the sum and difference of two values together, for example, and you get the difference of squares. Another pattern gives you the sum or difference of two cubes. These patterns are actually going to mean a lot more to you when you do the factoring of binomials, but for now, just practice with these patterns.
Here’s the rule: If you multiply a binomial times a particular trinomial — one that has the squares of the two terms in the binomial as well as the opposite of the product of them, such as (y – 5)(y2 + 5y + 25) — you get the sum or difference of two perfect cubes.
(a + b)(a2 – ab + b2) = a3 + b3 and (a – b)(a2 + ab + b2) = a3 – b3
Q. (y + 3)(y2 – 3y + 9) =
A. y3 + 27. If you don’t believe me, multiply it out: Distribute the binomial over the trinomial and combine like terms. You find all the middle terms pairing up with their opposites and becoming 0, leaving just the two cubes.
Q. (5y – 1)(25y2 + 5y + 1) =
A. 125y3 – 1. The number 5 cubed is 125, and –1 cubed is –1. The other terms in the product drop out because of the opposites that appear there.
21. (x – 2)(x2 + 2x + 4) =
Solve It
22. (y + 1)(y2 – y + 1) =
Solve It
23. (2z + 5)(4z2 – 10z + 25) =
Solve It
24. (3x – 2)(9x2 + 6x + 4) =
Solve It
Raising Binomials to Higher Powers
The nice pattern for cubing binomials, 1-3-3-1, gives you a start on what the coefficients of the different terms are — at least what they start out to be before simplifying the terms. Similar patterns also exist for raising binomials to the fourth power, fifth power, and so on. They’re all based on mathematical combinations and are easily pulled out with Pascal’s Triangle. Check out Figure 7-1 to see a small piece of Pascal’s Triangle with the powers of the binomial identified.
Figure 7-1: Pascal’s Triangle can help you find powers of binomials.
Q. Refer to Figure 7-1 and use the coefficients from the row for the fourth power of the binomial to raise (x – 3y) to the fourth power, (x – 3y)4.
A. x4 – 12x3y + 54x2y2 – 108xy3 + 81y4
(x – 3y)4 = (x + (–3y))4
Insert the coefficients 1-4-6-4-1 from the row in Pascal’s Triangle. (Actually, the 1’s are understood.)
= x4 + 4x3(–3y)1 + 6x2(–3y)2 + 4x1(–3y)3 + (–3y)4
= x4 + 4x3(–3y) + 6x2(9y2) + 4x(–27y3) + (81y4)
= x4 – 12x3y + 54x2y2 – 108xy3 + 81y4
Q. Raise (2x – 1) to the sixth power, (2x – 1)6.
A. 64x6 – 192x5 + 240x4 – 160x3 + 60x2 – 12x + 1
Insert the coefficients 1-6-15-20-15-6-1 from the row in Pascal’s Triangle. (Actually, the 1’s are understood.)
Algebra I Workbook For Dummies Page 7