EXPONENTS Three rules

WHEN A NUMBER is repeatedly multiplied by itself, we get the powers of that number (Lesson 1). Problem 1. What number is a) the third power of 2? 2 2 2 = 8 b) the fourth power of 3? = 81

c) the fifth power of 10? = 100,000

d) the first power of 8? = 8

Now, rather than write the third power of 2 as 2 2 2, we write 2 just once -- and place an exponent: 23. 2 is called the base. The exponent indicates the number of times to repeat the base as a factor. Problem 2. What does each symbol mean? a) x5 = xxxxx b) 53 = 5 5 5

c) (5a)3 = 5a 5a 5a

d) 5a3 = 5aaa

In part c), the parentheses indicate that 5a is the base. In part d), only a is the base. The exponent does not apply to 5. Problem 3. 34 = 81. a) Which number is called the base? 3 b) Which number is the power? 81 is the power of 3. c) Which number is the exponent? 4. It indicates the power. Problem 4. Write out the meaning of these symbols. a) aa3 = aa aaa b) (ab)3 = ab ab ab c) (a)3 = a a a

Problem 5. Write out the meaning of these symbols. In each one, what is the base?

a) a4 = aaaa. The base is a. b) a4 = aaaa. The base again is a. This is the negative of a4. c) (a)4 = (a)(a)(a)(a). Here, the base is (a). Problem 6. Evaluate. a) 24 = 16. b) 24 = 16. This is the negative of 24. The base is 2. See Problem 5b) above. c) (2)4 = +16, according to the Rule of Signs (Lesson 4). The parentheses indicate that the base is2. See Problem 5c). Example 1. Negative base. (2)3 = (2)(2)(2) = 8, again according to the Rule of Signs. Whereas, (2)4 = +16. When the base is negative, and the exponent is odd, then the product is negative. But when the base is negative, and the exponent is even, then the product is positive. Problem 7. Evaluate. a) (1) = 1 e) (1)100 = 1 b) (1)3 = 1 f) (1)253 = 1 c) (1)4 = 1 g) (2)4 = 16 d) (1)5 = 1 h) (2)5 = 32

Problem 8. Rewrite using exponents. a) xxxxxx = x6 b) xxyyyy = xy4 c) xyxxyx = x4y d) xyxyxy = x3y3 Problem 9. Rewrite using exponents. a) (x + 1)(x + 1) = (x + 1) b) (x 1)(x 1)(x 1) = (x 1)3

c) (x + 1)(x 1)(x + 1)(x 1) = (x + 1)(x 1)

d) (x + y)(x + y) = (x + y)3

Three rules

Rule 1. Same Base aman = am + n "To multiply powers of the same base, add the exponents." For example, aa3 = a5. Why do we add the exponents? Because of what the symbols mean. Problem 4a. Example 2. Multiply 3x 4x5 2x Solution. The problem means (Lesson 5): Multiply the numbers, then combine the powers of x : 3x 4x5 2x = 24x8 Two factors of x -- x -- times five factors of x -- x5 -- times one factor of x, produce a total of 2 + 5 + 1 = 8 factors of x : x8. Problem 10. Multiply. Apply the rule Same Base. a) 5x 6x4 = 30x6 d) 2x 3x 4x = 24x3 b) 7x3 8x6 = 56x9 c) x 5x4 = 5x5

e) x3 3x 5x = 15x6 f) x5 6x8y = 6x13y

g) 4x y 5x y3 = 20x3y4 h) 2xy 9x3y5 = 18x4y6 ab3a3b4 = a5b7 j) a2bc3bac = a3b3c4 l) apbqab = ap + 1bq + 1

i)

k) xmynxpyq = xm + pyn + q

Example 3. Compare the following: a) x x5 Solution. a) x x5 = x6 b) 2 25 = 26 Part b) has the same form as part a). It is part a) with x = 2. One factor of 2 multiplies five factors of 2 producing six factors of 2. Problem 11. Apply the rule Same Base. a) xx7 = x8 d) 10 105 = 106 b) 3 37 = 38 e) 3x 36x6 = 37x7 c) 2 24 25 = 210 2 2 = 4 is not an issue. b) 2 25

Problem 12. Apply the rule Same Base. a) xnx = xn + 2 e) x xn + 2 = xn + 3 b) xnx = xn + 1 f) xnxm = xn + m c) xnxn = x2n d) xnx1 n = x

g) x2nx2 n = xn + 2

Rule 2: Power of a Product of Factors (ab)n = anbn "Raise each factor to that same power." For example, (ab)3 = a3b3. Why may we do that? Again, according to what the symbols mean: (ab)3 = ab ab ab = aaabbb = a3b3. The order of the factors does not matter: ab ab ab = aaabbb. Problem 13. Apply the rules of exponents.

a) (xy)4 = x4y4

b) (pqr)5 = p5q5r5

c) (2abc)3 = 23a3b3c3

d) x3yz4(xyz)5 = x3yz4 x5y5z5 Rule 2,

= x8y7z9 Rule 1.

Rule 3: Power of a Power (am)n = amn "To take a power of a power, multiply the exponents." For example, (a)3 = a2 3 = a6. Why do we do that? Again, because of what the symbols mean: (a)3 = aaa = a3 2 = a6 Problem 14. Apply the rules of exponents. a) (x)5 = x10 b) (a4)8 = a32 c) (107)9 = 1063

Example 4. Apply the rules of exponents: (2x3y4)5 Solution. Within the parentheses there are three factors: 2, x3, and y4. According to Rule 2, we must take the fifth power of each one. But to take a power of a power, we multiply the exponents. Therefore, (2x3y4)5 = 25x15y20 Problem 15. Apply the rules of exponents. a) (10a3)4 = 10,000a12 d) (xy3z5) = xy6z10 f) (2a4bc8)6 = 64a24b6c48 b) (3x6) = 9x12 c) (2ab3)5 = 32a10b15

e) (5xy4)3 = 125x6y12

Problem 16. Apply the rules of exponents.

a) 2x5y4(2x3y6)5 = 2x5y4 25x15y30 = 26x20y34 b) abc9(ab3c4)8 = abc9 a16b24c32 = a17b25c41 Problem 17. Use the rules of exponents to calculate the following. a) (2 10)4 = 24 104 = 16 10,000 = 160,000 b) (4 10)3 = 43 106 = 64,000,000 c) (9 104) = 81 108 = 8,100,000,000

Example 5. Square x4. Solution. (x4)2 = x8. Thus to square a power, double the exponent. Problem 18. Square the following. a) x5 = x10 b) 8a3b6 = 64a6b12 c) 6x7 = 36x14 d) xn = x2n

Note: In part c): The square of a negative number is positive. (6)(6) = +36. Problem 19. Apply a rule of exponents -- if possible. a) xx5 = x7, Rule 1. b) (x)5 = x10, Rule 3.

c) x + x5 Not possible. The rules of exponents apply only to multiplication. In summary: Add the exponents when the same base appears twice: xx4 = x6. Multiply the exponents when the base appears once -- and in parentheses: (x)5 = x10. Problem 20. Apply the rules of exponents. a) (xn)n = xn n = xn b) (xn) = x2n

Problem 21. Apply a rule of exponents or add like terms -- if possible.

a) 2x + 3x4 Not possible. These are not like terms (Lesson 1). b) 2x 3x4 = 6x6. Rule 1. c) 2x3 + 3x3 = 5x3. Like terms. The exponent does not change. d) x + y Not possible. These are not like terms. e) x + x = 2x. Like terms. f) x x = x4. Rule 1. g) x y3 Not possible. Different bases. h) 2 26 = 27. Rule 1. i) 35 + 35 + 35 = 3 35 (Like terms) = 36.

MULTIPLYING OUT THE DISTRIBUTIVE RULE m(a + b) = ma + mb "To multiply a sum by a number, multiply each term of the sum." That is called the distributive rule. m multiplies a, then it multiplies b. We say that we have "distributed" m to a and b. (Compare Lesson 8 of Arithmetic.) Example 1. 2(x + y + z) = 2x + 2y + 2z.

We have distributed 2 to x, y, and z. We have "multiplied out." Example 2. That is, 3x4 x = 3x6, Rule 1 of exponents (Lesson 13) 3x4(x 5x + 1) = 3x6 15x5 + 3x4.

3x4 5x = 15x5, 3x4 1 = 3x4. Problem 1. 1(a b + c d)

What will be the effect of multiplying by 1? To see the answer, pass your mouse over the colored area. To cover the answer again, click "Refresh" ("Reload"). Do the problem yourself first! Every sign will change. 1(a b + c d) = a + b c + d It follows, then, that we may change all the signs on both sides of an equation. This equation x + a b = c implies this one: x a + b = c. Theoretically, we have multiplied both sides by 1. Problem 2. Multiply out. a) 5(x + 4) = 5x + 20 b) 5(x 4) = 5x 20 d) 2x(3x + 5x 6) = 6x3 + 10x 12x

c) x(x + 1) = x + x

e) 3x2(4x3 3x + 5x 8) = 12x5 9x4 + 15x3 24x2 f) 5x4(x3 4x + 2x 6) = 5x7 + 20x6 10x5 + 30x4 g) 2xy(x 3xy + y) = 2x3y 6xy + 2xy3 h) 4xy(x3y 6xy 2x + 3y + 1) = 4x4y3 + 24xy4 + 8xy 12xy3 4xy

Problem 3. Multiply out and simplify, that is, add the like terms. a) 2(4x + 5y) + 3(5x y) = 8x + 10y + 15x 3y

= 23x + 7y b) 4(2x 1) 5(x 2) = 8x 4 5x + 10

= 3x + 6 c) 3x(3x 2y) 2y(x y) = 9x 6xy 2yx + 2y

= 9x 8xy + 2y d) x(x 10x + 25) 5(x 10x + 25)

= x3 10x + 25x 5x + 50x 125

= x3 15x + 75x 125 e) a(a 2ab + b) b(a 2ab + b)

= a3 2ab + ab ba + 2ab b3

= a3 3ab + 3ab b3

A sum by a sum (a + b + c)(x + y + z) First distribute a to x, y, and z.

Then distribute b. Then distribute c. (a + b + c)(x + y + z) = ax + ay + az + bx + by + bz + cx + cy + cz Problem 4. Multiply (p q)(x y + z). Observe the Rule of Signs (Lesson 4). (p q)(x y + z) = px py + pz qx + qy qz Example 3. Multiply out (x 2)(x + 3). Simplify by adding the like terms. Solution. First distribute x, then distribute 2: (x 2)(x + 3) = x x + x 3 2 x 2 3 = x + 3x 2x 6 = x + x 6 The student should not have to write the first line, but should be able to write the second line -x + 3x 2x 6 -- immediately. Problem 5. Multiply out. Always simplify by adding the like terms.

a) (x + 5)(x + 2) = x + 2x + 5x + 10

= x + 7x + 10 b) (x + 5)(x 2) = x 2x + 5x 10

= x + 3x 10 c) (x 5)(x 2) = x 2x 5x + 10

= x 7x + 10 d) (2x 1)(x + 4) = 2x + 8x x 4

= 2x + 7x 4 e) (3x + 2)(4x 5) = 12x 15x + 8x 10

= 12x 7x 10 f) (5x 1) = (5x 1) (5x 1)

= 25x 5x 5x + 1

= 25x 10x + 1 g) (6x + 1)(6x 1) = 36x 6x + 6x 1

= 36x 1 Example 4. (x 4)(x + 3x 10) = x3 + 3x 10x

4x 12x + 40

= x3 x 22x + 40.

Notice: Upon distributing 4, we have anticipated the like terms by aligning them. However, that is not strictly necessary. Problem 6. Multiply out. a) (x + 2)(x + 4x 5) = x3 + 4x 5x

+ 2x + 8x 10

= x3 + 6x + 3x 10 Note: The effect of multiplying by x is simply to increase each exponent by 1.

b) (x 3)(x 6x + 9) = x3 6x + 9x

3x + 18x 27

= x3 9x + 27x 27 c) (3x 4)(x 7x 2) = 3x3 21x 6x

4x + 28x + 8

= 3x3 25x + 22x + 8 d) (x 1)(x3 + x + x + 1) = x4 + x3 + x + x

x3 x x 1

= x4 1 Note: Multiplication by 1 simply changes the signs. TO FACTOR A NUMBER or an expression, means to write it as a product of factors. Example 1. Factor 30. Solution. 30 = 2 15 = 2 3 5 If we begin 30 = 5 6, we still obtain -- apart from the order -- 30 = 5 2 3. Problem 1. Factor 50. To see the answer, pass your mouse over the colored area. To cover the answer again, click "Refresh" ("Reload"). Do the problem yourself first! 50 = 2 25 = 2 5 5 Factoring, then, is the reverse of multiplying. When we multiply, we write 2(a + b) = 2a + 2b. But if we switch sides and write 2a + 2b = 2(a + b), then we have factored 2a + 2b as the product 2(a + b). In the sum 2a + 2b, 2 is a common factor of each term. It is a factor of 2a, and it is a factor of 2b. This Lesson is concerned exclusively with recognizing common factors. Problem 2. Factor 3x 3y. 3x 3y = 3(x y) Problem 3. Rewrite each of the following as the product of 2x and another factor.

For example, 10x3 = 2x 5x. Rule 1 of exponents.

a) 8x = 2x 4

b) 6ax = 2x 3a

c) 2x = 2x x

d) 2x3 = 2x x

e) 4x10 = 2x 2x9

f) 6x5 = 2x 3x4

g) 2ax6 = 2x ax5

h) 2x = 2x 1

Example 2. Factor 10a 15b + 5. Solution. 5 is a common factor of each term. Display it on the left of the parentheses: 10a 15b + 5 = 5(2a 3b + 1) If we multiply the right-hand side, we will get the left-hand side. In that way, the student can always check factoring. Also, the sum on the left has three terms. Therefore, the sum in parentheses must also have three terms -- and it should have no common factors. Problem 4. Factor each sum. Pick out the common factor. Check your answer. a) 4x + 6y = 2(2x + 3y) b) 6x 6 = 6(x 1)

c) 8x + 12y 16z = 4(2x + 3y 4z)

d) 12x + 3 = 3(4x + 1)

e) 18x 30 = 6(3x 5)

f) 2x + ax = x(2 + a)

g) x + 4x = x(x + 4) Problem 5. Factor each sum.

h) 8x 4x = 4x(2x 1)

a) 2 + 6 + 10 + 14 + 18 = 2(1 + 3 + 5 + 7 + 9) b) 30 + 45 + 60 + 75 = 15(2 + 3 + 4 + 5)

Again, the number of terms in parentheses must equal the number of terms on the left . And the terms in parentheses should have no common factors.

Polynomials A monomial in x is a single term that looks like this: axn, where n is a whole number. The following are monomials in x: 5x8, 3x, 6.

(We say that the number 6 is a monomial in x, because as we will see in Lesson 21, 6 = 6x0 = 6 1.) A polynomial in x is a sum of monomials in x. 5x4 7x3 + 4x + 3x 2 When we write a polynomial, the style is to begin with the highest exponent and go to the lowest. 4, 3, 2, 1. (For a more complete definition of a polynomial, see Topic 6 of Precalculus.) The degree of a polynomial is the highest exponent. The polynomial above is of the 4th degree. The constant term is the term in which the variable does not appear. In other words, it is the number at the end. In that example, the constant term is 2. (It is called the constant term, because it does not depend on the variable, and therefore even though the value of the variable changes, the value of the constant term does not change.) Problem 6. Describe each polynomial in terms of the variable it is "in," and say its degree. a) x3 2x 3x 4 A polynomial in x of the 3rd degree. b) 3y + 2y + 1 A polynomial in y of the 2nd degree. c) x + 2 A polynomial in x of the 1st degree. d) z5 A polynomial in z of the 5th degree. e) 4w 8 A polynomial in w of the 1st degree.

Factoring polynomials If every term is a power of x, as in this example, x7 + 3x6 + 2x5 + x4 then the lowest power is the highest common factor. x7 +3x6 + 2x5 + x4 = x4(x3 + 3x + 2x + 1). For, lower powers are factors of higher powers . x7 = x4 x3 x6 = x4 x2 x5 = x4 x Rule 1 of exponents. The lowest power, x4 in this example, typically appears on the right. Again, when we write a polynomial, we begin with the highest exponent and go to the lowest. 7, 6, 5, 4. Once more, to say that we have factored the polynomial on the left -x7 +3x6 + 2x5 + x4 = x4(x3 + 3x + 2x + 1) -- means that we will obtain that polynomial if we multiply the factors on the right. The student should confirm that. Problem 7. Factor these polynomials. Pick out the highest common factor.

(How can you check your factoring? By multiplying!) a) x8 + x7 + x6 + x5 = x5(x3 + x + x + 1) b) 5x5 4x4 + 3x3 = x3(5x 4x + 3) c) x3 + x = x(x + 1) d) 6x5 + 2x3 = 2x3(3x + 1) e) 2x3 4x + x = x(2x 4x + 1)

f) 3x6 2x5 + 4x4 6x = x(3x4 2x3 + 4x 6) Problem 8. Factor each polynomial. Pick out the highest common numerical factor and the highest common literal factor. a) 12x + 24x 30 = 6(2x + 4x 5). There is no common literal factor. The sum in parentheses has no common factors. b) 16x5 32x4 + 24x3 = 8x3(2x 4x + 3) c) 36y15 27y10 18y5 = 9y5(4y10 3y5 2) d) 8z 12z + 20 = 4(2z 3z + 5) e) 16x 24x + 40 = 8(2x 3x + 5) f) 20x4 12x3 + 36x 4x = 4x(5x3 3x + 9x 1) g) 18x8 81x6 + 27x4 45x = 9x(2x6 9x4 + 3x 5) h) 12x10 6x3 + 3 = 3(4x10 2x3 + 1) Example 3. Factor xy3z4 + x4yz3. Solution. The highest common factor (HCF) will contain the lowest power of each letter. The HCF is xyz3. With that as the common factor, reconstruct each term: xy3z4 + x4yz3 = xyz3(yz + x) If we multiply the right-hand side, we will obtain the left-hand side. Problem 9. Factor. a) 3abc 4ab = ab(3c 4) b) 2xy 8xyz = 2xy(1 4z) c) xy3 x3y = xy(y x) d) 8ab3 + 12ab = 4ab(2b + 3a) e) a5b5 a8b = a5b(b3 a3) f) x6yz + xy4z3 x3y3z4 = xyz(x4 + y3z xyz)

In a polynomial, the leading term is the term with the highest exponent. Normally, it is the first term on the left. In this polynomial, 3x 6x + 9, the leading term in 3x. Now we like the leading term to be positive. (We will see that when we factor trinomials.) Therefore, if we have the following, 3x + 6x 9, we can make the leading term positive by writing 3x + 6x 9 = 3(x 2x + 3). We can remove a negative factor. Problem 10. Make the leading term positive. a) 3x 6 = 3(x + 2) b) 5x + 5x = 5x(x 1)

c) x5 x3 = x3(x + 1)

d) 2x + 6x 2 = 2(x 3x + 1)

e) 32x3 12x + 8x = 4x(8x + 3x 2) f) 9x5 + 30x4 3x3 = 3x3(3x 10x + 1) Problem 11. In each sum, remove the factor xn by displaying it on the left. For example, xn + xn + 2 = xn(1 + x). a) xn + 2 + xn + 3 = xn(x + x3) b) xn + 2 + xn + 4 = xn(x + x4)

c) xn + 3 xn = xn(x3 1)

d) xn + 1 + xn = xn(x + 1)

Problem 12. In each sum, factor out the lower power of x. For example, xn + 1 + xn + 2 = xn + 1(1 + x),

where xn + 1 is the lower power. On multiplying out, we would add the exponents (Lesson 13) and obtain the left-hand side. a) xn + 4 + xn + 1 = xn + 1(x3 + 1) b) xn + 2 + xn + 3 = xn + 2(1 + x)

c) xn xn 2 = xn 2(x 1)

d) xn 1 xn + 1 = xn 1(1 x2)

Problem 13. The Rule of Signs. By applying the definition of the negative of a number, prove that, if ab is positive, then (a)b is the negative of ab. That is, prove: Unlike signs produce a negative number: (a)b = ab. ab + (a)b = [a + (a)]b = 0 b = 0. Therefore, since a number has one and only one negative, (a)b = ab.

Example 4. x(x + 5) + 3(x + 5). What is the common factor? (x + 5) is the common factor. Therefore, x(x + 5) + 3(x + 5) = (x + 3)(x + 5) This is similar to adding like terms. In the first term, x is the coefficient of (x + 5). In the second term, 3 is its coefficient. We add the coefficients of (x + 5). And we preserve the common factor on the right. Problem 14. Add the common factors. Do not remove parentheses. a) x(x + 1) + 2(x + 1) = (x + 2)(x + 1) b) x(x 2) 3(x 2) = (x 3)(x 2) c) x(x + 1) (x + 1) = (x 1)(x + 1) d) x(x 5) + 4(x 5) = (x + 4)(x 5) Example 5. Factoring by grouping. Factor x3 5x + 3x 15.

Solution. Group the first and second terms -- find their common factor. Do the same with the third and fourth terms. x3 5x + 3x 15 = x(x 5) + 3(x 5) = (x + 3)(x 5). Problem 15. Factor by grouping. a) x3 + x + 3x + 3 = x(x + 1) + 3(x + 1)

= (x + 3)(x + 1) b) 2x3 6x + 5x 15 = 2x(x 3) + 5(x 3)

= (2x + 5)(x 3) c) 3x3 15x 2x + 10 = 3x(x 5) 2(x 5)

= (3x 2)(x 5) d) 12x3 + 2x 18x 3 = 2x(6x + 1) 3(6x + 1)

= (2x 3)(6x + 1) e) x3 + 2x x 2 = x(x + 2) (x + 2)

= (x 1)(x + 2) f) 12x3 6x 2x + 1 = 6x(2x 1) (2x 1)

= (6x 1)(2x 1) Problem 16. Show by factoring the left-hand side: (1 + x) + x(1 + x) = (1 + x)3. (1 + x) + x(1 + x) = (1 + x)(1 + x) (1 + x) is the common factor; = (1 + x)3

Example 6.

Solve for x (Lesson 9): px q = rx + s

1. Transpose the x's to the left and everything else to the right: px rx = s + q 2. Factor: x(p r) = s + q 3. Solve for x: x = Problem 17. Solve for x. ax + bx = c s+q pr

x(a + b) = c

x=

c a+b

Problem 18. Solve for x. ax + b = cx + d ax cx = d b x(a c) = d b db ac

x=

Problem 19. Solve for x. ax a = x ax x = a x(a 1) = a a a1

x=

QUADRATIC TRINOMIALS Products of binomials Vocabulary A binomial is a sum of two terms. a + b. A trinomial is a sum of three terms, while a multinomial is more than three. Quadratic is another name for a polynomial of the 2nd degree. For example, 2x 7x + 5 Problem 1. Which of the following is a quadratic?

To see the answer, pass your mouse over the colored area. To cover the answer again, click "Refresh" ("Reload"). Do the problem yourself first! a) 5x 2x + 4 Yes. 2 is the highest exponent. b) 4x 9 No. c) x3 + x + 1 No.

d) y + 8y + 10 Yes.

e) x + x5 No.

f) z Yes.

Products of binomials (2x + 3)(x + 5) Multiplying binomials come up so often that the student should be able to write the product quickly and easily. It is one of the skills of algebra. Therefore, let us multiply those binomials and see what results. First, we will distribute 2x, then we will distribute 3. (2x + 3)(x + 5) = 2x + 10x + 3x + 15 = 2x + 13x + 15. When we multiply two such binomials, then, what form is produced?

A quadratic trinomial.Now, the first term of the trinomial is no mystery: It is 2x x. And the last term is no problem either -- it is 3 5. And so the only question is: Where does the middle term 13x come from? The middle term is the sum of the like terms: 10x + 3x :

(2x + 3)(x + 5) = 2x + 10x + 3x + 15 = 2x + 13x + 15.

Therefore, if we say that when multiplying binomials, there are four steps,

then which steps produce the like terms? The 2nd plus the 3rd.

Or, as we often call them, the "Outers" plus the "Inners." It is skillful to be able to pick out the like terms quickly. Because in the next Lesson we will want to factor 2x + 13x + 15. Will it be factored as (2x + 5)(x + 3) ? Or as (2x + 3)(x + 5) ? The key lies in choosing the combination that correctly gives the middle term, 13x. In the first possibility, can we make 13x by combining the Outers plus the Inners: 6x with 5x ? No, we cannot. But in the second possibility we can: 10x + 3x = 13x. 2x + 13x + 15 will therefore be correctly factored as (2x + 3)(x + 5). To check that, the student should be skillfull in adding the like terms mentally. That is, to multiply (2x + 3)(x + 5) , look at 2x and x and write 2x.

Next, look at 2x 5 and 3 x together -- "10x + 3x" -- and write 13x. Finally, 3 5 = 15. (2x + 3)(x + 5) = 2x + 13x + 15. With practice, your eye will get used to picking out the Outers plus the Inners. You do algebra with your eyes. Those are the like terms. Example 1. (3x 1)(x + 2). Write only the sum of the like terms. Answer. 5x. That is,

6x x = 5x. Example 2. Multiply (3x 1)(x + 2). Answer. 3x + 5x 2

The first term of the trinomial is 3x x = 3x. The middle term we found to be 5x. And the third term is 1 2 = 2. Problem 3. Write only the sum of the like terms. a) (2x + 1)(x + 3). 7x b) (3x 2)(x + 4). 10x

c) (5x 3)(2x 1). 11x

d) (x + 2)(x + 3). 5x

e) (x + 4)(x 6). 2x

f) (x 5)(x 3). 8x

Problem 4. Write only the trinomial product -- do not write all four terms. a) (3x + 2)(2x + 1) = 6x + 7x + 2 b) (3x + 2)(2x 1) = 6x + x 2

c) (3x 2)(2x + 1) = 6x x 2 c) (3x 2)(2x 1) = 6x 7x + 2 Problem 5. Write only the trinomial product. a) (7x 2)(5x + 4) = 35x + 18x 8 b) (x 1)(3x + 8) = 3x + 5x 8 c) (5x 4)(x 1) = 5x 9x + 4 d) (2x + 3)(2x + 5) = 4x + 16x + 15 e) (6x + 7)(2x 3) = 12x 4x 21 f) (4x 3)(5x 2) = 20x 23x + 6

Example 3. 1 the coefficient of x. (x + 2)(x + 3) When 1 is the coefficient of the x's, the multiplication is especially simple:

When we add the outers plus the inners, the coefficient of x is the sum of the two numbers. The constant term, as always, is their product. Example 4. (x 1)(x + 5) = x + 4x 5. The outers plus the inners give 4x. More simply, the coefficient of x is 1 + 5. The constant term is 1 5. Problem 6. Write only the trinomial product. a) (x + 2)(x + 4) = x + 6x + 8 b) (x + 2)(x 4) = x 2x 8 c) (x 2)(x 4) = x 6x + 8 d) (x + 1)(x 8) = x 7x 8

e) (x 1)(x + 8) = x + 7x 8 f) (x 1)(x 8) = x 9x + 8 g) (x + 4)(x + 5) = x + 9x + 20 h) (x + 2)(x 8) = x 6x 16 i) (x 3)(x 7) = x 10x + 21 Example 5. Multiply 2(x + 3)(x 1) Solution. We can multiply only two factors at a time. First multiply the binomials, then distribute 2: 2(x + 3)(x 1) = 2(x + 2x 3) = 2x + 4x 6. Example 6. (x + 4)(x 5) = (x x 20)

= x + x + 20. Multiply the binomials. Then remove the parentheses. Example 7. (x 4)(x + 5)(x 2) Multiply two of the binomials. Then multiply that product with third. (Lesson 14, Example 4.) (x 4)(x + 5)(x 2) = (x 4)(x + 3x 10) = x3 + 3x 10x 4x 12x + 40 = x3 x 22x + 40. Problem 7. Multiply.

a) 4(x 1)(x + 3) = 4(x + 2x 3) = 4x + 8x 12 b) x(x 2)(3x + 4) = x(3x 2x 8) = 3x3 2x 8x c) (x + 1)(x 2) = (x x 2) = x + x + 2

d) (x + 1)(x + 2)(x + 3) = (x + 1)(x + 5x + 6)

= x3 + 5x + 6x + x + 5x + 6

= x3 + 6x + 11x + 6 e) (x 2)(x + 4)(x 5) = (x 2)(x x 20)

= x3 x 20x 2x + 2x + 40

= x3 3x 18x + 40

FACTORING TRINOMIALS 2nd Level: Positive leading term Quadratics in different arguments

FACTORING IS THE REVERSE of multiplying. Skill in factoring, then, depends upon skill in multiplying: Lesson 16. As for a quadratic trinomial -2x + 9x 5

-- it will be factored as a product of binomials: (? ?)(? ?) Now, how will 2x be produced? There is only one way: 2x x : (2x ?)(x ?) And how will 5 be produced? Again, there is only one way: 1 5. But does the 5 go with 2x or with x ? (2x 5)(x 1) or (2x 1)(x 5) ?

Notice: We have not yet placed any signs! How shall we decide between these two possibilities? It is the combination that will correctly give the middle term, 9x : 2x + 9x 5. Consider the first possibility: (2x 5)(x 1) Is it possible to produce 9x by combining the outers and the inners: 2x 1 with 5x ? No, it is not. Therefore, we must eliminate that possibility and consider the other: (2x 1)(x 5) Can we produce 9x by combining 10x with x ? Yes -- if we choose +5 and 1: (2x 1)(x + 5) (2x 1)(x + 5) = 2x + 9x 5 Skill in factoring depends on skill in multiplying -- particularly in picking out the middle term Problem 1. Place the correct signs to give the middle term. a) 2x + 7x 15 = (2x 3)(x + 5) b) 2x 7x 15 = (2x + 3)(x 5)

c) 2x x 15 = (2x + 5)(x 3) d) 2x 13x + 15 = (2x 3)(x 5) Note: When the constant term is negative, as in parts a), b), c), then the signs in each factor will be different. But when that term is positive, as in part d), the signs will be the same. Usually, however, that happens by itself. Nevertheless, can you correctly factor the following? 2x 5x + 3 = (2x 3)(x 1) Problem 2. Factor these trinomials. a) 3x + 8x + 5 = (3x + 5)(x + 1) b) 3x + 16x + 5 = (3x + 1)(x + 5) c) 2x + 9x + 7 = (2x + 7)(x + 1) d) 2x + 15x + 7 = (2x + 1)(x + 7) e) 5x + 8x + 3 = (5x + 3)(x + 1) f) 5x + 16x + 3 = (5x + 1)(x + 3) Problem 3. Factor these trinomials.

a) 2x 7x + 5 = (2x 5)(x 1) b) 2x 11x + 5 = (2x 1)(x 5) c) 3x + x 10 = (3x 5)(x + 2 ) d) 2x x 3 = (2x 3)(x + 1) e) 5x 13x + 6 = (5x 3)(x 2) f) 5x 17x + 6 = (5x 2)(x 3) g) 2x + 5x 3 = (2x 1)(x + 3) h) 2x 5x 3 = (2x + 1)(x 3) i) 2x + x 3 = (2x + 3)(x 1)

j) 2x 13x + 21 = (2x 7 )(x 3) k) 5x 7x 6 = (5x + 3)(x 2) i) 5x 22x + 21 = (5x 7)(x 3)

Example 1. 1 the coefficient of x. Factor x + 3x 10. Solution. The binomial factors will have this form: (x a)(x b) What are the factors of 10? Let us hope that they are 2 and 5: x + 3x 10 = (x 2)(x 5). We must now choose the signs so that the coefficient of the middle term -- the sum of the outers plus the inners -- will be +3. Choose +5 and 2. x + 3x 10 = (x 2)(x + 5). Note: When 1 is the coefficient of x, the order of the factors does not matter. (x 2)(x + 5) = (x + 5) (x 2). Example 2. Factor x x 12. Solution. We must find factors of 12 whose algebraic sum will be the coefficient of x : 1. Choose 4 and + 3: x x 12 = (x 4 )(x + 3). Problem 4. Factor. Again, the order of the factors does not matter. a) x + 5x + 6 = (x + 2)(x + 3) b) x x 6 = (x 3 )(x + 2) c) x + x 6 = (x + 3 )(x 2) d) x 5x + 6 = (x 3)(x 2 ) e) x + 7x + 6 = (x + 1)(x + 6 )

f) x 7x + 6 = (x 1)(x 6 ) g) x + 5x 6 = (x 1)(x + 6 ) h) x 5x 6 = (x + 1)(x 6 ) Problem 5. Factor. a) x 10x + 9 = (x 1 )(x 9) b) x + x 12 = (x + 4)(x 3) c) x 6x 16 = (x 8)(x + 2) d) x 5x 14 = (x 7)(x + 2) e) x x 2 = (x + 1)(x 2) f) x 12x + 20 = (x 10 )(x 2) g) x 14x + 24 = (x 12 )(x 2) Example 3. Factor completely 6x8 + 30x7 + 36x6. Solution. To factor completely means to first remove any common factors (Lesson 15). 6x8 + 30x7 + 36x6 = 6x6(x + 5x + 6).

Continue by factoring the trinomial:

= 6x6(x + 2)(x + 3). Problem 6. Factor completely. First remove any common factors. a) x3 + 6x + 5x = x(x2 + 6x + 5) = x(x + 5)(x + 1) b) x5 + 4x4 + 3x3 = x3(x2 + 4x + 3) = x3(x + 1)(x + 3) c) x4 + x3 6x = x(x + x 6) = x(x + 3)(x 2) d) 4x 4x 24 = 4(x x 6) = 4(x + 2)(x 3)

e) 2x3 14x 36x = 2x(x2 7x 18) = 2x(x + 2)(x 9) f) 12x10 + 42x9 + 18x8 = 6x8(2x + 7x + 3) = 6x8(2x + 1)(x + 3).

2nd Level Example 4. Factor by making the leading term positive. x + 5x 6 = (x 5x + 6) = (x 2)(x 3). Problem 7. Factor by making the leading term positive. a) x 2x + 3 = (x + 2x 3) = (x + 3)(x 1) b) x + x + 6 = (x x 6) = (x + 2)(x 3) c) 2x 5x + 3 = (2x + 5x 3) = (2x 1)(x + 3) Quadratics in different arguments Here is the form of a quadratic trinomial with argument x : ax + bx + c. The argument is whatever is being squared. x is being squared. x is called the argument. The argument appears in the middle term. a, b, c are called constants. In this quadratic, 3x + 2x 1, the constants are 3, 2, 1. Now here is a quadratic whose argument is x3: 3x6 + 2x3 1. x6 is the square of x3. (Lesson 13: Exponents.) But that quadratic has the same constants -- 3, 2, 1 -- as the one above. In a sense, it is the same quadratic only with a different argument. For it is the constants that distinguish a quadratic. Now, since the quadratic with argument x can be factored as

3x + 2x 1 = (3x 1)(x + 1), then the quadratic with argument x3 is factored in the same way: 3x6 + 2x3 1 = (3x3 1)(x3 + 1). Whenever a quadratic has constants 3, 2, 1, then for any argument, the factoring will be (3 times the argument 1)(argument + 1).

Example 5.

z 3z 10 = (z + 2)(z 5).

x8 3x4 10 = (x4 + 2)(x4 5). The trinomials on the left have the same constants 1, 3, 10 but different arguments. That is the only difference between them. In the first, the argument is z. In the second, the argument is x4. (The square of x4 is x8.) Each quadratic is factored as (argument + 2)(argument 5). Every quadratic with constants 1, 3, 10 will be factored that way. Problem 8. a) Write the form of a quadratic trinomial with argument z. az + bz + c b) Write the form of a quadratic trinomial with argument x4. ax8 + bx4 + c c) Write the form of a quadratic trinomial with argument xn. ax2n + bxn + c

Problem 10. Multiply out each of the following, which have the same constants, but different argument. a) (z + 3)(z 1) = z + 2z 3 c) (y6 + 3)(y6 1) = y12 + 2y6 3 d) (x5 + 3)(x5 1) = x10 + 2x5 3 Problem 11. Factor each quadratic. a) x 6x + 5 = (x 1)(x 5) b) z 6z + 5 = (z 1)(z 5) c) x8 6x4 + 5 = (x4 1)(x4 5) d) x10 6x5 + 5 = (x5 1)(x5 5) e) x6y6 6x3y3 + 5 = (x3y3 1)(x3y3 5) f) sinx 6 sin x + 5 = (sin x 1)(sin x 5). sinx -- "sine squared x" -- means (sin x). Problem 12. Factor each quadratic. a) x4 x 2 = (x 2)(x + 1) b) y6 + 2y3 8 = (y3 + 4)(y3 2) c) z8 + 4z4 + 3 = (z4 + 1)(z4 + 3) d) 2x10 + 5x5 + 3 = (2x5 + 3)(x5 + 1) e) x4y 3xy 10 = (xy + 2)(xy 5) f) cosx 5 cos x + 6 = (cos x 3)(cox x 2) b) (y + 3)(y 1) = y + 2y 3

PERFECT SQUARE TRINOMIALS

The square of a binomial The square numbers The square of a binomial 2nd level (a + b) The square of a trinomial Completing the square

LET US BEGIN by learning about the square numbers. They are the numbers 1 1 2 2 3 3 and so on. The following are the first ten square numbers -- and their roots. Square numbers 1 4 9 16 25 36 49 64 81 100 Square roots 1 2 3 4 5 6 7 8 9 10

1 is the square of 1. 4 is the square of 2. 9 is the square of 3. And so on. The square root of 1 is 1. The square root of 4 is 2. The square root of 9 is 3. And so on. In a multiplication table, the square numbers lie along the diagonal.

The square of a binomial Let us square the binomial (x + 5): (x + 5) = (x + 5)(x + 5) = x + 10x + 25. (See Lesson 16: Quadratic trinomials.) x + 10x + 25 is called a perfect square trinomial. It is the square of a binomial.

The square of a binomial come up so often that the student should be able to write the trinomial product quickly and easily. Therefore, let us see what happens when we square any binomial, a +b: (a + b) = (a + b)(a + b) = a + 2ab + b The square of any binomial produces the following three terms: 1. The square of the first term of the binomial: a 2. Twice the product of the two terms: 2ab 3. The square of the second term: b The square of every binomial -- every perfect square trinomial -- has that form: a + 2ab + b. To recognize that is to know the "multiplication table" of algebra. (See Lesson 8 of Arithmetic: How to square a number mentally, particularly the square of 24, which is the "binomial" 20 + 4.) Example 1. Square the binomial (x + 6). Solution. (x + 6) = x + 12x + 36

x is the square of x. 12x is twice the product of x 6. (x 6 = 6x. Twice that is 12x.) 36 is the square of 6. Example 2. Square the binomial (3x 4). Solution. (3x 4) = 9x 24x + 16

9x is the square of 3x. 24x is twice the product of 3x 4. (3x 4 = 12x. Twice that is 24x.) 16 is the square of 4. Note: If the binomial has a minus sign, then the minus sign appears only in the middle term of the trinomial. Therefore, using the double sign ("plus or minus"), we can state the rule as follows: (a b) = a 2ab + b

This means: If the binomial is a + b, then the middle term will be +2ab; but if the binomial is a b, then the middle term will be 2ab Example 3. (5x3 1) = 25x6 10x3 + 1 25x6 is the square of 5x3. (Lesson 13: Exponents.) 10x3 is twice the product of 5x3 1. (5x3 1 = 5x3. Twice that is 10x3.) 1 is the square of 1. Example 4. Is this a perfect square trinomial: x + 14x + 49 ? Answer. Yes. It is the square of (x + 7). x is the square of x. 49 is the square of 7. And 14x is twice the product of x 7. In other words, x + 14x + 49 could be factored as x + 14x + 49 = (x + 7) Note: If the coefficient of x had been any number but 14, this would not have been a perfect square trinomial. Example 5 Is this a perfect square trinomial: x + 50x + 100 ? Answer. No, it is not. Although x is the square of x, and 100 is the square of 10, 50x is not twice the product of x 10. (Twice their product is 20x.) Example 6 Is this a perfect square trinomial: x8 16x4 + 64 ? Answer. Yes. It is the perfect square of x4 8. Problem 1. Which numbers are the square numbers? To see the answer, pass your mouse over the colored area. To cover the answer again, click "Refresh" ("Reload"). Do the problem yourself first! 1, 4, 9, 16, 25, 36, 49, 64, etc. These are the numbers 1, 2, 3, and so on. Problem 2. a) State in words the formula for squaring a binomial.

The square of the first term. Twice the product of the two terms. The square of the second term. b) Write only the trinomial product: (x + 8) = x + 16x + 64 c) Write only the trinomial product: (r + s) = r + 2rs + s Problem 3. Write only the trinomial product. a) (x + 1) = x + 2x + 1 b) (x 1) = x 2x + 1

c) (x + 2) = x + 4x + 4

d) (x 3) = x 6x + 9

e) (x + 4) = x + 8x + 16

f) (x 5) = x 10x + 25

g) (x + 6) = x + 12x + 36

h) (x y) = x 2xy + y

Problem 4. Write only the trinomial product. a) (2x + 1) = 4x + 4x + 1 b) (3x 2) = 9x 12x + 4

c) (4x + 3) = 16x + 24x + 9

d) (5x 2) = 25x 20x + 4

e) (x3 + 1) = x6 + 2x3 + 1

f) (x4 3) = x8 6x4 + 9

g) (xn + 1) = x2n + 2xn + 1 Problem 5. Factor: p + 2pq + q.

h) (xn 4) = x2n 8xn + 16

p + 2pq + q = (p + q) The left-hand side is a perfect square trinomial. Problem 6. Factor as a perfect square trinomial -- if possible. a) x 4x + 4 = (x 2) b) x + 6x + 9 = (x + 3)

c) x 18x + 36 Not possible.

d) x 12x + 36 = (x 6)

e) x 3x + 9 Not possible.

f) x + 10x + 25 = (x + 5)

Problem 7. Factor as a perfect square trinomial, if possible. a) 25x + 30x + 9 = (5x + 3) b) 4x 28x + 49 = (2x 7)

c) 25x 10x + 4 Not possible. d) 25x 20x + 4 = (5x 2)

e) 1 16y + 64y = (1 8y )

f) 16m 40mn+ 25n = (4m 5n)

g) x4 + 2xy + y4 = (x + y)

h) 4x6 10x3y4 + 25y8 Not possible.

Problem 8. Without multiplying out a) explain why (1 x) = (x 1). Because (1 x) is the negative of (x 1). And (a) = a for any quantity a. b) explain why (1 x) = (x 1). (a) = a for any quantity a. * The following problems show how we can go from what we know to what we do not know. Problem 9. Use your knowledge of (a + b) to multiply out (a + b)3.

[Hint: (a + b)3 = (a + b)(a + b)] To see the answer, pass your mouse over the colored area. To cover the answer again, click "Refresh" ("Reload"). Do the problem yourself first!

(a + b)(a + b) = (a + b)(a + 2ab+ b) = a3 + 2ab + ab + ab + 2ab + b3 (a + b)3 = a3 + 3ab + 3ab + b3 Problem 10. Multiply out (x + 2)3. (x + 2)(x + 2) = (x + 2)(x + 4x+ 4) = x3 + 4x + 4x + 2x + 8x + 8 (x + 2)3 = x3 + 6x + 12x + 8 Problem 11. Multiply out (x 1)3. (x 1)(x 1) = (x 1)(x 2x + 1) = x3 2x + x x + 2x 1 (x 1)3 = x3 3x + 3x 1 Problem 12. The square of a trinomial. Use your knowledge of (a + b) to multiply out (a + b + c). [Hint: Treat as a binomial with as the first term.]

Show that it will equal the sum of the squares of each term, plus twice the product of all combinations of the terms. ( + c) = (a + b) + 2(a + b)c + c

= a + 2ab + b + 2ac + 2bc + c = a + b + c + 2ab + 2ac + 2bc Problem 13. Can you generalize the result of the previous problem? Can you immediately write down the square of (a + b + c + d)? (a + b + c + d) = a + b + c + d + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd

Completing the square x + 8x + _?_ = (x + _?_) When the coefficient of x is 1, as in this case, then to complete a perfect square trinomial, we must add a square number. What square number must we add? We must add the square of half of coefficient of x. The trinomial will then be the square of x plus half that coefficient. x + 8x + 16 = (x + 4) We add the square of half the coefficient of x -- which in this case is 4 -- because when we multiply (x + 4), the coefficient of x will be twice that number.

Problem 14.

a) How do we indicate half of any number b? b) How do we indicate half of any fraction ? (Skill in Arithmetic, Lesson 26.)

b 2

p p q 2q

Example 7. Complete the square: x 7x + ? = (x ?) 7 Solution. We will add the square of half of 7, which we write as . 2

x 7x +

7 49 = (x ) 4 2

And since the middle term of the trinomial has a minus sign, then the binomial also must have a minus sign. Problem 15. Complete the square. The trinomial is the square of what binomial? a) x + 4x + ? x + 4x + 4 = (x + 2) b) x 2x + ? x 2x + 1 = (x 1) c) x + 6x + ? x + 6x + 9 = (x + 3) d) x 10x + ? x 10x + 25 = (x 5) e) x + 20x + ? x + 20x + 100 = (x + 10)

f) x + 5x + ? x + 5x +

25 5 = (x + ) 4 2 81 9 = (x ) 4 2

g) x 9x + ? x 9x +

h) x + bx + ? x + bx +

b b = (x + ) 4 2b b = (x + ) 4a 2a

i) x + x + ?

b a

x + x +

b a

THE DIFFERENCE OF TWO SQUARES Summary of Multiplying/Factoring 2nd level: The form (a + b)(a b) Factoring by grouping The sum or difference of any powers

WHEN THE SUM of two numbers multiplies their difference --

(a + b)(a b) -- then the product is the difference of their squares: (a + b)(a b) = a bFor, the like terms will cancel. (Lesson 16.) Symmetrically, the difference of two squares can be factored: x 25 = (x + 5)(x 5) x is the square of x. 25 is the square of 5. Example 1. Multiply (x3 + 2)(x3 2). Solution. Recognize the form: (a + b)(a b). The product will be the difference of two squares: (x3 + 2)(x3 2) = x6 4. x6 is the square of x3. 4 is the square of 2. When confronted with the form (a + b)(a b), the student should not do the FOIL method. The student should recognize immediately that the product will be a b. Also, the order of factors never matters: (a + b)(a b) = (a b)(a + b) = a b. Problem 1. Write only final product.. a) (x + 9)(x 9) = x 81 b) (y + z)(y z) = y z

c) (6x 1)(6x + 1) = 36x 1

d) (3y + 7)(3y 7) = 9y 49

e) (x3 8)(x3 + 8) = x6 64

f) (xy + 10)(xy 10) = xy 100

g) (xy z3)(xy + z3) = xy4 z6 Problem 2. Factor. a) x 100 = (x + 10)(x 10)

h) (xn + ym)(xn ym) = x2n y2m

b) y 1 = (y + 1)(y 1)

c) 1 4z = (1 + 2z)(1 2z) d) 25m 9n = (5m + 3n)(5m 3n)

e) x6 36 = (x3 + 6)(x3 6)

f) y4 144 = (y + 12)(y 12)

g) x8 y10 = (x4 + y5) (x4 y5) Problem 3. Factor completely. a) x4 y4 = (x + y)(x y)

h) x2n 1 = (xn + 1)(xn 1)

= (x + y)(x + y)(x y) b) 1 z8 = (1 + z4)(1 z4)

= (1 + z4)(1 + z)(1 z)

= (1 + z4)(1 + z)(1 + z)(1 z) Problem 4. Completely factor each of the following. First remove a common factor. Then factor the difference of two squares. a) xy xz = x(y z) = x(y + z)(y z)

b) 8x 72 = 8(x 9) = 8(x + 3)(x 3) c) 64z z3 = z(64 z) = z(8 + z)(8 z) d) rs3 r3s = rs(s r) = rs(s + r)(s r) e) 32mn 50n3 = 2n(16m 25n) = 2n(4m + 5n)(4m 5n) f) 5x4y5 5y5 = 5y5(x4 1) = 5y5(x + 1)(x + 1)(x 1) The Difference of Two Squares completes our study of products of binomials. Those products come up so often that the student should be able to recognize and apply each form. Summary of Multiplying/Factoring In summary, here are the four forms of Multiplying/Factoring that characterize algebra.

1. Common Factor

2(a + b) = 2a + 2b

2. Quadratic Trinomial

(x + 2)(x + 3) = x + 5x + 6

3. Perfect Square Trinomial

(x 5) = x 10x + 25

4. The Difference of Two Squares

(x + 5)(x 5) = x 25

Problem 5. Distinguish each form, and write only the final product. a) (x 3) = x 6x + 9. Perfect square trinomial.

b) (x + 3)(x 3) = x 9. The difference of two squares. c) (x 3)(x + 5) = x + 2x 15. Quadratic trinomial. d) (2x 5)(2x + 5) = 4x 25. The difference of two squares. e) (2x 5) = 4x 20x + 25. Perfect square trinomial. f) (2x 5)(2x + 1) = 4x 8x 5. Quadratic trinomial.

Problem 6. Factor. (What form is it? Is there a common factor? Is it the difference of two squares? . . . ) a) 6x 18 = 6(x 3). Common factor. b) x6 + x5 + x4 + x3 = x3(x3 + x + x + 1). Common factor. c) x 36 = (x + 6)(x 6). The difference of two squares. d) x 12x + 36 = (x 6). Perfect square trinomial. e) x 6x + 5 = (x 5)(x 1). Quadratic trinomial. f) x x 12 = (x 4)(x + 3) g) 64x 1 = (8x + 1)(8x 1) h) 5x 7x 6 = (5x + 3)(x 2) i) 4x5 + 20x4 + 24x3 = 4x3(x + 5x + 6) = 4x3(x + 3)(x + 2) The difference of two squares: 2nd Level Back to Section 1 The form (a + b)(a b) Factoring by grouping The sum or difference of any powers

Example 2. The form (a + b)(a b). The following has the form (a + b)(a b):

(x + y + 8)(x + y 8) x + y is the first term; 8 is the second. Therefore, it will produce the difference of two squares: (x + y + 8)(x + y 8) = (x + y) 64 = x + 2xy + y 64 -- upon applying the rule for the square of a binomial. Problem7. Each of these will produce the difference of two squares. Multiply out.

a) (p + 3q + 2)(p + 3q 2) = (p + 3q) 4

= p + 6pq + 9q 4 b) (x y 1)(x y + 1) = (x y) 1

= x 2xy + y 1 Example 3. Factoring by grouping. x3 + 2x 25 x 50. Let us factor this by grouping (Lesson 15) -- and then recognize the difference of two squares: x3 + 2x 25x 50 = x(x + 2) 25(x + 2) = (x 25)(x + 2) = (x + 5)(x 5)(x + 2) Problem 8. Factor by grouping. a) x3 + 3x 4x 12 = x(x + 3) 4(x + 3)

= (x 4)(x + 3)

= (x + 2)(x 2)(x + 3) b) x3 4x 9x + 36 = x(x 4) 9(x 4)

= (x 9)(x 4)

= (x + 3)(x 3)(x 4) c) 2x3 3x 50x + 75 = x(2x 3) 25(2x 3)

= (x 25)(2x 3)

= (x + 5)(x 5)(2x 3) d) 3x3 + x 3x 1 = x(3x + 1) (3x + 1)

= (x 1)(3x + 1)

= (x + 1)(x 1)(3x + 1) The sum or difference of any powers a5 + b5 = (a + b)(a4 a3b + ab ab3 + b4) a5 b5 = (a b)(a4 + a3b + ab + ab3 + b4) Look at the form of these. A factor of a5 + b5 is (a + b), while a factor of a5 b5 is (a b).

To produce a5, the second factor begins a4. The exponent of a then decreases as the exponent of b increases -- but the sum of the exponents in each term is 4. (We say that the degree of each term is 4.) In the second factor of a5 + b5, the signs alternate. (If they did not, then on multiplying, nothing would cancel to produce only two terms.) In the second factor of a5 b5, all the signs are + . (That insures the canceling.) By multiplying out, the student can verify that these are the factors of the sum and difference of 5th powers. (For a proof based on the Factor Theorem, see Topic 13 of Precalculus.) In particular, xn 1 can always be factored for any positive integer n, because 1 = 1n, and all powers of 1 are 1.

Problem 9. Factor the following.

a) x5 1 = (x 1)(x4 + x3 + x + x + 1) b) x5 + 1 = (x + 1)(x4 x3 + x x + 1) Problem 10. Factor. a) a3 + b3 = (a + b)(a ab + b) b) a3 b3 = (a b)(a + ab + b) In practice, it is these, the sum and difference of 3rd powers, that tend to come up. Problem 11. Factor. a) x3 + 8 = x3 + 23

= (x + 2)(x x 2 + 2)

= (x + 2)(x 2x + 4) b) x3 1 = x3 13

= (x 1)(x + x 1 + 1)

= (x 1)(x + x + 1) The difference of even powers So much for the sum and difference of odd powers. As for the sum and difference of even powers, only their difference can be factored. (If you doubt that, then try to factor a2 + b2 or a4 + b4. Verify your attempt by multiplying out.) If n is even, then we can always recognize the difference of two squares: a4 b4 = (a2 + b2)(a2 b2). But also when n is even, an bn can be factored either with (a b) as a factor or (a + b). a4 b4 = (a b)(a3 + a2b + ab2 + b3) a4 b4 = (a + b)(a3 a2b + ab2 b3) [If n is odd, then an bn can be factored only with the factor (a b).] For, with n even and the factor (a + b), the right-hand factor will have an even number of terms. And since those terms alternate in sign, the final term will be bn1. Therefore, bn will be correctly produced upon multiplication with +b. But when n is odd, the right-hand factor will have an odd number of terms. Therefore the final term would be +bn1. Hence it will be impossible to produce bn upon multiplication with +b.] Problem 12. Factor x4 81 with (x + 3) as a factor. x4 81 = x4 34

= (x + 3)(x3 x2 3 + x 32 33)

= (x + 3)(x3 3x2 + 9x 27) ALGEBRAIC FRACTIONS The principle of equivalent fractions Reducing to lowest terms 2nd Level

FRACTIONS IN ALGEBRA are often called rational expressions. (See Topic 18 of Precalculus.) We begin with the principle of equivalent fractions, which appears as follows:

x ax = y ay "We may multiply both the numerator and denominator by the same factor." x ax and are y ay

Both x and y have been multiplied by the factor a.

called equivalent fractions. This principle is the single most important fact about fractions. Problem 1. Write the missing numerator. To see the answer, pass your mouse over the colored area. To cover the answer again, click "Refresh" ("Reload"). Do the problem yourself first! 6 18 = n 3n The denominator has been multiplied by 3; therefore the numerator will also be multiplied by 3. Problem 2. Write the missing numerator.

4 4x = x x The denominator has been multiplied by x; therefore the numerator will also be multiplied by x. Problem 3. Write the missing numerator. m 8xm = 3 x 8x The denominator has been multiplied by 8x; therefore the numerator will also be multiplied by 8x. The student will see that the original denominator on the left will be a factor of the new denominator on the right. It must be a factor, because to produce that new denominator it was multiplied Problem 4. Write the missing numerator. ("The denominator has been multiplied by _____. Therefore the numerator will also be multiplied by ____.") a 5a = b 5b 3 6 = x 2x 5 5y = y y

a)

b)

c)

d)

8 8y = x xy

e)

a 2xa = x 2x3

f)

b bxy = y xy

g)

p prs = q qrs ? b

h)

2 2ac = b abc

i)

4 4(x + 1) = x x(x + 1)

Example 1.

a=

Solution. To explain the solution, we will write a as a ab = 1 b Since 1 has been multiplied by b, then so will a.

a . 1

The numerator ab however is simply the product of a times b. It is a kind of cross-multiplying, and the student should not have to write the denominator 1. a= You do algebra with your eyes. Problem 5. Write the missing numerator. 3x 3 x x 2ab ab 2x + 2 x+1 x x x 1 x1 ab b

a) x = d) 1 =

b) 2 = e) 2 =

c) x =

f) x + 1 =

Part f) is The Difference of Two Squares. There will be more problems of this type at the 2nd Level.

Reducing to lowest terms The numerator and denominator of a fraction are called its terms. Since we may multiply both terms, then, symmetrically, we may divide both terms.

ax x = ay y "We may divide both the numerator and denominator by a common factor." When we do that, we say that we have reduced the fraction to its lowest terms. 5x . 5y

Example 2. Reduce Answer. 5x x = . 5y y

5 is a common factor of the numerator and denominator. Therefore we may divide each of them by 5. We say that we have "canceled" the 5's.

Example 3. Reduce

5+x . 5+y

Answer. This can not be reduced. 5 is not a factor of either the numerator or the denominator. It is a term. You cannot cancel terms. You can divide both the numerator and denominator only when they have a common factor. * The word term does double duty in algebra. We speak of the terms of a sum and also the terms of a fraction, which are the numerator and denominator. A fraction is in its lowest terms when the numerator and denominator have no common factors. Problem 6. Reduce to lowest terms. 3a a = 3b b = 8xy 2y = 12x 3 56y 8 = 77xy 11x

a)

b)

c)

d) 2x + 3 4x + 9

No canceling! The numerator and denominator are made up of terms. You cannot cancel terms. And those terms have no common factors.

See Example 7 below. Example 4. Reduce Answer. 4x = 4. x 4x . x

We may think of this as 4x divided by x.

Example 5. Reduce Answer. x 1 = . 4x 4

x . 4x

When the numerator cancels completely, we must write 1. For, x = x 1. x 1 1 = . 4x 4

Example 6. Reduce Answer.

x3 . 6(x 3)

x3 1 = . 6(x 3) 6

We can view x 3 as a factor of the numerator, because x 3 = (x 3) 1 Again, when the numerator cancels completely, we must write 1. Problem 7. Reduce. 2a =2 a 5(x 2) =5 x2 a 1 = ab b x+1 1 = 2(x + 1) 2 2x 1 = 8xy 4y 3(x + 2)x 1 = 6(x + 2)xy 2y

a) d)

b) e)

c) f)

Example 7. Reduce

3a + 6b + 9c . 12d

Answer. When the numerator or denominator is made up of terms, then if every term has a common factor, we may divide every term by it. In this example, every term in both the numerator and denominator has a factor 3. Therefore, upon dividing every term by 3, we can write immediately: 3a + 6b + 9c a + 2b + 3c = 12d 4d There is no more canceling. The numerator and denominator no longer have a common factor. This example illustrates the following principle: To divide a sum -- such as 3a + 6b + 9c -- by a number, we must be able to divide every term by that number.

Example 8. Reduce

3a + 6b + 8c . 12d

Answer. Not possible The numerator and denominator have no common factor. That fraction is in its lowest terms.

Example 9. Reduce

8x . 8x + 10

Answer. 2 is a factor of every term in both the numerator and denominator. Therefore, 8x 4x = . 8x + 10 4x + 5 There is no more canceling. We cannot cancel the 4x's, because 4x is not a common factor of the denominator. 4x appears only as the first term.

Example 10. Reduce Answer. Not possible

15x . 5x 3 The numerator and denominator have no common factor. x x 6 . x 4x + 3

Example 11. Reduce

Answer. In its present form, there is no canceling -- because there are no common factors. But we can make factors: x x 6 (x 3)(x + 2) x+2 = = x 4x + 3 (x 3)(x 1) x1 (x 3) is shown to be a common factor. We can cancel it. And when we do, the numerator and denominator no longer have a common factor. The end.

Example 12. Reduce:

4x 9x 4x + 6x

.

Answer. The only common factor is x. And we could display it by factoring both the numerator and denominator: x(4x 9) 4x 9 4x 9x = 2x(2x + 3) = 2(2x + 3)

4x + 6x The fraction is now in its lowest terms. No common factors. Problem 8. Reduce. 5x 5x x = = 10x + 15 5(2x + 3) 2x + 3 3x 12 3(x 4) x4 = = 3x 3x x 12x 18y + 21z 4x 6y + 7z = , 6y 2y

a)

b)

c)

upon dividing every term by their common factor, 3.

d)

2m 2m 2 = = m 2m m(m 2) m2 x x x(x 1) = = x1 x x 12x 12x 3 = = 3 3 16x 20x 4x(4x 5) 4x 55

e)

f)

g)

x+3 x+3 1 = = 4x + 12 4(x + 3) 4 2x 8 2(x 4) = = 2 x4 x4 2x 2y 3x 3y 2(x y) 2 = 3(x y) 3

h)

i)

=

Problem 9. Make factors, and reduce. x 2x 3 (x + 1)(x 3) x 3 = = x x 2 (x + 1)(x 2) x 2

a)

b)

x + x 2 (x + 2)(x 1) x 1 = = x x 6 (x + 2)(x 3) x 3 x 2x + 1 (x 1) x1 = = x 1 (x + 1)(x 1) x + 1 x 100 (x + 10)(x 10) = = x 10 x + 10 x + 10 x+3 x+3 1 = = x + 6x + 9 (x + 3) x + 3 x + 4x _ x(x + 4) x = = x + x 12 (x 3)(x + 4) x 3

c)

d)

e)

f)

Problem 10. Simplify by canceling -- if possible.

a)

3+x Not possible. The numerator and denominator have no common factors. 3x 8a + b Not possible. Again, no common factors. 2ab 8a + 2b 2(4a + b) 4a + b = = 2ab 2ab ab Not possible. The numerator and denominator have no common factors. 3 is not a factor of either the numerator or denominator. It is a factor only of the first term.

b)

c)

d) 6a + b 3a + b e)

6(a + b) =2 3(a + b) 2x + 4y + 6z x + 2y + 3z = Divide every term by 2. 10 5 2x + 4y + 5z 10 (x + 1) + (x + 2) (x + 1)(x + 3) (x + 1)(x + 2) x + 2 = (x + 1)(x + 3) x + 3 Not possible. The numerator and denominator have no common factor.

f)

g)

h)

Not possible. The numerator is not made up of factors.

i)

j)

ab + c Not possible. The numerator and denominator have no common factors. abc ab + ac a(b + c) b + c = = abc abc bc x x 12 (x + 3)(x 4) x 4 = = x + x 6 (x + 3)(x 2) x 2

k)

l)

2nd Level

NEGATIVE EXPONENTS Power of a fraction Subtracting exponents Negative exponents Section 2 Exponent 0 Scientific notation

Power of a fraction

"To raise a fraction to a power, raise the numerator and denominator to that power."

Example 1. For, according to the meaning of the exponent, and the rule for multiplying fractions:

Example 2. Apply the rules of exponents: Solution. We must take the 4th power of everything. But to take a power of a power -multiply the exponents:

Problem 1. Apply the rules of exponents. To see the answer, pass your mouse over the colored area. To cover the answer again, click "Refresh" ("Reload"). Do the problem yourself first!

a)

=

x y =

b)

=

8x 27

c)

=

d)

e)

=

x 2x + 1 x + 2x + 1

Perfect Square Trinomial

Subtracting exponents In the previous Lesson we saw the following rule for canceling: ax x = ay y "If the numerator and denominator have a common factor, it may be canceled." Consider these examples of canceling: 2 2 2 2 2 = 2 2 2

2 2 ___2 2___ __1__ = 2 2 2 2 2 2 2 2 If we write these examples with exponents, then = 23

22

2 1 = 3 25 2 In each case, we subtract the exponents. But when the exponent in the denominator is larger, we write 1-over their difference.

Example 3.

x3

= x5

x8 Here is the rule:

=

1 x5

Problem 2. Simplify the following. (Do not write a negative exponent.) x 1 = x5 x3 = x4 x 1 = x5 x4 = 1 x

a)

= x3 x = x x

b)

c)

d)

e)

f)

Problem 3. Simplify each of the following. Then calculate each number.

a)

= 23 = 8 2 = 2 2

b)

2 1 1 5 = 3 = 2 2 8 = 24 = 16

c)

2 1 1 5 = 4 = 2 2 16 = 1 1 = 2 4

d)

e)

f)

Example 4. Simplify by reducing to lowest terms: Solution. Consider each element in turn:

Problem 4. Simplify by reducing to lowest terms. (Do not write negative exponents.

a)

=

y 5x 3z_ 5x4y3

b)

=

8a 5b

c)

=

d)

=

c 16

e)

(x + 1) (x 1) (x + 1) = (x 1) (x + 1) (x 1)

Negative exponents We are now going to extend the meaning of an exponent to more than just a positive whole number. We will do that in such a way that the usual rules of exponents will hold. That is, we will want the following rules to hold for any exponents: positive, negative, 0 -- even fractions aman = am + n Same Base

(ab)n = anbn

Power of a Product

(am)n = amn

Power of a Power

We begin by defining a number with a negative exponent to be the reciprocal of that power with a positive exponent. an = an is the reciprocal of an. 1 an

Example 5.

23 =

1 1 = 23 8

The base, 2, does not change. The negative exponent becomes positive -- in the denominator. Example 6. Compare the following. That is, evaluate each one: 32 32 (3)2 (3)3 1 1 = 32 9

Answers.

32 = Next,

32 is the negative of 32. The base is still 3. 32 = 1 9

As for (3)2, the parentheses indicate that the base is 3: (3)2 = 1 1 2 = (3) 9

Finally, (3)3 = 1 1 3 = (3) 27

A negative exponent, then, does not produce a negative number. Only a negative base can do that. And the exponent must be odd

Example 7. Simplify a .

a5 Solution. Since we have invented negative exponents, we can now subtract any exponents as follows: a = a2 5 = a3 a5 That is, we now have the following rule for any exponents m, n:

In fact, we defined a n as to hold. We want

1 because we want that rule an

= a3 But = 1 a3 1 . a3

Therefore, we define a3 as

Example 8.

a1 =

1 a

a1 is now a symbol for the reciprocal, or multiplicative inverse, of any number a. It appears in the following rule (Lesson 5): a a1 = 1 Problem 5. a) (log 2)(log 2)1 = 1 b) (x 7x + 5) (x 7x + 5)1 = 1

3 2 c) ( )1 = 3 2Example 9. Use the rules of exponents to evaluate (23 104)2.

Solution.

(23 104)2 = 26 108 Power of a power

=

=

_64_ 100,000,000

Problem 6. Evaluate the following.

a) 24 =

1 1 = 24 16

b) 52 =

1 1 = 52 25

c) 101 =

1 1 = 101 10

d) (2)3 = e) (2)4 =

1 1 1 = = (2)3 8 8 1 1 = (2)4 16 f) 24 = 1 1 = 24 16

g) ()1 = 2. 2 is the reciprocal of . Problem 7. Use the rules of exponents to evaluate the following. a) 10 104 = 102 4 = 102 = 1/100. b) (23) = 26 = 1 1 6 = 2 64 34 81 = 28 256

c) (32 24)2 = 34 28 =

d) 22 2 = 22+1 = 21 =

1 2

Problem 8. Rewrite without a denominator. x = x25 = x3 x5 = x3y4 y = y16 = y5 y6 = a1b6c7

a)

b)

c)

d)

e)

1 = x1 x (x + 1) = (x + 1)x1 x

f)

1 3 3 = x x

g)

h)

(x + 2)2 = (x + 2)4 (x + 2)6

Example 10.

Rewrite without a denominator and evaluate:

Answer. The rule for subtracting exponents --

-- holds even when an exponent is negative.

Therefore, = 103 + 5 2 + 4 = 104 = 10,000.

Exponent 2 goes into the numerator as 2; exponent 4 goes there as +4. Problem 9. Rewrite without a denominator and evaluate. 10 = 102 + 2 = 104 = 10,000

a)

2 = 22 + 3 = 25 = 32

b)

23

102

c)

= 102 5 4 + 6 = 101 =

1 10

d)

= 25 6 + 9 7 = 21 = 2

The reciprocal of an. Reciprocals come in pairs. The reciprocal of an is an : 1 = an. an And the reciprocal of an is an : 1 = an. an That implies: Factors may be shifted between the denominator and the numerator by changing the sign of the exponent.

Example 11. Rewrite without a denominator:

Answer. The exponent 3 goes into the numerator as 3; the exponent 4 goes there as +4. Problem 10. Rewrite with positive exponents only.

a)

x = xy y2

b)

=

c)

=

d)

=

e)

=

Problem 11.

Apply the rules of exponents, then rewrite with positve exponents.

a)

=

=

b)

=

=