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Unit 6 Exponents and Polynomials Introduction Symbols are very important in Algebra. There was a time when mathematicians did not have a standard set of symbols. Without the use of variables, symbols and all kinds of notation, mathematics looks much different and is much harder. For instance, a problem written without any symbols or variables would similar to the example below. Find the sum of the square of the number plus the number multiplied by a factor of two increased by one all divided by the difference of the square of the number and one. For most people this problem, when written without the variables and symbols, is very hard to interpret or solve. Below the same problem is written with our variables and symbols. Simplify x 2 2x 1 x2 1 This representation with variables and symbols can be understood and simplified by most Algebra I students when they finish the next two units. Symbols, variables, and notation make it possible for us to use the properties of real numbers, the Properties of Equality, solve equations, and do many other things. In this unit we will learn to manipulate expressions, as well as perform operations with polynomials. Additionally, we will learn to represent and use exponents in new ways. These abilities are essential to Algebra and mathematics. Unit 6 Vocabulary and Concepts Base The number that is going to be raised to a power. Binomial A binomial is an expression with exactly two terms. Coefficient A numerical or constant quantity placed before and multiplying the variable in an algebraic expression. Constant A constant is a number on its own, or sometimes a letter such as ‘a’, ‘b’, or ‘c’ to stand for a fixed number. Degree The degree of a term is the sum of the exponents in the term. Exponent An exponent is a raised value or variable to the right of the base that tells us how many times to use a factor. Factors Factors are numbers we multiply together. Monomial A monomial is an expression with exactly one term. Order The order of a polynomial is the highest degree in the polynomial. Polynomial A polynomial is an expression with one or more terms. Trinomial A trinomial is an expression with exactly three terms. Term A term is part of an expression separated by a plus or minus sign. Key Concepts for Exponents and Polynomials Adding and Subtracting Monomials To add or subtract monomials we combine like terms. Multiplying Monomials When we multiply monomials we should multiply coefficients and add exponents. Dividing Monomials When we divide monomials we should reduce coefficients and subtract exponents in the denominator from the exponents in the numerator. Negative Exponents A negative exponent means the factor belongs in the denominator of a fraction. Power of Zero Any value or variable raised to the zero power is 1. Common Factoring Common factoring is reversing the Distributive Property. This is done by putting the GCF on the outside of parentheses and finding the terms that belong inside the parentheses. Unit 6 Section 1 Objective The student will add and subtract monomials. The most basic type of expression is called a monomial. To understand what a monomial is we need to break the word down into its parts. The root word in monomial is “nomial”. This word means “term” in Algebra. There are several prefixes that can be attached to the root. The prefixes are: poly, tri, bi, and mono. The meanings of the prefixes are listed below. “poly” “tri” “bi” “mono” means many means three means two means one Putting the root word and the prefixes together we get the following vocabulary. A monomial is an expression with exactly one term. A binomial is an expression with exactly two terms. A trinomial is an expression with exactly three terms. A polynomial is an expression with one or more terms. Polynomial is really the name for the overarching category of expressions. Monomials binomials and trinomials are all subcategories of expressions. The Venn diagram below shows these relations ships. Polynomials monomials binomials trinomials Examples of each of the categories are given below. Monomials Binomials Trinomials 2xy -11a2bc 4 2y + 3z 4x2 – 25y2 6x + 1 4a – 2b + 7 3x2 - 16x + 5 5az3 – 9az2 – 2az Manipulating expressions in Algebra begins with adding and subtracting monomials. This skill is actually something we already know how to do. Another name for adding and subtracting monomials is combining like terms. To add or subtract monomials we must remember the definition of “like” terms. Terms are “like” when they have the same variables with the same exponents. The way to combine like terms is to add or subtract the coefficients of the like terms. Several examples follow. Example A 5z – y + z + 4y 3y + 6z The like terms in this example are 4y with -1y and 5z paired with 1z. So we add their coefficients. We should usually put our variable terms in alphabetical order. Example B 4x – x2 + 7x + 6x2 The like terms here are 4x with 7x and -1x2 paired 5x2 + 11x with 6x2. We have two pairs of like terms because the x2 and x are not like due to the different exponents. We should usually put the higher order exponents first. Example C 6 + 2a – 7a + a – 1 The like terms in this expression are those with ‘a’ -4a + 5 and the constants or plain numbers. We usually list the constant at the end of the expression. Example D x+5+y–y+3 x+8 When we combine the like terms here 1y and -1y add up to 0 and so we have 0y or 0 but we don’t need to write the 0y. There is nothing to combine with the ‘x’ so it simply stays in the expression. When we are working with monomial expressions we will be asked to determine the degree of a term. The degree of a term is the sum of the exponents in the term. The examples that follow show a sample of monomials and the degree of the monomial. Example A 4x2 The monomial is of degree 2. Example B -2x3yz The monomial is of degree 5. The ‘y’ and ‘z’ have understood exponents of 1, so 3 + 1 + 1 is 5. Example C a2bc4 The monomial is of degree 7. The order of a polynomial is the highest degree within the polynomial. The examples show us how to determine the order of a polynomial. Example A 2x2 – y3 – 1 The polynomial is of order 3 since the 2nd term is of degree 3 and that is the highest in the polynomial. Example B 5x4y + x3y3z The polynomial is of order 7 since the sum of the exponents in the 2nd term is 7 and this is the highest degree term. Whenever possible polynomials should be written in standard form. The general rule for putting a polynomial into standard form is that the highest degree terms come first followed by the lower degree terms in descending order. For instance: 9x + 4x4 – 5x2 + 7 4x4 - 5x2 + 9x + 7 Should be written as: VIDEO LINKS: Khan Academy Adding and Subtracting Polynomials 1 Khan Academy Adding and Subtracting Polynomials 2 Khan Academy Terms, Coefficients, and Exponents in a Polynomial Exercises Unit 6 Section 1 Define the following 1. Term 2. Exponent 3. Coefficient 4. Constant 5. Base 6. Monomial Given the expression 2v + w4 + 6x3 + yz + 9 answer the following questions 7. How many terms are in the expression? 8. What is the coefficient of the third term? 9. What is the coefficient of w? 10. What is the constant? 11. What is the base of the second term? 12. What is the exponent of the third term? 13. What is the exponent of the first term? Complete the following 14. Two terms are "like" when they have _____________________________________ 15. If we want to add monomials then we should ______________________________ ___________________________________________________________________ Simplify add or subtract as indicated and write your answers in standard form. 16. 4z + 3z - z 18. -7x + 7 + x + 5 + 11x 2 2 2 20. 8wx + 3x - 2wx + 6x - 10wx + 4x 2 2 2 17. -5a + 9a - a 19. 3x + 8 - 16x + 9 - x 3 3 21. -6xy + 5a - 6 + 7a + 6 - 3xy + 9xy 22. 15x + 8a + 3a - 15x 24. -7x + 7 + x - 7 + 14x 2 2 2 23. 7z - 5z - 2z 25. 5x + 8x2 - (-4x) + 9 - x2 3 3 26. 3x + 4z - 2x + 6z - x + (-10z ) 27. 6y + 4a - (-4) + 9a + 6 + -3y - 10 28. x + x + x + x + a + a - x + a - x + a 29. 2(x + 3y) + 5x + 6y 30. 3(a2 - 5) + a2 – 4 31. -2(3x + 4y - 6z) + x - 2y + z 32. 4(-2x + 4z) + 5(6 - x) 33. 3x + 8y + 7z + 11 34. 1.5x + 7.2x - 3.4x 35. 2a - 4a2 + a + 6a 36. 1 3 a 2a a 2 2 List the degree of each monomial. 37. 4x 38. -2ab 39. x2y3 40. 11ab4c2 41. 42. 9xayb 5 List the order of each polynomial. 43. 3x + x2 – 6 + 8x3 44. x2yz4 + x5y - 10xyz2 Unit 6 Section 2 Objective The student will multiply monomials. Multiplying monomials requires that we understand how to use exponents in products. We will review how exponents and factors can be used before we look at the rule for multiplying monomials. Definition Factors are values or variables that we multiply together. The monomial 2abc when broken down into its factors can be written as (2)(a)(b)(c) or 2 . a . b . c. When factors are repeated we can use exponents. A repeated factor is called a base. Definition An exponent is a raised value or variable to the right of the base that tells us how many times to use a factor. Below are some examples of products written as factors and in exponential form. x3 = x . x . x w5 = w . w . w . w . w a2b4 = a . a . b . b . b . b xy2z = x . y . y . z The understood exponent is 1 (one). The expression xy2z could have been written as x1y2 z1 because there is only one ‘x’ in the product, additionally there is only one ‘z’ in the product so both variables can have the understood exponent of 1. The examples below show us how multiplying monomials can affect exponents. Example A x3 . x4 = (x . x . x) . (x . x . x . x) = x . x . x . x . x . x . x = x7 In this example we can use the Associative Property of Multiplication to remove the parentheses and then count the x’s to find the resulting exponent. Example B a2 . a6 = (a . a) . (a . a . a . a . a . a) = a . a . a . a . a . a . a . a = a8 In this example we can use the Associative Property of Multiplication to remove the parentheses and then count the ‘a’ s to find the resulting exponent. Example C a2b . a4b3 = (a . a . b) . (a . a . a . a) . (b . b . b) = a . a . b . a . a . a . a . b . b . b = a6b4 We could use this technique to do all of our multiplying; however, we can find a better way. If we summarize the examples we may see the better method. Example A Example B Example C x 3 . x 4 = x7 a2 . a6 = a8 a2b . a4b3 = a6b4 The questions we need to ask ourselves are: How How How How can can can can we we we we use use use use the the the the values values values values 3 2 2 1 and 4 and 6 and 4 and 3 to get to get to get to get 7? 8? 6? 4? The arithmetic we need to do is addition! This should make sense because x3 . x4 has a total of 7 x’s multiplied together so the new exponent should be 7. Our rule then for multiplying monomials is: When we multiply monomials we should multiply coefficients and add exponents. Below are some examples of applying this rule. Example A Example B 3ab . 5a2b (-2xy3z)(4y2z5) 15a3b2 -8xy5z6 In this example we could have written the problem as: 3a1b1 . 5a2b1 After putting in the understood exponent, we add the exponents. VIDEO LINK: In this example we have a variable ‘x’ that is only in one of the monomials. Even though we don’t add any exponents for ‘x’, it must still be in our answer. Khan Academy Multiplying Monomials Example C (-wx)(3w2)(-7wx6)(2wxy) 42w5x8y In example C we have four monomials. We must still follow our rule and multiply the coefficients and add exponents. The first monomial could have a coefficient of -1. Exercises Unit 6 Section 2 Fill in the blanks 1. An expression with one term is called a _________________________________ 2. Part of an expression separated by plus or minus is called a ___________________ 3. The number to the right of a variable/value that tells us how often to use a factor is called the _____________________ 4. A plain number is called a _________________________ 5. Two terms are "like" when they have _____________________________________ 6. If we want to add monomials then we should ______________________________ Simplify and add or subtract as indicated. 7. 3z + 7z - 8z 8. 9. -8x + 9 + x - 5 + 12x 2 10. 2 2 2 2 2 5a - 9a - 2a 13x + 8 - 16x + (-8) + 3x 3 3 11. 5wz + 3z - 8wz - (-5x ) - wz + z 12. -a + 5a - 2 + a + 8 + -3a + 9 13. 5x - 8a + 3a - 15x - 5a 14. 15. 5x + 8x2 - (-4x) + 9 - x2 16. x + 4z - 5x + 11z - (-2z) + z 3 7a - 5b - 2c 2 2 3 17. -6y + 4a - (-4) + 11a + z + (-3y) - 10 + 9y 18. x + x + x + a + a - x + a - x + a + a + a 19. 5(x + 3y) + 3x - 6y 20. -4(a2 - 5) + a2 - 1 21. 7(3x - 2y - 5z) + x - 2y + 2z 22. -(-3x + 4z) + 5(-2 + x) 1 3 24. - a 4a a 2 2 23. 3.5x + 4.2x - 6.2x Complete the sentence 25. If we want to multiply monomials then we should ________________________________ ________________________________________________________________________ Multiply 26. x 29. 5 . 4 x 2 . a 3 27. x 7. . . 5 b b b a 4 31. (-2a b2c)(-5a bd) 5 . 2. x 3 . x 30. (-3x)(-2x)(-2x)(x)(x) 4 3 6. 28. a a 3 2 3 2 32. (-x )(-5y z)(-x z ) (-x z ) a 33. y3 . y7 34. 3x2 . 36. (1.5w3)(2.4w2)(w) 37. 1 2 z 2 39. (-.2a2b)(.4a)(-.5ab) 40. 3 3 a 2 -x . . . -4x5 1 5 z 2 1 - az 5 35. z . a5 . z 3 38. - z5 4 1 z 2 . . - . 4 5 z 3 4 5 2 . az z 3 Solve the following equations 41. 6x + 5 = 37 42. 43. 8 = -4(a - 5) 44. 45. 3(x - 7) = 3x - 21 46. 47. |x – 3| = 12 48. 3a + 7 = a - 25 2(a + 7) = 3(2a + 10) + 24 |x| = 7 2|x + 4| + 3 = 25 49. Which of the values below is the solution to the equation 6(x + 4) – 1 = 47 a) 4 b) -4 c) 8.3 d) -8.3 50. Which of the values below is a solution to the equation a) {4, -20} b) {-4, 20} -2|x + 8| = -24 c) {8, -12} d) {12, -12} Multiply and simplify the following. 51. (xa)(xb) 54. (xa)(x) 52. (xa)(xa) 55. (x2)(xb) 53. (xa)(x-a) 56. (x2a+1)(xa-4) Unit 6 Section 3 Objective The student will divide monomials with natural number exponents. We know how to add, subtract, and multiply monomials. In this section we will learn how to divide monomials. To understand the dividing of monomials we need to review how to factor and reduce fractions. Factoring to reduce a fractional expression is a technique that will be very useful in the future. The method consists of factoring the numerator and denominator, and dividing out the common factors. Some examples are given below. Example A Example B . . 2 5 10 5 = = In this example we turned the 10 into 2 times 5 14 7 2 7 and the 14 into 2 times 7; we divided out the two’s and get 5 over 7. One of the important features of this method is that the numbers never get larger, only smaller. . . . . 1 2 2 3 12 2 = = 18 2 3 3 3 We factored both numbers to primes, and the negative one for the sign. When we divided out the common factors we were left with -1, 2, and 3. . The same method for reducing fractions can be used to reduce monomial expressions in fractions. The examples below show us how to divide monomials. Example A Example B Example C x5 x2 a4 a3 6a 3 b 2 2ab 2 . . . . . a a a a a a a x x x x x x x x3 . . . . . a1 or a . . . . . .b . . . 2 3 a a a b 2 a b b 3a 2 This method of factoring and dividing out the common factors is easy but the larger the monomials, the more time it will take. We need to develop a rule that will make the process quicker. To find our rule we need to see how we can use arithmetic to find the new exponents. We have to combine the 5 and 2 to get 3 in example A. In example B, we combine the 4 and 3 to get 1. And in the last example, the 3 and the understood exponent of 1 gives us 2. In each case, we can subtract the exponent in the denominator from the exponent in the numerator. Our rule then for dividing monomials is: When we divide monomials we should reduce coefficients and subtract exponents in the denominator from the exponents in the numerator. The examples below show us how to use the rule. Example A Example B x7 x3 16a 5 2a 3 x4 8a 2 In this example 7–3=4 so our answer has an exponent of 4. Example C 10 x 5 y 6 z 2y2z 5x 5 y 4 In this example 5–3=2 so our answer has an exponent of 2. In this example 6–2=4 1 -1 = 0 so our answer has an exponent of 4 and no ‘z’. Example C also illustrates a fact that will be very important in the future. The variable divides out in example C. When we say “divides out”, this is not the same thing as “cancels out”. When terms cancel out during addition they add up to zero and are gone. When common factors “divide out”, they gives us a “1” and we do not have to write the “1” because one times any number does not change the number. The example below illustrates this. x4 x4 =1 x4 = x0 x4 This must be true since any number divided by itself is 1. This must be true if we use our rule and subtract exponents. In conclusion then we know x0 = 1. IN words this means: Any number raised to the zero power is 1. The examples below illustrate this rule. ao = 1 50 = 1 -2.50 = 1 (2x)0 = 1 We should also be able to divide monomials when the problem is written horizontally as well as vertically. The problems below are in horizontal form. To do these problems the divisor’s (2nd monomial) exponents are subtracted from the dividend’s (1st monomial) exponents. Example A Example B Example C x8 ÷ x6 = x2 a4b5c ÷ a3b2c = ab3 -15xy7z2 ÷ 5yz2 = -3xy6 VIDEO LINK: Youttube: Dividing Monomials Exercises Unit 6 Section 3 Complete the following 1. To Divide monomials we should ___________________________________ Divide the following 2. x7 x3 6. ax 5 ax 2 7. 9. 12wx 7 3x 3 10. 21a 2 x 4 y 3ax 4 3. a6 a w8 w7 4. a 4 x5 y a4x 5. 8. y3 y3 w3 x 4 z x4 z 11. 16w 2 x 3 z 4w 2 x 3 z 12. x7 ÷ x5 13. a4 ÷ a4 14. x2y4 ÷ xy 15. 24x6 ÷ 3x2 16. -5a3b ÷ 5a3b 17. -18wx3y9 ÷ -6x2y8 Complete the following 18. To multiply monomials we should ___________________________________ Multiply the following 19. x 3 . 4 x . 20. x x 5. x 21. x . x . x . x . x . x 22. x . x . y . x . y . x . x 23. (-5ab)(3b)(a)(-a)(-b3) 25. (.3w2)(.2w5)(w) 26. 1 3 a 3 . 2 5 a 3 . 1 a 2 24. (-4a2bc)(-9a4b3d) 3 27. - w 5 . 3 4 w 2 Complete the following 28. To add monomials then we should ______________________________ Simplify and add or subtract as indicated. 2z + 5z - 9z 30. 6w3 – 4w3 – 2w3 31. 5x + 3 + x - 8 - 4x 32. 14a + 2 – 17a + (-7) + 3x 29. 2 2 2 33. 6ab + 5a – 12ab - (-5a ) - ab + a 34. -11 + 14a4 - 2 + a4 + 4 + -7a2 + 9 + a2 Match each term with its definition. 35. 36. 37. 38. _____ _____ _____ _____ relation range function domain A. B. C. D. the set of x-values we can use the set of y-values that can be generated as answers a set of ordered pairs or points a relation in which any 'x' value is paired with at most one 'y' value. 39. The important or special feature of a function is that a) b) c) d) if we substitute one number for 'y' we get more than one number for 'x'. if we substitute one number for 'y' we get one number for 'x'. if we substitute one number for 'x' we get more than one number for 'y'. if we substitute one number for 'x' we get one number for 'y'. 40. In the equation "g(x) = 2x + 4" the "g(x)" could be replaced by ________. 41. Decide which of the following rosters represent a function and which are nonfunctions. ( Write yes for a function and no for a non-function. ) a. b. c. d. {(3, 7), (7, 3), (-4, ½) (½, -4)} {(1.5, 7), (-4,2), (1.5, -½) (3, 8)} {(4, 2), (-1.5,-2), (-9, 7), ( , 5)} {(-2, -1), (3,-1), (4.5, -1) (0, -1)} Divide and simplify 42. xa xb 43. xa x 44. xb xb 45. w 4a wa 46. w 2 a 3 w a 7 Unit 6 Section 4 Objective The student will divide monomials with integer exponents. In the previous section the numerator’s exponents were greater than or equal to the denominator’s exponents. We need to be able to divide when the denominator has the larger exponents. We will follow the same rule for division that we found in Section 3. When we divide monomials we should reduce coefficients and subtract exponents in the denominator from the exponents in the numerator. When we subtract a larger number from a smaller number we get a negative value. So our answers for this type of problem will be negative. The examples below illustrate this. Example A Example B Example C x2 x7 15a 3a 4 14 x 2 yz 3 4y6 z x-5 5a-3 - We must subtract 2–7 So we get -5 We must subtract 1–4 So we get -3 7 2 -5 2 xy z 2 The coefficients were reduced. The ‘x’ was only in the numerator and did not change. For the ‘y’ variable the exponent is 1 – 6 = -5. For the ‘z’ variable the exponent is 3 – 1 = 2. We need to understand exactly what a negative exponent means. To investigate this we will go back to our process of dividing out common factors. We will go back to Example A. When factors divide out we x x 1 x2 Example A = = 5 are left with 1’s. So we have 7 x x x x x x x x x a 1 in the numerator. . . . . . . . When we use our rule for division we get this means that 1 = x-5. x5 x2 = x-5 7 x This reworking of example A on the previous page tells us that a negative exponent means the factor belongs in the denominator of a fraction. We should be able to convert between negative exponents and fractional representations as illustrated below. A negative exponent means the factor should be in the denominator of a fraction. x-n = Example A a-2 = Example B 1 a2 x-4y-3 = 1 xn Example C 1 4 3 x y 5b-5c = 5c b5 Example D 4y3 4 1 3 6 x y z = 3 3 xz 6 We should also be able to start with a fractional representation and convert it to an expression with negative exponents. Example A 3 w5 = Example B 2a 4 = 2a4b-3c-1 b 3c 3w-5 Example C 3 3xy3 = w-1xy3z -2 2 5 5wz Finally, we should be able to divide monomials and list our answer in either or both the fractional form, or with negative exponents. Example A 5y 30 xy3 = 5x-3y = 3 4 2 x 6x y Example B 20a 2 bc 4 5a 2 c 3 5 2 -4 3 -1 = ab cd = 3 3b 4 d 12b 5 cd Example C w2 y w 2 yz = z -1 = 1 z VIDEO LINKS: Youtube: Simplifying to Negative Exponents Youtube: Brightstorm: Zero and Negative Exponents Exercises Unit 6 Section 4 Complete the Following 1. To divide monomials we should __________________________________. Simplify the following expressions 2. x5 3. x3 6. x6 4. x 20a 3x 5y 7. 4ax 5 x8 5. x7 6ax 4 2x 2 5w 2 x 3z x 3z Simplify the following expressions and list your answer in fractional form ( do not use negative exponents ) 8. x5 x7 9. x 10. x5 x7 x8 Simplify the following expressions and all answers should be listed with negative exponents where appropriate. 11. x5 x7 12. x x5 13. x7 x8 14. A negative exponent tells us that the variable or factor should be placed in the ___________________ of a fraction. ( hint: compare problems 8-10 with 11-13 ) Write each fraction with negative exponents where appropriate. 15. 18. 21. 1 x7 1 a 4b a2 16. 19. 22. 1 a5 1 a 4b 3 c a 5b 24. Any base raised to a zero power is ___________ 17. 20. 23. 2 z8 3 xy 5 z 8 ax yz 3 Simplify 25. b0 26. 50 27. x3 x3 Write each monomial as a fraction. 28. a-4 29. x-1 30. x-2y-5 31. a-4b 32. w6x-1y-3 33. 2x-2 34. -5a-4b3c-2 35. -x-2 36. 5-3 Divide and list your answer as both a fraction and with negative exponents. 24a 3 x 2 y 6ax 5 y 37. 6ab 2 c 4 39. 10a 3 bc 4 4ab 2 c 4 20b 5 cd 38. Perform each operation and list your answer with negative exponents where needed. 40. (x3)(x-6) 41. 7-3 . 72 43. (ax4)(ax-2) 44. 46. x 2 x 47. 1 49. When we evaluate the expression a. -27 b. 9 (a 3 )( a 2 ) a7 50. The expression a. a 2 51. The expression b. 1 a 2 (a 4 )( a 2 ) a 4 42. x-2 . x2 9 2 96 45. 92 x 5 x 3 48. (x-2)(x-8) __________ 5 x x4 (a )( a 5 ) with a = -3 the result is a2 c. -243 d. 81 in its completely simplified for is c. 1 a2 d. 1 a in its completely simplified form is 52. Which of the following expression is equivalent to a. x2 x2 b. 2 x c. x5 3x 2 d. 53. Which of the following expression is equivalent to a. x3 b. x-3 c. 5x 3 x 1 1 . 2 + for x ≠ 0 2 x x x x2 x5 x6 (5 x 2 ) 0 x 3 for x ≠ 0 d. 0 54. Given the expression 3-2x4y-1z a. Write the expression in a simplified form without negative exponents. b. Explain how you found the coefficient of the simplified form. Unit 6 Section 5 Objective The student will add or subtract polynomials. We have learned to perform operations with monomials in section 4. Now we must move on to operations with polynomials. The first two operations we will deal with are addition and subtraction of polynomials. The process of adding polynomials is simplified by the Associative Property of Addition. The problem below demonstrates how we add polynomials. (6x3 + x2 + 11x – 12) + (-2x3 + 7x2 – 4x + 7) The Associative Property of Addition means that we can switch or rearrange the parentheses any way we need to. In this operation we can remove the parentheses and combine like terms. 6x3 + x2 + 11x – 12 + -2x3 + 7x2 – 4x + 7 So our answer is: 4x3 + 8x2 + 7x – 5 To add polynomials, we remove the parentheses and combine like terms. Some examples are given below. Example A (5a3 + a – 2) + (a2 – a + 11) 5a3 + a – 2 + a2 – a + 11 5a3 + a2 + 9 Example B (-6x2 + 3xy – 9y2) + (-x2 - 2xy + 14y2) -6x2 + 3xy – 9y2 + -x2 - 2xy + 14y2 -7x2 + xy + 5y2 Subtraction can be done with several approaches. One of the easiest methods is to use the Distributive Property. The example below shows how this is done. Subtraction is the same thing as (10y3 + 4y2 + 3y – 11) − (-4y3 + 6y2 – 9y + 3) addition of the opposite. This (10y3 + 4y2 + 3y – 11) + -(-4y3 + 6y2 – 9y + 3) means we can change the problem to addition. We can insert the understood (10y3 + 4y2 + 3y – 11) + -1(-4y3 + 6y2 – 9y + 3) coefficient of 1 and the use the 10y3 + 4y2 + 3y – 11 + 4y3 – 6y2 + 9y – 3 Distributive Property. Then remove the parentheses. Our final answer is 14y3 – 2y2 + 12y – 14 We combined like terms. The effect of multiplying by negative one is to change all the signs in the second parentheses. Below are some examples of subtracting the polynomials. Example A (-3a3 + 9a2 + a – 13) − (4a3 + 7a2 – 8a + 4) -3a3 + 9a2 + a – 13 + -4a3 – 7a2 + 8a – 4 -7a3 + 2a2 + 9a – 17 Example B (11m2 – 5mn + n2) – (-5m2 – 3n2) 11m2 – 5mn + n2 + 5m2 + 3n2 16m2 – 5mn + 4n2 Example C (-3y + 7x + 14) – (-3y + 7x – 14) -3y + 7x + 14 + 3y – 7x + 14 28 VIDEO LINK: Youtube: Simplifying with Negative Exponents Exercises Unit 6 Section 5 Simplify the following 1. (4x – 3) + (7x + 8) 2. (3b + 4) + (-2b – 6) 3. (5m + 8) + (4m + 3) 4. (-2t – 7) + (6t – 3) 5. (5k – 6n + 4) + (-5k + 8n + 2) 6. (6x2 - 2xy + 3y2) + (4x2 – xy – y2) 7. (4x – 4) – (3x + 8) 8. (5x – 2) – (4x + 1) 9. (2m2 - 3mn – 5n) – (-8m2 – n2) 10. (5a2 – 6ab) – (-2a2 + 9ab – b2) 11. (5x – 3y + z) – (2x + 5y – z) 12. (4x + 5y – z) – (4x – 3y – z) 13. Given the rectangle below find the expression for its perimeter. 3x - 5 2x + 1 14. Given the diagram to the right which of the expressions below represents the perimeter of the triangle. x2 + 13x 3x2 + 11 -4x - 4 4x2 a) + 17x + 15 2 b) 3x + 9x + 7 c) 4x2 - 17x + 7 d) 4x2 + 9x + 7 15. The building to the right has a parking garage underneath it ( shown in black ). The distance from the bottom of the parking garage to the top of the building is 3x2 + 5x – 1. And the distance from the parking garage floor to the floor on the first level of the building is x2 + x + 1. The expression for the height of the building is a) b) c) d) 2x2 + 4x – 2 4x2 + 6x 2x2 + 4x 4x2 + 4x + 2 Write each fraction with negative exponents where appropriate. 16. 1 x6 17. 3 a4 18. 5 z 19. 5x y3 20. w xy 7 21. 2ab 3 9 yz 5 Write each monomial as a fraction. 22. a-7 23. x-8 24. x-1 25. a-1b 26. -9wx-4y-2 27. 2-1x-2 Divide and list your answer as both a fraction and with negative exponents. 28. 28ax 3 y 4 4ax 5 y 29. 2b 4 cd 20ab 5 c 30. 16ab 2 c 4 10a 9 b 2 c 31. Given the triangle to the right with the sides as labeled a. Find the expression for the perimeter of the triangle and explain how you found it. 2x + 3 x – x2 x2 + x + 5 b. If the perimeter is 52 cm find the value of x Show your work algebraically and explain how you found your answer. c. Find the lengths of the three sides and explain how you found your answer. 32. Given the triangle to the right with the sides as labeled a. Find the expression for the area and explain how you found it. b. If the area of the triangle is 144 in2 find the value of x and explain how you found the answer. c. Find the base and height of the triangle. x 2x Unit 6 Section 6 Objective The student will multiply a monomial times a polynomial. Polynomials can have one or more terms. So this means we will multiply a monomial times a binomial, a trinomial, or higher number of terms. The method for doing the multiplication is to use the Distributive Property. The Distributive Property tells us we can multiply through a parentheses term by term. The examples below show us how to use the Distributive Property. Example A 2x(x3 – 9x) 2x4 – 18x2 Example B Example C -3a2b(2a – 4b + 1) -6a3b + 12a2b2 – 3a2b -(7ab – 11b3) -1(7ab – 11b3) -7ab + 11ab3 In example C we have insert the understood exponent of 1 and then multiply. This has the effect of changing all the signs. VIDEO LINKS: Khan Academy Multiplying Monomials by Polynomials Exercises Unit 6 Section 6 Use the Distributive Property to simplify the following: 1. x(2x - 4) 2. 5a(7a2 – 9a – 1) 3. -3y(5y - 11) 4. 2ab(a2 + 3ab - 4b2) 5. -6a2(4a3 + 8ax - 2x2) 6. –(-4x2 + 6x – 2) 7. 8a2x3(3a2x + ax - 3x4) 8. y3(x2y – xy + y2) 9. a(8a + 5ab) 10. a-2(4a2 - 3a + 6a-1) Simplify and add or subtract as indicated. 11. 2 2 2 5z + 11z - 12z 12. -3a - 8a - 2a 13. -8x + 9 + x - 5 + 12x 14. 15. x(x + 3y) + 4x2 - 6xy 16. -4(a2 - 5) + 4a2 - 20 2x - 5x2 - (-7x) + 3 - x2 Complete the following 17. If we want to multiply monomials then we should ____________________ Multiply 18. x 3 . 4 x 3 19. x 7 3 . 5. x . x 3 21. (-5a b2c)(-3a bd) . 20. a5 a3 a 4 3 2 2 22. (x )(-5y z)(-3x z ) (-xz ) Simplify the following expressions 23. x7 x3 24. x8 x 25. 16a 4 x 6 y 4ax 5 y Simplify the following expressions and list your answer in fractional form ( do not use negative exponents ) 26. x5 x9 27. x x6 x5 x6 28. Write each fraction with negative exponents where appropriate. 29. 1 x8 30. 2 3 x yz 4 31. x-2y-1 34. 4a-5b3c-2d 5b 3 c4 Write each monomial as a fraction. 32. a-5 33. Simplify 35. a0 36. -70 37. a4 a4 Perform each operation and list your answer with negative exponents where needed. 38. (x-2)(x-6) 39. x-2 . x2 40. 42. Given the rectangle to the right with the sides as labeled find the expression for the area and explain how you found it. x 4 x 1 41. x 3 x 3 4x2 - 5x + 2 7x2 43. What is the product of 6x5 – 2x4 + x3 – 9x2 and -5x-2 a. -30x3 + 10x2 – 5x + 45 b. -30x3 + 10x2 – 5x c. -30x7 + 10x6 – 5x5 + 45x4 d. 30x3 – 10x2 + 5x – 45 Unit 6 Section 7 Objective The student will find the Greatest Common Factor of monomials. To find the greatest common factor (GCF) of monomials we need to review the method for finding the GCF of any two numeric values. While there are several ways to do this the best method involves factoring the numbers to primes. A prime number is divisible only by one and itself. The factoring is best shown by factor trees. Below are examples of factor trees. Example A Example B 12 2 2 . Example C 42 . 6 . 2 2 3 2 . . 3 -45 21 . 7 -1 . 5 . 9 . 3 . -1 5 . 3 We can use these factor trees to find the greatest common factor for sets of numbers. Once we have the factor trees for our values then we look to use the pairs of numbers that the two trees have in common and multiply these factors together. Below are examples of finding the CGF for pairs of values. Example A 12 Example B 18 -30 -20 4 . 2 . 2 . 3 2 . 3 . 3 -1 . 2. 3 . 5 2 . 2 . 3 2 . 3 . 3 -1 . 5 . 3 =6 3 2 GCF = 2 . 9 -1 -1 . 2 . 2 . 15 2 .3 . -1 . 2 -1 . . 2 . 10 2 . . 2 5 . 5 GCF = -1 . 2 . 5 = -10 In these examples once we factored the numbers, we circle and joined the pairs of common factors. We also listed them in the GCF and then we multiplied. We must also find the greatest common factors of variable terms as well as numbers. The process can be similar to working with plain numbers. We factor the terms completely and then find what the two terms have in common. The examples below show us how to find the GCF. Example A Example B x5 x2 y3 y4 x ∙ x ∙ x ∙ x∙ x x ∙ x y ∙ y ∙ y y ∙ y ∙ y ∙ y x ∙ x ∙ x ∙ x∙ x x ∙ x y ∙ y ∙ y y ∙ y ∙ y ∙ y GCF = x ∙ x = x2 GCF = y ∙ y ∙ y = y3 The conclusion we can draw from these examples is that the greatest common factor of variable terms can be found by using the lowest exponent from the variable. To find the greatest common factor of variable terms, use the lowest exponent for each variable the two terms have in common. Below are some examples of applying this conclusion. Example A a7 a4 GCF = a4 Example B b Example C b5 x 4 y2 GCF = b Example D xy5 a2bc6 GCF = xy2 bc3d GCF = bc3 We must use the prime factorization for the coefficients, and then find the GCF for the variables using the lowest exponents for each variable when we work with monomials. The examples below show use how to use this process. Example A 14ab3 21a2b3 GCF = 7ab3 VIDEO LINK: Example B -15x2y3 -12y2z5 GCF = -3y2 Example C 8ax3 12ax -16a4x2 GCF = 2ax Khan Academy Monomial Greatest Common Factor Exercises Unit 6 Section 7 1. What does the abbreviation GCF stand for? Use factor trees to find the GCF for each set of numbers. ( Show the work ) 2. 20 12 3. 28 42 4. 45 30 5. -24 -18 6. 98 -42 7. 55 44 7 9. 11 8. 21 13 Find the GCF for each set of variable terms 10. abc bcd 11. a2b4c 12. x5y x5y3 13. ab a3bc5 xyz2 Find the GCF for each set of monomials 14. 16. 12a2 -6x3y4z5 18. 100mn4 20. 32ab 14a2 15. 18ab3 -15x2y5z 17. 8ab 16a5b 30n5 19. -9x3y6 -21x2yz 21. 5xy 8xy 12a3b5 7wx Put the missing factor in the blank. 22. 4 25. . ____ = 12 23. 3 . ____ = 18 24. 5 . ____ = 25 . ____ = -40 2 . ____ = 24 26. -2 . ____ = -16 27. 4 28. -7 . ____ = 28 29. 9 . ____ = -9 30. 5a 33. x 31. 4a2 . ____ = 24a5 32. ax . ____ = a5x3 34. 3x2 . ____ = 18x4 35. -2b3 . ____ = 2b6c . . ____ = 15a3 ____ = x2y Unit 6 Section 8 Objective The student will perform common factoring on polynomials. Common factoring is a way to reverse the Distributive Property. To reverse the Distributive Property, we first have to find the greatest common factor between the terms of the polynomial. This GCF will be placed on the outside of a parenthetical expression. We then need to figure out the terms that belong in the parentheses. This process is illustrated below. Example A 4x2 + 3x x x( ) x(4x ) x(4x + 3) Example B The given binomial This is the GCF. Which we use as the factor outside the parentheses. Since x times 4x is 4x2 the 1st term will be 4x. Because x times 3 is 3x the 2nd term will be 3. 10ax2 + 14ax The given binomial 2ax This is the GCF. 2ax( ) Which we use as the factor outside the parentheses. 2ax(5x ) Since 2ax times 5x is 5ax2 the 1st term will be 5x. 2ax(5x + 7) Because 2ax times 7 is14ax the 2nd term will be 7. Example C -9x3 – 6xy -3x -3x( ) 2 -3x(3x ) 2 -3x(3x + 2y) The given binomial This is the GCF. Which we use as the factor outside the parentheses. Since -3x times 3x2 is -9x3 the 1st term will be 3x2. Because -3x times 2y is -6xy the 2nd term will be 2y. Example D 7a3 – 14a2b – 21ab 7a 7a( ) 2 7a(a ) 7a(a2 – 2ab ) 2 7a(a – 2ab – 3b) The given trinomial This is the GCF. Which we use as the factor outside the parentheses. Since 7a times a2 is 7a3 the 1st term is a2. Because 7a times -2ab is -14a2b the 2nd term is -2ab. Finally 7a times -3b is -21ab the 3rd term is -3b. We can check to see if our answer is correct by using the Distributive Property and multiplying through the parentheses. We will check Example B below. 2ax(5x + 7) 10ax2 + 14ax The binomial is the same as what we started with so the factoring is correct. VIDEO LINK: Kahn Academy: Factoring and the Distributive Property 2 Exercises Unit 6 Section 8 Perform common factoring on the following polynomials (reverse the Distributive Property). If the polynomial cannot be factored write “prime”. 2 1. 12a - 15b 2. -4x – 10 3. x - x z 4. 28ab + 7a 5. 4aby - 12by2 6. 8y2z + 2yz 8. 2a2c4 – 10a3c3 9. 2x2 - 9xyz 11. 3ax2 – 5ax + 2a 12. -40a c - 10ac4 14. 5cd -10c2d + 15cd2 15. 11x + 13y 3 7. 18x + 12x2 2 10. 4x2 + 20x z 2 13. 6x - 10x - 18 3 2 2 3 3 2 16. -12x - 2x – 4x 17. 6x y - 22x y + 2xy3 18. x2y + x3y2 – x4y 19. 16v2 – 24vw 20. 24x4 – 18x3 + 36x2 + 12x Write each fraction with negative exponents where appropriate. 21. 1 z4 22. 4 y8 23. 1 a 24. 3x 5 y2 25. a b 3c 26. 7 ax 2 3 yz Write each monomial as a fraction. 27. v-9 28. bx-8 29. u-1 30. a-1b-2 31. -4wx-3y 32. (5x)-1 Divide and list your answer as both a fraction and with negative exponents. 33. 27 axy 9a 2 x 5 y 34. 5bc 5 d 20b 5 c 36. The best way to factor 18x6y2 – 12x3y8 a. 6x3y2(3x2y – 4xy4) b. 3(6x6y2 – 4x3y8) c. 6x3(3x3y2 – 2y8) d. 6x3y2(3x3 – 4y6) 35. 16a 3 b 7 c 4 14a 3 b 2 c Unit 6 Section 9 Objective The student will multiply binomials of the form (x + a)(x + b) where a and b are constants. Multiplying binomials is really applying the Distributive Property twice. To see how this works we will use constants and try the multiplying. Example A 43 x 25 215 86 1075 In example A we use the traditional algorithm for multiplying two digit numbers. Example B 40 + 3 20 + 5 200 + 15 800 + 60 1000 + 75 1075 In example B we use a place value representation for multiplying. We do 5 x 3 to get 15. 5 x 40 to get 200 20 x 3 to get 60 20 x 400 to get 800 When we add them up we get 1075. Example C (40 + 3)(20 + 5) 800 + 200 + 60 + 15 1075 In example C we still use place value representation for multiplying. We are doing the multiplication with the Distributive Property. The 40 must be multiplied times both the 20 and 5. The 3 must be multiplied times the 20 and 5 as well. We get the same answer from all three representations. The difference is that if we use the method in Example C we can do the multiplication with variables as well as constants. Example C has arrows that show us how to use the Distributive Property. The first product was from (40)(20), the second was (40)(5). In this way we multiplied the first term times both terms in the second binomial. Then we used the 3 and multiplied it by the 20, and then the 3 times 5. Now that we have used the Distributive Property with both the terms in the first binomial we can add to find our answer. The order in which we do the multiplying is critical to much of what we will do in the future. Below is an illustration of this order. Given the binomials (x + a)(x + b) We start with (x + a)(x + b) We multiply the Then we multiply (x + a)(x + b) We multiply the OUTER terms in each binomial. Next comes (x + a)(x + b) We multiply the INNER terms in each binomial. Last we multiply (x + a)(x + b) We multiply the LAST terms in each binomial. This gives us x2 + bx + ax + ab as an answer. FIRST terms in each binomial. This way of multiplying binomials is actually the exact same method we saw in example C above. When we have variables, we can’t add all the terms up. Instead we will get a polynomial as an answer. There is an abbreviation we can use to help us remember how to multiply the binomials. This order for multiplying binomials is F.O.I.L. for First, Outer, Inner, and Last. Below are some examples of using the FOIL method to multiply binomials. Example A Given (x + 5)(x + 7) find the product. (x + 5)(x + 7) First terms give us x2 Outer terms give us 7x Inner terms give us 5x Last terms give us 35 The answers written horizontally are x2 + 7x + 5x + 35 We can combine like terms and get our final answer as x2 + 12x + 35 Example B Given (x – 4)(x – 3) find the product. (x – 4)(x – 3) The answer from FOIL written horizontally is x2 – 3x – 4x + 12 We can combine like terms and get our final answer as x2 – 7x + 12 Example C Given (a – 7)(a + 2) find the product. (a – 7)(a + 2) The answer from FOIL written horizontally is a2 + 2a – 7a – 14 We can combine like terms and get our final answer as a2 – 5a – 14 Example D Given (a – 6b)(a + 9b) find the product. (a – 6b)(a + 9b) The answer from FOIL written horizontally is a2 + 9ab – 6ab – 54b2 We can combine like terms and get our final answer as a2 + 3ab – 54b2 VIDEO LINK: Youtube: Using FOIL Exercises Unit 6 Section 9 Complete the following. 1. If we are multiplying binomials then we must apply the ____________________ Property twice. 2. The abbreviation we can use to tell us how to multiply binomials is _______________ 3. The abbreviation FOIL stands for _________________________________________ Multiply the following binomials: 4. (x + 7)(x + 3) 5. (x + 10)(x - 6) 6. (x + 9)(x + 2) 7. (a - 8)(a - 5) 8. (y + 7)(y + 4) 9. (a + 10)(a - 3) 10. (x - 5z)(x - 5z) 11. (x - 6)(x + 6) 12. (a + 1)(a - 9) 13. (a + 8b)(a - 8b) 14. (y + 7)(y + 7) 15. (x - 4y)(x – 4y) 16. (x – 11)(x + 4) 17. (z – y)(z – 6y) 18. (b + 12)(b + 1) 19. (w – 2z)(w + 9z) 20. (a + 1)(a + 2) 21. (y – 1)(y – 1) 22. (t + 4s)(t + 5s) 23. (y + 10)(y – 12) 24. (v – 3w)(v – 3w) 25. (a – 8)(a + 7) 26. Given the rectangle to the right as labeled Find the expression for the area and explain how you found it. x + 15 x-2 Unit 6 Section 10 Objective The student will multiply binomials of the form (ax + b)(cx + d) where a,b,c and d are constants. This section still involves binomial multiplication. The first terms in each binomial variable can have coefficients, which does not change the process, although it will change our product. Below are examples of binomial multiplication with these coefficients. Example A Example B (2x + 5y)(3x – y) (4x2 – 5z)(-7x2 + 3z) 6x2 – 2xy + 15xy – 5y2 -28x4 + 12x2z + 35x2z – 15z2 6x2 +13xy – 5y2 -28x4 + 47x2z – 15z2 Example C Example D (2a3 – 9)(5a3 – 4) (3y2 + 2z3)(4y2 + z3) 10a6 – 8a3 – 45a3 + 36 12y4 + 3y2z3 + 8y2z3 + 2z6 10a6 – 53a3 + 36 12y4 + 11y2z3 + 2z6 In these examples we see that the resulting trinomial has a leading coefficient. We can also see that the variables in the binomials have exponents other than the understood one. We must be careful how we apply the rule for multiplication and add the exponents when appropriate. VIDEO LINK: Youtube: Using FOIL Exercises Unit 6 Section 10 Multiply the following binomials: 1. (2x + 5)(x + 5) 2. (6z – 1)(7z – 2) 3. (x + 7)(4x – 3) 4. (-2x + 1)(3x – 8) 5. (3x + 2y)(5x – 2y) 6. (a – 8b)(9a - b) 7. (5y2 + 7)(y2 + 4) 8. (9a2 + b2)(3a2 – 4b2) 9. (x - 2)(x - 11) 10. (x - 8)(x + 2) 11. (a + 4)(a - 5) 12. (a + 12)(a - 12) 13. (y + z)(y + z) 14. (x - 6y)(x – 10y) 15. (x – 9)(x + 5) 16. (z + 15y)(z – y) 17. (b + 7)(b + 4) 18. (ab – 3z)(ab + z) 19. (a – 1.5)(a + 1.5) 20. (a – 1 2 )(a + ) 3 3 Multiply 21. x 24. 5 . 3 . a 6 x 22. x 2. . . 4 b b b a 4 . 2. x . . 23. a2 a2 a2 x 4 25. (-2x5y2z)(7w xy) 26. (-3x)(-4x)(2x)(x)(x)(x) Use the Distributive Property to simplify the following: 27. 2(3a - 11) 28. 4z(3z2 – z – 1) 29. -8y(2y2 – 10y) 30. 2ab(a2b+ 5ab2 - ab) 31. a2(a3 + 8a2 – 2a) 32. –(x2 – 12x + 5) Perform common factoring on the following polynomials (reverse the Distributive Property). If the polynomial cannot be factored write “prime”. 33. 14z - 16 2 36. 5x2 + 25x y 34. -6x2 – 18x 35. x3y – x2y2 + xy3 37. 3ax2 – 5ax3 + 2ax4 38. -4a3c - 18a 39. Given the rectangle to the right as labeled a. Find the expression for the area and explain how you found it. 5x - 2 3x - 4 b. If x = 5 cm use the expression you found in part ‘a’ to find the area. Show your work algebraically. c. Calculate the lengths of the sides in the rectangle with x = 5 cm and then find the area. d. Compare your two answers and explain the result. Unit 6 Section 11 Objective The student will multiply special binomial products. There are two special binomial products. The first one we will study is a binomial square. To understand how we should multiply out a binomial square we need to review how exponents work. An exponent of two, means to write down the factor twice and multiply as shown below. x2 = x ∙ x or (3x)2 = 3x ∙ 3x When we have a parentheses raised to a second power we write down the expression in the parentheses twice and multiply. The binomial will work the same way. (x + 4)2 = (x + 4)(x + 4) One way to multiply out the binomial square then is to write the binomial twice and use the FOIL process. Examples of this process are given below. Example A Example B (a – 5)2 (a – 5)(a – 5) 2 a – 5a – 5a + 25 a2 – 10a + 25 (3x + 7y)2 (3x + 7y)(3x + 7y) 9x2 + 21xy + 21xy + 49y2 9x2 + 42xy + 49y2 To multiply binomial squares we write the binomials twice and use the FOIL process. The second special binomial product is called “sum and difference binomials. The corresponding terms in the binomials are the same except that one of the binomials is a sum and the other binomial is a difference. The examples below are sum and difference binomials. (x + 7)(x – 7) (2w – 11y)(2w + 11y) (5z2 + 3)( 5z2 – 3) In each example above the binomials are the same except for the operation, addition versus subtraction. We can use the FOIL process to multiply out the sum and difference binomials. The examples below show us how to do the multiplication. Example A Example B (x + 9)(x – 9) (2w – 5)(2w + 5) x2 – 9x + 9x - 81 4w2 + 10w – 10w – 25 x2 – 81 4w2 – 25 In example A we can see that the middle terms in the product were opposite in sign and added up to zero. We don’t have to write the 0 so we get a binomial answer. In example B we can see that the middle terms in the product were opposite in sign and added up to zero. We don’t have to write the 0 so we get a binomial answer. Example C (3a2 + 4b)(3a2 – 4b) 9a4 – 12a2b+ 12a2b – 16b2 9a4 – 16b2 In example C we can see that the middle terms in the product were opposite in sign and added up to zero. We don’t have to write the 0 so we get a binomial answer. In each of these examples the middle terms added up to zero and we did not need to write anything because they cancelled out. This means that there is a short cut for finding the product of sum and difference binomials. Because the middle terms will always cancel out, the FOIL process can become FL or just First and Last. The terms from multiplying the Outer and Inner parts of the binomials will always add up to zero. Some examples of sum and difference binomial multiplication are given below. Example A Example B Example C (a – 10)(a + 10) (2x + 11)(2x – 11) (3y2 +5z)(3y2 – 5z) a2 – 100 4x2 – 121 9y4 – 25z2 VIDEO LINK: Khan Academy Special Products of Binomials Exercises Unit 6 Section 11 Examine each problem below and decide if it is a “sum and difference” problem, a “binomial square” or a problem which can be worked just with “FOIL”. Do not work the problem, just list the problem type. ( Use the quoted words in the instructions to identify the problems. ) 1. (x + 4)(x – 4) 2. (y + 5)(y – 7) 3. (3x – y)(2x + y) 4. (z + 8)2 5. (9a – 2b)(9a + 2b) 6. (4w2 – 3v)2 7. (q + u)(q – u) 8. (a – 10)(a – 9) 9. (10 – 6z)2 Multiply the following sum and difference binomials. 10. (x + 3)(x – 3) 11. (2w – 5z)( 2w + 5z) 12. (7a + 9)(7a – 9) 13. (a + b)(a – b) 14. (x3 – 4)(x3 + 4) 15. (8 – a)(8 + a) 16. (11m2 + 3n2)(11m2 – 3n2) 18. (w + 12)(w – 12) 17. (6ab – 10c)(6ab + 10c) 19. (1.5x + .7)(1.5x – .7) 20. ( 1 1 x + 2)( x – 2) 2 2 Multiply out the following binomial squares. 21. (x + 11)2 22. (a – 6)2 23. (a + 3b)2 24. (2b + 9)2 25. (-3y + 5)2 26. (a + b)2 27. (x + 6)(x – 11) 28. (a + 2)(a + 15) 29. (b – 7)(b – 8) 30. (y – 9z)(2y – 3) 31. (5n + 8)(n + 1) 32. (z + 2)(7z – 4) 33. (2w + 8)(3w – 1) 34. (6x + 1)(3x – 1) 35. (d + 1)(d – 1) 36. (w + 8y)2 37. (3y – 10)2 38. (2a – .5)(2a + .5) Multiply the following 39. Given the rectangle to the right with the dimensions as labeled. Find the area of the rectangle using the dimensions. a. b. c. d. 3x2 – x – 4 3x2 + x + 4 3x2 + x – 4 4x – 3 3x - 4 x+1 40. Given the triangle to the right with the dimensions as labeled. Find the area of the triangle using the dimensions. a. b. c. d. 4x2 – 11x – 3 4x2 + 11x + 3 4x2 + 11x – 3 8x2 – 22x - 6 2x + 6 4x - 1 41. The expression for the side of a square is given as 2x + 7. Find the expression for the area of the square and explain how you found the answer. Unit 6 Section 12 Objective The student will multiply polynomials. We will at times need to multiply a binomial times a trinomial or perhaps two trinomials. We should be able to multiply any polynomial times any polynomial. To multiply any polynomials we simply need to use the Distributive Property. Each term from the first polynomial must be multiplied times all the terms in the second polynomial. The example below illustrates this process. Example A Multiply (x + 4)(x2 – 5x + 7) (x + 4)(x2 – 5x + 7) We begin by multiplying the ‘x’ times the second polynomial we get x3 – 5x2 + 7x. (x + 4)(x2 – 5x + 7) We continue by multiplying ‘4’ times the second polynomial we get 4x2 – 20x + 28 When we combine all the products we get x3 – 5x2 + 7x + 4x2 – 20x + 28 Now we can combine the like terms. The final answer is x3 – x2 – 13x + 28 Example B Expand the following (x – 5)3 ( expand means to multiply out ) This means (x – 5)(x – 5)(x – 5) So we multiply one set of binomials and get x2 – 10x + 25 The problem then becomes (x – 5)(x2 – 10x + 25) (x – 5)(x2 – 10x + 25) We begin by multiplying the ‘x’ times the second polynomial we get x3 – 10x2 + 25x. (x – 5)(x2 – 10x + 25) We continue by multiplying ‘-5’ times the second polynomial we get -5x2 + 50x – 125 When we combine all the products we get x3 – 10x2 + 25x – 5x2 + 50x – 125 Now we can combine the like terms. The final answer is x3 – 15x2 + 75x – 125 VIDEO LINK: Khan Academy Multiplying Polynomials Example Exercises Unit 6 Section 12 Multiply the following polynomials 1. (x+ 2)(x2 – 7x + 4) 2. (2x + 1)(x2 + 6x – 3) 3. (a – 4)(a2 + 5a + 9) 4. (a + b)(a2 + 8ab + 12b2) 5. (3y – 4)(y3 + 2y2 + y) 6. (a +2b + 1)(2a – b + 3) 7. (x + 3)3 8. (2a – 1)3 Multiply the following binomials 9. (x + 7)(x + 10) 10. (a – 2)(a + 9) 11. (z + 3)(z – 11) 12. (m – 6)(m – 8) 13. (2x + 7)(x + 12) 14. (a + 4b)(a – 4b) 15. (n + 6)2 16. (x + y)(3x -8y) 17. (2z – 5)(2z + 5) 18. (3n – 1)2 19. (a2 + 5)(a2 – 8) 20. (5 – 9a)(5 + 9a) Perform common factoring in the problems below. If the polynomial cannot be factored write “prime”. 21. 16x – 12y 22. -9x2 + 3x 23. x3y – x2z 24. 15a2c4 – 10a3c3 25. 2w2 - 9xyz 26. 27. 14ax3 – 21ax2 + 28ax 28. 6x - 21x - 15 2 3x2 + 27x2z 3 Divide and list your answer as both a fraction and with negative exponents. 30. 32ax 5 y 4a 4 x 5 y 31. bc 5 b 5 c 1 18ab 2 c 4 32. 9ab 7 c 33. Any number or variable raised to the zero power is _______ x5 34. What does 5 x simplify to? 2 29. -16x - 8x – 6x 35. The expression for the side of a cube is given as 3x – 1. Find the expression for the volume of the cube and explain how you found it. 36. Given the box to the right as labeled Find the expression for the volume of the cube and explain how you found it. 3x + 5 2x + 1 2x - 1 37. A fish tank has a width of x inches, a length of 3x inches and a height of x + 10 inches. Find the expression for the volume of the tank and explain how you found it.