Section P.1. Properties of Exponents. Chapter P – Polynomials

Section P.1: Properties of Exponents Chapter P – Polynomials #1 – 14: Identify the base and the exponent and the coefficient in the following expressions 1) 5x3 Solution: Coefficient 5, Base x, exponent 3 3) 3x4 5) -y6 I need to think of this problem as -1y6 Solution: Coefficient -1, Base y, exponent 6 7) -11x 9) 3(9x-4)2 Solution: Coefficient 3, Base 9x-4, exponent 2 11) -(2x-1)5 I need to think of this problem as -1(2x-1)5 Solution: Coefficient -1, base 2x – 1, exponent 5 13) (2x-3)2 #15 – 32: Simplify 15) x3*x2 = x3+2 Solution: x5 17) a*a3 19) 3x2*2x5 = 3*2*x2x5 (multiply the 3 and 2 and add the exponents of the x’s) Solution: 6x7 21) (-2y)(3y4) (multiply the -2 and 3 and add the exponents of the y’s) Solution: -6y5 23) xy3*x2y4 25) abc*a2b4c 27) (3ab2)(-4a3b) = 3(-4)aa3b2b Remember that the letters without exponents written have exponents of 1. 3(-4)a1a3b2b1 Solution: -12a4b3 29) (3a2b)(4b) 31) (8st)(s4t) #33-52: Simplify, assume no denominator equals 0 33) 𝑥4 𝑥 Subtract the exponents, think of this as 𝑥4 𝑥1 Solution: x3 35) 𝑧3 𝑧2 Subtract the exponents. Solution: z 37) 25 22 Subtract the exponents to get 23, this would be a correct answer, but it would be better to write the answer without an exponent. Solution: 8 39) 74 72 41) 𝑥 3𝑦4 𝑥 2𝑦3 Just subtract the exponent of each letter separately. Solution: xy 43) 𝑥 5𝑦7 𝑥 2𝑦3 Just subtract the exponent of each letter separately. Solution: x3y4 45) 20𝑥 4 5𝑥 Divide the 20 and 4, subtract the exponents of the x’s Solution: 5x3 47) 18𝑥𝑦 3 12𝑦 49) (5𝑥)4 (5𝑥)3 Here the (5x) is the base, I will just subtract the exponents. I shouldn’t divide the 5’s. When I use the rules of exponents the bases don’t change only the exponents change. This reduces to: (5x)1 I don’t need to write the exponents, nor do I need the parenthesis. Solution: 5x 51) (2𝑥𝑦)8 (2𝑥𝑦)3 Here the (2xy) is the base, I will just subtract the exponents. I shouldn’t divide the 2’s. When I use the rules of exponents the bases don’t change only the exponents change. This reduces to: (2xy)5 This is an okay answer. It would be better to clear the parenthesis. = 25x1*5y1*5 Solution: (2xy)5 or 32x5y5 #53-76: Simplify 53) (x3)2 This is a one base / 2 exponent problem. Reduce by multiplying the exponents. Solution: x6 55) (z4)3 57) (x2y3)2 Multiply each of the exponents by 2. Solution: x4y6 59) (yz3)7 Multiply each of the exponents by 7. Remember to think of this as (y1z3)7 Solution: y7z21 61) (3x2)4 63) (4x3y)2 Square the 4 and multiply the exponents of the letters. = 42x3*2y1*2 Solution: 16x5y2 65) (5xy)2 𝑥 3 67) (𝑦2 ) To clear the parenthesis multiply each exponent by 3. Keep the fraction. 𝒙𝟑 𝒚𝟔 Solution: 𝑥𝑧 2 4 69) ( 𝑦2 ) To clear the parenthesis multiply each exponent by 4. Don’t forget the x inside the parenthesis has an exponent of 1. Keep the fraction. Solution: 𝑥 4𝑧8 𝑦8 2𝑥 4 71) (𝑦2 ) Raise the 2 to the 4th power and multiply the exponents of the x and y. = 24 𝑥 1∗4 𝑦 2∗4 𝟏𝟔𝒙𝟒 𝒚𝟖 Solution: 7𝑧 2 2 73) (5𝑦2 ) Raise the 5 and the 7 to the 2nd power and multiply the exponents of the x and y. Solution: 𝟒𝟗𝒛𝟒 𝟐𝟓𝒚𝟒 5𝑥 3 ) 6𝑦 2 75) ( Section P.1: Properties of Exponents Chapter P – Polynomials #77 – 86: Simplify 77) (xy2)3(x2y) Clear the first parenthesis first by multiplying the exponents by 3. = (x3y6)(x2y) Now add the exponents of the x and y’s. Solution: x5y7 79) (3xy3)(4x2y)3 Simplify the second parenthesis first. =(3xy3)(42x2*3y1*3) =(3xy3)(16x6y3) Now multiply the numbers and add the exponents of the x and y. Solution: 48x7y6 81) (2xy3)4(3x2y)2 Simplify both parenthesis on the first step. (24x1*4y3*4)(32x2*2y1*2) =(16x4y12)(9x4y2) Multiply the numbers add the exponents of the x and y Solution: 144x8y14 83) (5x)2(2x3)2 85) (6y3z)(5y)2 Section P.2: Properties of Negative and Zero Exponents Chapter P – Polynomials #1-42: Simplify 1) a) 𝑥2 𝑥2 Anything divided by itself is equal to 1. You may also subtract the exponents to get x0 Solution: 1 b) x0 If you solved part (a) by subtracting the exponents you would get x2-2 which is equal to x0. x0 must equal to the answer to part a. Solution: 1 3) a) 34 34 Anything divided by itself is equal to 1. You may also subtract the exponents to get 30 Solution: 1 b) 30 If you solved part (a) by subtracting the exponents you would get 34-4 which is equal to 30. 30 must equal to the answer to part a. Solution: 1 5) a) 2*30 You can check this on your calculator since it only has numbers. By hand the 30 is equal to 1. = 2*1 Solution: 2 b) (2*3)0 Again, you can check this on your calculator. The 2 is in the parenthesis, and this really equals 20*30 which is 1*1 Solution 1 7) a) -1*30 If a number is not in a parenthesis, it doesn’t turn to 1 when you apply the 0 exponent rule. = -1*1 (only the 30 turns to a 1) Solution: -1 b) (-1*3)0 Since the -1 is in a parenthesis, the entire -1*3 turns to a 1. Solution: 1 9) a) -30 If a number is not in a parenthesis, it doesn’t turn to 1 when you apply the 0 exponent rule. Think of this as = -1*30 (only the 30 turns to a 1) = -1*1 Solution: -1 b) (-30) everything is in a parenthesis, this will equal 1. Solution: 1 11) a) -1*y0 b) (-1*y)0 13) a) -y0 b) (-y)0 15) a) 5*60 = 5*1 Solution: 5 b) (5*6)0 everything is in a parenthesis, this will equal 1. Solution: 1 17) a) 2x0 b) (2x)0 19) a) 3xy0 Only the y0 turns into a 1 = 3x*1 Solution: 3x b) (3xy)0 21) a) 3x0y b) (3x0y)0 Since the entire 3x0y is in the parenthesis, it all turns to 1 when the 0 exponent rule is applied. Solution: 1 23) x0 25) 30 27) -30 I can simplify this using my calculator. By hand, I need to think of this as -1*30, and only the 30 turns into a 1. = -1*30 = -1*1 Solution_ -1 29) (-2)0 I can simplify this using my calculator. By hand, since the entire -2 is in a parenthesis the entire problem simplifies to 1. Solution: 1 31) -w0 33) (-w)0 35) 2c0 37) (2x)0 39) 2bc0 41) (2xy0)0 Section P.2: Properties of Negative and Zero Exponents Chapter P – Polynomials #43 - 148: Simplify the expression. Write the answer with positive exponents only. 43) a) 3−2 This can be done using a calculator. By hand I need to make the exponent positive by creating a fraction with a 1 in the numerator. 1 = 32 Solution: 𝟏 𝟗 b) x-2 I need to make the exponent positive by creating a fraction with a 1 in the numerator. Solution: 𝟏 𝒙𝟐 45) a) 3-3 b) z-3 47) 4-3 49) x-7 51) a) b) 1 3−2 1 𝑥 −2 53) a) 2 3−2 This can be done using a calculator. By hand I need to make the exponent positive by moving the 3-2 up to the numerator. = 2*32 Now order of operations requires me to do the 32 first. = 2*9 Solution: 18 b) 2 𝑥 −2 I need to make the exponent positive by moving the x-2 up to the numerator. Solution: 2x2 55) 7 5−1 57) 2 𝑥 −5 59) 𝑥3 𝑥 −5 I need to make all of the exponents positive. I can do this by moving the x-5 up to the numerator. My answer will not be a fraction as I will have taken the only thing out of the denominator away from the denominator. = x3*x5 Solution: x8 61) 𝑥2 𝑥 −6 63) 𝑥3 𝑥 −5 65) a) 3−2 2−3 I can simplify this on my calculator. By hand move the 3-2 down to the denominator, and the 2-3 up to the numerator. 23 = 32 𝟖 Solution: 𝟗 b) 𝑥 −2 𝑦 −3 67) 2−3 5−1 69) 𝑥 −3 𝑦 −1 By hand move the x-3 down to the denominator, and the y-1 up to the numerator. Solution: 71) 𝑥 −3 𝑥 −1 73) 𝑥 −5 𝑥 −1 75) a) 𝒚 𝒙𝟑 3−2 33 I need to move the 3-2 down to the denominator. I will leave a 1 in the numerator as I can’t write a fraction without a numerator. 1 = 33 ∗32 now I will add the exponents 1 = 35 I will use my calculator to simplify 35 Solution: 𝟏 𝟐𝟒𝟑 𝑥 −2 𝑥6 b) I need to move the x-2 down to the denominator. I will leave a 1 in the numerator as I can’t write a fraction without a numerator. 1 = 𝑥6𝑥2 Solution: 77) 𝟏 𝒙𝟖 𝑥 −2 𝑥5 79) a) 2∗3−2 5−2 I will move the 3-2 down and the 5-2 up. = 2∗52 32 = 2∗25 9 Solution: b) 𝟓𝟎 𝟗 4𝑥 −2 𝑦 −3 I will move the x-2 down and the y-3 up. Solution: 81) 3𝑥 −1 𝑦 −3 83) 3𝑥 −1 𝑥 −3 𝟒𝒚𝟑 𝒙𝟐 Section P.2: Properties of Negative and Zero Exponents 85) a) Chapter P – Polynomials 3∗22 5−2 I will move the 5-2 up to the numerator. I won’t need to write a fraction since I have taken everything away from the denominator. = 3*22*52 =3*4*25 Solution: 300 b) 4𝑥 2 𝑦 −3 87) a) 3∗2−2 52 I will move the 2-2 down to the denominator. Nothing else has a negative exponent. Nothing else moves. 3 = 52 ∗22 3 = 25∗4 Solution: b) 𝟑 𝟏𝟎𝟎 4𝑥 −2 𝑦3 I will move the x-2 down to the denominator. Nothing else has a negative exponent. Nothing else moves. Solution: 𝟒 𝒙𝟐 𝒚𝟑 89) 2𝑥 −3 𝑦 91) 2𝑥 3 𝑦 −5 93) 6𝑥 −2 𝑦 −6 21𝑥 5 𝑦 −4 Move the x-2, y-6 and y-4 6𝑦 4 = 21𝑥 5 𝑥2 𝑦6 6 Reduce the 21, then subtract the exponents on the letters. The y winds up in the denominator since the bigger exponent of the y’s is in the denominator. The x winds up in the denominator since both x’s are in the denominator. Add the exponents of the x’s since they are on the same level of the fraction. Solution: 𝟐 𝟕𝒙𝟕 𝒚𝟐 95) 9𝑥 2 𝑦 6 30𝑥 −5 𝑦 −4 Only move the x-5 and y-4 = 9𝑥 2 𝑥 5 𝑦 6 𝑦 4 30 9 Reduce the 30 and add the exponents of the x and y. Solution: 97) 2𝑥 −3 𝑥 99) 2𝑥 3 𝑥 −5 101) a) 𝟑𝒙𝟕 𝒚𝟏𝟎 𝟏𝟎 23 25 Subtract the exponents, leave the 2 in the denominator as that is where the larger exponent is. I have to leave 1 in the numerator. = 1 22 Solution: b) 𝟏 𝟒 𝑥3 𝑥5 Subtract the exponents, leave the x in the denominator as that is where the larger exponent is. I have to leave 1 in the numerator. 1 = 𝑥 5−3 Solution: 103) 𝑦2 𝑦6 105) 𝑥𝑦 3 𝑥2𝑦 𝟏 𝒙𝟐 Subtract the exponents of the x and y. The x winds up in the denominator, the y in the numerator. When you divide with positive exponents you subtract the exponents and leave the letter where the larger exponent starts. 𝑦 3−1 = 𝑥 2−1 Solution: 𝒚𝟐 𝒙 107) a) 2*3-2 Create a fraction and move the 3-2 to the denominator. Leave the 2 in the numerator since it does not have a negative exponent. 2 = 32 Solution: 𝟐 𝟗 b) 2x-2 Create a fraction and move the x-2 to the denominator. Leave the 2 in the numerator since it does not have a negative exponent. Solution: 𝟐 𝒙𝟐 109) 5x-3 111) 2x-1 113) x3x-4 Create a fraction and move the x-4 to the denominator. Leave the x3 in the numerator since it does not have a negative exponent. 𝑥3 𝑥4 now subtract the exponents and leave the x in the denominator Solution: 𝟏 𝒙 115) y-2y3 117) a) 3*4-1*5 Move the 4-1, but leave the 3 and the 5. = 3∗5 41 Solution: 𝟏𝟓 𝟒 b) 2𝑦 −5x Move the y-5, but leave the 2 and the x. Solution: 𝟐𝒙 𝒚𝟓 119) a) 3*2*5-2 Move the 5-2, but leave the 3 and the 2. 3∗2 = 52 Solution: 𝟔 𝟐𝟓 b) 4xy-3 Move the y-3, but leave the 4 and the x. Solution: 𝟒𝒙 𝒚𝟑 121) a) 3-1*4*5-1 Move both the 3-1 and 5-1, leave the 4. 4 = 31 ∗51 4 =3∗5 Solution: 𝟒 𝟏𝟓 b) 4-1xy-3 Move both the 4-1 and y-3, leave the x. 𝑥 = 41 𝑦3 Solution: 𝒙 𝟒𝒚𝟑 123) 6xy-4 125) 6-1xy-3 Move both the 6-1 and y-3, leave the x. 𝑥 = 61 𝑦 3 Solution: 𝒙 𝟔𝒚𝟑 127) 6x-2y-3 129) 2-1x-2y-3 131) 6x-2x-3 Section P.2: Properties of Negative and Zero Exponents Chapter P – Polynomials 133) 2-1x-2x-4 Move everything to the denominator. I will need to leave a 1 in the numerator. 1 = 21 𝑥 2 𝑥 4 Now, add the exponents of the x’s Solution: 𝟏 𝟐𝒙𝟔 34) 7-1x-2x-5 2 −4 135) a) (3) Take the reciprocal of the fraction and make the exponent positive. 3 4 = (2) Now raise the 3 and 2 to the 4th power. 34 = 24 𝟖𝟏 Solution: 𝟏𝟔 𝑥 −4 b) (𝑦) Take the reciprocal of the fraction and make the exponent positive. 𝑦 4 (𝑥 ) I need to multiply the exponents by 4. I will write the exponents in, but it is not necessary. 𝑦1 4 = (𝑥 1 ) = 𝑦 1∗4 𝑥 1∗4 𝒚𝟒 Solution: 𝒙𝟒 2 −2 ) 3∗52 137) 𝑎) ( Take the reciprocal of the fraction and make the exponent positive. 2 3∗52 ) 2 =( I can simplify the inside of the parenthesis next. 75 2 =(2) Now I need to square each number Solution: 𝟓𝟔𝟐𝟓 𝟒 −2 𝑥 b) (3𝑦2 ) Take the reciprocal of the fraction and make the exponent positive. 2 3𝑦 2 ) 𝑥 =( = 32 𝑦 2∗2 𝑥 1∗2 𝟗𝒚𝟒 𝒙𝟐 Solution: −2 3∗4−1 ) 5 139) a) ( I can simplify the inside of the parenthesis first. −2 3 = (5∗41 ) 3 −2 20 =( ) Now I can take the reciprocal to make the exponent positive. 20 2 =( 3 ) 202 = 32 Solution: 𝟒𝟎𝟎 𝟗 −2 3∗𝑥 −1 ) 5 b) ( I can simplify the inside of the parenthesis first. 3 −2 = (5𝑥) Now I can take the reciprocal to make the exponent positive. 5𝑥 2 = (3) = 52 𝑥 1∗2 32 Solution: 𝟐𝟓𝒙𝟐 𝟗 −1 3𝑥 −3 ) 𝑦 141) ( 2𝑥 −2 143) (𝑦−4 ) Simplify the inside of the parenthesis first. = (2𝑥𝑦 4 )−2 Now I can take the reciprocal to make the exponent positive. 1 = (2𝑥𝑦4 )2 1 = 22 𝑥 1∗2 𝑦4∗2 Solution: 𝟏 𝟒𝒙𝟐 𝒚𝟖 2 3𝑥 −3 ) 𝑦 145) ( 2𝑥 3 ) 𝑦 −4 147) ( Section P.3: Introduction to Polynomials Chapter P – Polynomials Definition of a monomial: A monomial is a variable, a real number, or a multiplication of one or more variables and a real number with whole-number exponents. A monomial may contain a numerical fraction, but it may not have a variable in the denominator of a fraction. #1-16: Classify the following terms as monomials or not monomials 1) 5x3 Solution: this is a monomial 3) 3x-4 Solution: this is not a monomial (because of the negative exponent) 5) -y6 Solution: this is a monomial 7) -11xy3 Solution: this is a monomial (it is okay to have more than one letter) 9) 2 𝑎𝑏 2 𝑐 3 Solution: this is a monomial (it is okay to have a number as a fraction, you just can’t have a letter in the denominator of a fraction) 11) 1 𝑦𝑥 −4 2 13) 2 5𝑥 Solution: this is not a monomial (because of the x in the denominator of a fraction) 15) 2 𝑥+5 #17 – 28: Determine the degree and coefficient of each of the following monomials 17) 8x2 Solution: this is a 2nd degree polynomial. The coefficient is the 8. 19) z3 21) 2 𝑥 3 23) 3x2y6 I need to add the exponents of the x and the y to get the degree of the term. Solution: The degree is 8. The coefficient is 3. 25) -x4y5 27) -xyz Polynomial definition: A polynomial is a monomial or the sum or difference of monomials. Each monomial is called a term of the polynomial. Important!:Terms are separated by addition signs and subtraction signs, but never by multiplication signs A polynomial with one term is called a monomial A polynomial with two terms is called a binomial A polynomial with three terms is called a trinomial #29-44: Classify the following terms as polynomials or not polynomials. If the expression is a polynomial classify it as a monomial, binomial, trinomial or other. 29) 4x + 5 Solution: this is a polynomial it is also a binomial 31) x2 + 5x +6 Solution: this is a polynomial it is also a trinomial 33) 3x-4 Solution: this is not a polynomial because of the negative exponent 35) x3 - 5 𝑥 Solution: this is not a polynomial because of the x in the denominator of a fraction 37) 𝑥+3 𝑥−4 Solution: this is not a polynomial because of the x in the denominator of a fraction 39) xy3 + 4x + 5y –7 Solution: this is a polynomial (but it is not a monomial / binomial nor trinomial) 41) 5x4 + 3xy Solution: this is a polynomial it is also a binomial 3 43) 5 + 𝑥+𝑦 Solution: this is not a polynomial because of the x and y in the denominator of a fraction Section P.3: Introduction to Polynomials Chapter P – Polynomials The degree of a polynomial is the highest of the degrees of all its terms. The leading term of a polynomial is the term with the highest degree The leading coefficient of a polynomial is the coefficient of the term with the highest degree. #45-56: Find the leading term of each polynomial, then state the degree of the polynomial and the leading coefficient. 45) 5x3 + 3y Solution: the 5x3 is the leading term, the polynomial is degree 3. 47) 3x3 + 5xyz3 The 3x3 term has degree 3, while the 5xyz3 term has degree 5 (add the exponents on each of the letters) Solution: the 5xyz3 is the leading term (because it has the highest degree), the polynomial is degree 5 49) 8x + 5x2 – x4 51) -xy5 + 2yz – 3x2y 53) 7 If I need to, I can think of this as 7x0 as 7x0 = 7*1 = 7 Solution: the 7 is the only term, thus it is the leading term. The polynomial is degree 0. 55) −1 2 #57 – 68: Evaluate each polynomial using x = 2, y = -3 and z = 4 57) 5x3 + 3y = 5(2)3 + 3(-3) = 5(8) + 3(-3) = 40 – 9 Solution 31 (it would be wrong to write x = 31, as x equals 2 in this problem) 59) 3x3 + 5xyz3 61) 8x + 5x2 – x4 63) -xy5 + 2yz – 3x2y I might think of this as -1xy5 + 2yz – 3x2y = -1(2)(-3)5 + 2(-3)(4) – 3(2)2(-3) = -1(2)(-243) + 2(-3)(4) – 3(4)(-3) = 486 – 24 + 36 Solution: 498 65) 7 Solution 7 (this is a constant function, since it doesn’t have any letters and it will always equal 7 regardless of the values of the letters) 67) −1 2 Section P.4 Addition and Subtraction of Polynomials Chapter P – Polynomials Like terms are monomials that contain the same variables raised to the same powers. #1- 16: Combine like terms. Write all answers in descending order. 1) 3x2 + 5x – 4 – 2x2 + 7x +8 = 3x2 – 2x2 + 5x + 7x – 4 + 8 Solution: x2 + 12x - 4 3) 4x3 + 5x – 2x3 + 8x + 4x3 – 9x 5) 3x + 5x2 + 11x – 5 7) 2 𝑥 3 1 1 2 1 3 + 3𝑥 − 2 + 2 1 3 Add the 3 𝑥 + 3 𝑥 then combine − 2 + 2 3 3 = 𝑥+ 2 2 Solution: x + 1 9) 4 2 𝑥 3 3 + 5 𝑥 2 + 5𝑥 − 4𝑥 + 1 Combine the x2 terms by adding the fractions, and also combine the 5x – 4x. 𝟕 𝟐 𝒙 𝟓 Solution: +𝒙+𝟏 11) 8xy + 2xy2 +3xy – 5x2y = 8xy + 3xy + 2xy2 – 5x2y (the 2xy2 and the -5x2y can’t be combined, they are not like terms) Solution: 11xy + 2xy2 – 5x2y 13) 12a2b + 5ab2 + 4a2b + 7a2b = 12a2b + 4a2b + 7a2b + 5ab2 (the 5ab2 can’t be added to the other three terms) Solution: 23a2b + 5ab2 15) 1 𝑥𝑦 2 2 2 1 + 3 𝑥𝑦 2 − 3 𝑥 2 𝑦 Only the first two terms can be added. I need to get a common denominator of 6, but only for the first two terms. 3 3 1 2 2 2 2 3 1 3 = ∗ 𝑥𝑦 2 + ∗ 𝑥𝑦 2 − 𝑥 2 𝑦 3 6 4 6 1 3 = 𝑥𝑦 2 + 𝑥𝑦 2 − 𝑥 2 𝑦 Solution: 7 𝑥𝑦 2 6 1 − 3 𝑥2𝑦 #17 – 26: Add 17) (3x + 5) + (2x – 4) = 3x + 5 + 2x – 4 Solution: 5x + 1 19) (x2 + 6x – 4) + (-5x2 – 3x + 1) = x2 + 6x – 4 + (-5x2) – 3x + 1 = x2 – 5x2 + 6x – 3x – 4 + 1 Solution: -4x2 + 3x – 3 21) (x3 + 5x2 – 3x) + (4x3 – 3x2 + 1) 23) (8xy + 4x2y + 3) + (17xy – 3x2y – 3) = 8xy + 4x2y + 3 + 17xy – 3x2y – 3 = 8xy + 17xy + 4x2y – 3x2y + 3 – 3 Solution: 25xy + x2y 25) (3ab2 – 4a2b) + (5a2b – 7ab2) #27 – 38: Rewrite each problem as a vertical addition problem, then add. 27) (3x2 + 5 – 2x) + (x – 4 +x2) 3𝑥 2 − 2𝑥 + 5 𝑥2 + 𝑥 − 4 ______________ Solution: 4x2 – x + 1 29) (x2 + 6x – 4) + (-5x2 – 3x + 1) 𝑥 2 + 6𝑥 − 4 −5𝑥 2 − 3𝑥 + 1 ______________ Solution: -4x2 + 3x - 3 31) (x2 + 5x3 – 3x) + (4x3 – 3x2 + 5x) 5𝑥 3 + 𝑥 2 − 3𝑥 4𝑥 3 − 3𝑥 2 + 5𝑥 ______________ Solution: 9x3 – 2x2 + 2x 33) (5x – 2) + (x2 + 6x – 4) 5𝑥 − 2 𝑥 2 + 6𝑥 − 4 ______________ Solution: x2 + 11x - 6 35) (y2 + 3y + 6y3) + (2y3 – 3y + 4) 6𝑦 3 + 𝑦 2 + 3𝑦 2𝑦 3 − 3𝑦 + 4 __________________ Solution: 8y3 + y2 + 4 37) (x2 + 3x + 1) + ( 5x – 3) 𝑥 2 + 3𝑥 + 1 5𝑥 − 3 ______________ Solution: x2 + 8x - 2 Section P.4 Addition and Subtraction of Polynomials #39 – 50: Find the opposite of each polynomial 39) 3x – 5 = -(3x – 5) Solution: -3x + 5 41) 4x + 1 = -(4x + 1) Solution: -4x – 1 43) -8x + 9 = -(-8x + 9) Solution: 8x - 9 45) -2x – 5 47) -3x2 + 5x – 4 = -(-3x2 + 5x – 4) Solution: 3x2 – 5x + 4 49) 2x3 + 5x – 5 51) x4 + 3x3 – 2x2 + 5x – 1 = -( x4 + 3x3 – 2x2 + 5x – 1) Solution: -x4 – 3x3 + 2x2 – 5x + 1 #53-60: Simplify 53) -(3x – 4) Change the signs when you drop the parenthesis Solution 3x + 4 55) -(5x2 – 3x + 2) Solution: 5x2 + 3x – 2 57) -(-2x2 – 3x +1) Solution: 2x2 + 3x - 1 59) -(-x2 + 5x +8) Chapter P – Polynomials #61 – 70: Subtract 61) (3x + 5) – (2x – 4) = 3x + 5 – 2x + 4 = 3x – 2x + 5 + 4 Solution: x + 9 63) (x2 + 6x – 4) – (-5x2 – 3x + 1) = x2 + 6x – 4 + 5x2 + 3x – 1 = x2 + 5x2 + 6x + 3x – 4 – 4 Solution: 6x2 + 9x - 8 65) (x3 + 5x2 – 3x) – (4x3 – 3x2 + 1) 67) (8xy + 4x2y + 3) – (17xy – 3x2y – 3) = 8xy + 4x2y + 3 – 17xy + 3x2y + 3 = 4x2y + 3x2y + 8xy – 17xy + 3 + 3 Solution: 7x2y - 9xy + 6 69) (3ab2 – 4a2b) – (5a2b – 7ab2) #71 – 82: Rewrite each problem as a vertical subtraction problem, then subtract. 71) (3x2 + 5 – 2x) – (x – 4 +x2) Step 1 3𝑥 2 − 2𝑥 + 5 −(𝑥 2 + 𝑥 − 4) _______________ Step 2 3𝑥 2 − 2𝑥 + 5 −𝑥 2 − 𝑥 + 4 _______________ Solution: 2x2 – 3x + 9 73) (x2 + 6x – 4) – (-5x2 – 3x + 1) Step 1 𝑥 2 + 6𝑥 − 4 −(−5𝑥 2 − 3𝑥 + 1) ___________________ Step 2 𝑥 2 + 6𝑥 − 4 5𝑥 2 + 3𝑥 − 1 __________________ Solution: 6x2 + 9x - 5 75) (x2 + 5x3 – 3x) – (4x3 – 3x2 + 5x) Step 1 5𝑥 3 + 𝑥 2 − 3𝑥 −(4𝑥 3 − 3𝑥 2 + 5𝑥) ______________________ Step 2 5𝑥 3 + 𝑥 2 − 3𝑥 4𝑥 3 + 3𝑥 2 − 5𝑥 _________________ Solution: 9x3 + 4x2 – 8x 77) (5x – 2) – (x2 + 6x – 4) Step 1 5𝑥 − 2 −(𝑥 2 + 6𝑥 − 4) ___________________ Step 2 5𝑥 − 2 −𝑥 − 6𝑥 + 4 ___________________ 2 Solution: -x2 – x + 2 79) (y2 + 3y + 6y3) – (2y3 – 3y + 4) Step 1 6𝑦 3 + 𝑦 2 + 3𝑦 −(2𝑦 3 − 3𝑦 + 4) _______________________ Step 2 6𝑦 3 + 𝑦 2 + 3𝑦 −2𝑦 3 + 3𝑦 − 4) _______________________ Solution: 4y3 + y2 +6y - 4 81) (x2 + 3x + 1) – ( 5x – 3) Step 1 𝑥 2 + 3𝑥 + 1 −( 5𝑥 − 3) _________________ Step 2 𝑥 2 + 3𝑥 + 1 − 5𝑥 + 3 _________________ Solution: x2 – 2x + 4 Section P.5 Multiplication of Polynomials Chapter P – Polynomials #1-16: Multiply the following monomials 1) 5(4x5) Just multiply the 5 and the 4, the x5 doesn’t change. Solution: 20x5 3) (3x2)(-4) Just multiply the 3 and the -4, the x2 doesn’t change. Solution: -12x2 5) (3x)(4x3) Multiply the 3 and 4 and add the exponents of the x’s. Remember the first x has an exponent of 1. Solution: 12x4 2 6 7) (3 𝑥 2 ) (5 𝑥) The hard part is the fraction multiplication: 2 6 2 2 (3) (5) = (1) (5) cancel the 3 with the 6 to get 4 5 After multiplying the fractions, add the exponents of the x’s. Solution: 𝟒 𝟑 𝒙 𝟓 9) (-a4)(a5) Think of this as (-1a4)(a5) The -1 will remain, and I just need to add the exponents of the a’s. Solution: -1a9 which can be written as –a9 (it’s better not to write the 1, but is okay to write it.) 2 7 11) (3 𝑥𝑦 2 ) (4 𝑥𝑦) 13) (a2b)(ab3) 7 2 15) (−3𝑥) ( 𝑥) 7 First multiply the (−3) (2) this is done by multiplying the -3 and the 7. 7 2 (−3) ( ) = Solution: −21 2 , next add the exponents of the x’s. −21 2 𝑥 2 #17-36: Multiply the following monomial / polynomial products 17) 3x(2x – 5) = (3x)(2x) + (3x)(-5) Solution: 6x2 – 15x 19) 7(-2a + 4) = (7)(-2a) + (7)(4) Solution: -14a + 28 21) (x – 5)(3x) 23) (-3a + 5)7a 25) -2(5x + 8) 27) -2x(5x2 + 3x -1) 29) 2 (4𝑥 3 − 9) 2 3 2 3 = ( ) (4𝑥) + ( ) (−9) I will show the multiplication by the -9 as it is harder. First I make the -9 a fraction, then I cancel the -9 and the 3 8 2 −9 =3 𝑥 + (3) ( 1 ) 8 2 −3 = 3 𝑥 + (1) ( 1 ) 8 =3 𝑥 + (−6) Solution: 31) 𝟖 𝒙− 𝟑 −1 (6𝑥 2 𝟔 − 9) 33) 3y(4y3 – 3y2 + 5y –4) = (3y)(4y3) + (3y)(-3y2) + (3y)(5y) + (3y)(-4) Solution: 12y4 – 9y3 + 15y2 – 12y 3 35) 4y(8y3 – 3y2 + 6y –16) #37 – 52: Multiply the following binomials using FOIL, write the answer in descending order 37) (3x – 4)(2x+5) Firsts (3x)(2x) Outers (3x)(5) Inners (-4)(2x) lasts (-4)(5) = 6x2 + 15x - 8x – 20 Solution: 6x2 + 7x - 20 39) (3x + 1)(-2x –9) 41) (3 – 2x)(5x + 3) 43) (4 – 9y)(8y + 7) 1 2 3 2 45) ( 𝑥 + 6) ( 𝑥 − 5) Firsts Outers Inners lasts 1 3 1 3 (6)(-5) ( 𝑥) ( 𝑥) ( 𝑥) (−5) (6) ( 𝑥) 2 2 2 2 3 5 = 4 𝑥2 − 2 𝑥 + Solution: 𝟑 𝟐 𝒙 𝟒 18 𝑥 2 + − 30 𝟏𝟑 𝒙− 𝟐 3 4 𝟑𝟎 1 4 47) ( 𝑦 2 + 5𝑦) ( 𝑦 − 2) 49) (3x + 4)(3x–4) 1 4 1 4 51) (𝑥 + ) (𝑥 − ) Firsts (x)(x) Outers Inners −1 1 (𝑥) ( ) ( ) (𝑥) 4 4 1 1 1 = 𝑥 2 − 4 𝑥 + 4 𝑥 − 16 𝟏 Solution: 𝒙𝟐 − 𝟏𝟔 lasts 1 −1 (4) ( 4 ) #53 – 70: Multiply, write the answer in descending order 53) (2x–3)(x2 + 4x – 2) = (2x)(x2) + (2x)(4x) + (2x)(-2) + (-3)(x2) + (-3)(4x) + (-3)(-2) = 2x3 + 8x2 – 4x – 3x2 – 12x + 6 = 2x3 + 8x2 – 3x2 – 4x – 12x + 6 Solution: 2x3 + 5x2 – 16x + 6 55) (x – 4)(3x2 +x – 2) 57) (5y + 1)(3y2 + 4y –3) 59) (4z –3)(3z2+5z+4) ` 61) (4x2 +5x+1)(2x–7) 63) (6x2 –7x–1)(3x–4) 65) (x2 + 6x +2)(x2 + 5x +2) = (x2)(x2) + (x2)(5x) + (x2)(2) + (6x)(x2) + (6x)(5x) + (6x)(2) + (2)(x2) + (2)(5x)+2(2) = x4 + 5x3 + 2x2 + 6x3 + 30x2 + 12x + 2x2 + 10x + 4 = x4 + 5x3 + 6x3 + 2x2 + 30x2 + 2x2 + 12x + 10x + 4 Solution: x4 + 11x3 + 34x2 + 22x + 4 67) (5x2 + 3x + 1)(3x2 + 7x –1) 69) (-7x2 + 2x –4)(3x2 –5x –3) = (-7x2)(3x2) + (-7x2)(-5x) + (-7x2)(-3) + (2x)(3x2) + (2x)(-5x) + (2x)(-3) + (-4)(3x2) + (-4)(-5x) + (-4)(-3) = -21x4 + 35x3 + 21x2 + 6x3 – 10x2 – 6x – 12x2 + 20x + 12 = -21x4 + 35x3 + 6x3 + 21x2 – 10x2 – 12x2 – 6x + 20x + 12 Solution: -21x4 + 41x3 – x2 + 14x + 12 Section P.5 Multiplication of Polynomials Chapter P – Polynomials #71 – 88: Multiply the “differences of squares”, try to be efficient in your work 71) (x + 3)(x–3) I only need to do Firsts and lasts as the outers and inners will cancel. = (x)(x) + (3)(-3) Solution x2 - 9 73) (3x + 5)(3x–5) 75) (4x3 + 7)(4x3 -7) 2 2 77) (𝑥 + 3) (𝑥 − 3) 2 −2 = (x)(x) + (3) ( 3 ) Solution: 𝒙𝟐 − 𝟒 𝟑 2 2 79) (3𝑦 + 7) (3𝑦 − 7) 2 −2 = (3𝑦)(3𝑦) + (7) ( 7 ) Solution: 𝟗𝒚𝟐 − 𝟒 𝟒𝟗 81) (5 + 2y3)(5-2y3) = (5)(5) + (2y3)(-2y3) Solution: 25 – 4y6 83) (x2 – 5)(x2+5) 85) (3x – 4)(3x+4) 87) (5 – 2x)(5+2x) #89 – 106: Multiply the squares, write the answer in descending form 89) (x + 3)(x+3) Firsts (x)(x) Outers (x)(3) Inners (3)(x) lasts (3)(3) = x2 + 3x + 3x + 9 Solution: x2 + 6x + 9 91) (3x + 5)(3x+5) 93) (4x3 – 7)(4x3 –7) 2 3 2 3 95) (𝑥 − ) (𝑥 − ) 2 2 97) (3𝑦 + 7) (3𝑦 + 7) Firsts Outers Inners lasts 2 2 2 2 (3𝑦)(3𝑦) (3𝑦) ( ) ( ) (3𝑦) (7) (7) 7 7 6 6 4 = 9𝑦 2 + 7 𝑦 + 7 𝑦 + 49 Solution: 𝟗𝒚𝟐 + 𝟏𝟐 𝒚 𝟕 𝟒 + 𝟒𝟗 99) (5 – 2y3)(5–2y3) 101) (x2 + 5)(x2+5) 103) (3x – 4)(3x–4) = (3x)(3x) + (3x)(-4) + (-4)(3x) + (-4)(-4) = 9x2 – 12x – 12x + 16 Solution: 9x2 – 24x + 16 105) (5 +2x)(5+2x) #107 – 120: Simplify, write the answer in descending form 107) (3x + 5)2 = (3x+5)(3x+5) = 9x2 + 15x + 15x + 25 Solution: 9x2 + 30x + 25 109) (4x +1)2 = (4x+1)(4x+1) = 16x2 + 4x + 4x + 1 Solution: 16x2 + 8x + 1 111) (3x – 5)2 113) (4x –1)2 115) (3 + 7y)2 =(3 + 7y)(3+7y) = 9 + 21y + 21y + 49y2 Solution: 49y2 + 42y + 9 117) (2 + x)2 119) (3 – 7y)2 = (3 – 7y)(3 – 7y) = 9 – 21y – 21y + 49y2 Solution: 49y2 – 42y + 9 121) (2 – x)2 #123-138: Multiply and simplify 123) (2x + 3y)(4x – 2y) = (2x)(4x) + (2x)(-2y) + (3y)(4x) + (3y)(-2y) = 8x2 – 4xy + 12xy – 6y2 Solution: 8x2 +8xy – 6y2 125) 3xy(2x + 3y) = (3xy)(2x) + (3xy)(3y) Solution: 6x2y + 9xy2 127) 8x2(4xy – 3y) 129) (x2y3 + 2)(x2y3 -2) = (x2y3)(x2y3) + (x2y3)(-2) + 2(x2y3) + (2)(-2) = x4y6 – 2x2y3 + 2x2y3 – 4 Solution: x4y6 – 4 131) (3x – 4y)2 = (3x- 4y)(3x – 4y) = (3x)(3x) + (3x)(-4y) + (-4y)(3x) + (-4y)(-4y) = 9x2 – 12xy – 12xy + 16y2 Solution: 9x2 – 24xy + 16y2 133) (7 – 2xy)2 135) (5x – 2y)(3xy) 137) (5ab + 1)(3 + 2ab) = (5ab)(3) + (5ab)(2ab) + (1)(3) + 1(2ab) = 15ab + 10a2b2 + 3 + 2ab Solution: 10a2b2 + 17ab + 3 Section P.6: Division of Polynomials #1-16: Divide and Check (Monomial division) 1) = 6𝑥 2 +3𝑥+9 3 6𝑥 2 3 + 3𝑥 3 9 +3 Solution: 2x2 + x + 3 3) 4𝑦 3 +6𝑦 2 −4𝑦 4𝑦 5) 18𝑥𝑦 2 −3𝑥 2 𝑦+12𝑥𝑦 6𝑥𝑦 = 18𝑥𝑦 2 6𝑥𝑦 − 3𝑥 2 𝑦 6𝑥𝑦 + 12𝑥𝑦 6𝑥𝑦 𝟏 Solution: 𝟑𝒚 − 𝟐 𝒙 + 𝟐 7) 𝑥 3 𝑦 4 +12𝑥𝑦 3 −3𝑥 2 𝑦 5 6𝑥𝑦 2 9) (12𝑥 2 + 6𝑥 + 9) ÷ (3) = 12𝑥 2 3 + 6𝑥 3 + 9 3 Solution: 4x2 + 2x + 3 11) (4𝑥 3 − 8𝑥 2 + 6𝑥) ÷ (2𝑥) 13) (12𝑥 2 𝑦 3 − 6𝑥 2 𝑦 2 ) ÷ (−2𝑥𝑦) = 12𝑥 2 𝑦 3 −2𝑥𝑦 − 6𝑥 2 𝑦 2 −2𝑥𝑦 Solution: -6xy2 + 3xy 15) (24𝑥 2 + 3𝑥 − 9) ÷ (−3) = 24𝑥 2 −3 3𝑥 9 + −3 − −3 Solution: -8x2 – 1 + 3 Chapter P – Polynomials Section P.7: Test Review Chapter P – Polynomials You don’t have to write solutions for this section, just create answers. #1-2: Simplify 1) (3xy2)(2x2y4) 2) (-8st)(2s5t) #3-4: Simplify, assume no denominator equals 0 3) 15𝑥 4 20𝑥 4) 24𝑥𝑦 3 12𝑥𝑦 #5-8: Simplify 5) (3x4)3 6) (5x2y)2 7) (4xy3)3(2x5y)2 8) (7xy3)2(2x)3 #9-12: Simplify the expression 9) -50 10) (-5)0 11) 5x0 12) (5x)0 #13 – 17: Simplify the expression. Write the answer with positive exponents only. 13) 2-4xy-3 14) 24𝑥𝑦 3 36𝑥 −4 𝑦 2𝑥 −3 15) (𝑦−5 ) 3 −2 16) (4) 17) 𝑥 −3 𝑥2 #18-19: Evaluate each polynomial using x = -2, y = 4 and z = 3 18) -xy2 + 2yz – 2x3y 19) 2y3 – 2z2 + 6xy – 1 #20-21: Add 20) (3y2 + y – 6y3) + (5y3 – 3y + 4y2) 21) (8z2 + 5z3 + 4) + (6z3 + 2z – 7) #22-23: Subtract 1 2 3 4 22) (2 𝑥 + 3) − (4 𝑥 + 3) 23) (4xy – 2x2y +3x) – (5x – 6xy + 7x2y) #24-31: Multiply 2 10 24) (5 𝑥𝑦 4 ) ( 4 𝑥𝑦) 5 6 25) (3 𝑦 + 2) (7 𝑦 − 3) 26) -3b(2b2 –5b –7) 3 27) 2y(4y3 – 8y2 – 6y +12) 28) (2x + 8)(2x–8) 29) (3x + 2)2 30) (5x + 2y)(3x – 2y) 31) (x – 3y)2 #32-33: Divide and Check 32) 18𝑠4 𝑡 3 −12𝑠3 𝑡 2 +2𝑠𝑡 6𝑠𝑡 2 33) (12𝑥 4 − 16𝑥 3 + 6𝑥) ÷ (2𝑥) #34 – 35: Divide, if the remainder is 0 check your work 34) (6𝑥 2 + 7𝑥 − 20) ÷ (2𝑥 + 5) 35) 3𝑥 2 −8𝑥−3 𝑥−3