QA #5 Factors and Multiples

Questions and Answers Sheet 5 Factors and Multiples Question #1 Write as the sum and or difference of logarithms. Express powers as factors. 𝑥+6 log 4 ( 𝑥 7 ) A. B. C. D. 𝑙𝑜𝑔4 (𝑥 + 6) − 𝑙𝑜𝑔4 𝑥 7 log 4 𝑥 − log 4 (𝑥 + 6) 𝑙𝑜𝑔4 (𝑥 + 6) − 7𝑙𝑜𝑔4 𝑥 𝑙𝑜𝑔4 (𝑥 + 6) + 7𝑙𝑜𝑔4 𝑥 Answer: Step 1 Given, log 4 ( 𝑥+6 𝑥7 ) Step 2 Property used: 𝑚 log 𝑎 ( 𝑛 ) = log 𝑎 𝑚 − log 𝑎 𝑛 𝑙𝑜𝑔 _{𝑎}𝑚^{𝑛} = 𝑛 𝑙𝑜𝑔 _{𝑎}𝑚 Step 3 Apply the above property, we get ⇒ log 4(𝑥 + 6) − log4 (𝑥 7 ) ⇒ log 4(𝑥 + 6) − 7 log4 (𝑥 ) Question #2 Write the expression as a sum and/or difference of logarithms. Express powers as factors. 𝑙𝑛(𝑥 15 √7 − 𝑥), 0 < 𝑥 < 7 𝑙𝑛(𝑥 15 √7 − 𝑥) =? Answer: Step 1 Using formula: 1) log(𝐴, 𝐵) = log 𝐴 + log 𝐵 2) log 𝐴𝑛 = 𝑛 log 𝐴 1 root means power = 2 √𝑝 = 𝑝1/2 Step 2 𝑙𝑛(𝑥 15 √7 − 𝑥), 0 < 𝑥 < 7 = 𝑙𝑛(𝑥 15 ) + 𝑙𝑛(√7 − 𝑥) = 𝑙𝑛𝑥 15 + 𝑙𝑛(7 − 𝑥 )1/2 1 = 15𝑙𝑛𝑥 + 2 𝑙𝑛(7 − 𝑥 ) 1 hence, 𝑙𝑛(𝑥 15 √7 − 𝑥) = 15𝑙𝑛𝑥 + 2 𝑙𝑛(7 − 𝑥 ) Question #3 Let u, v, and w be any three vectors from a vector space V. Determine whether the set of vectors {𝑣 − 𝑢, 𝑤 − 𝑣, 𝑢 − 𝑤} is linearly independent or linearly dependent. Answer: Step 1 Given data: The given space vector is: 𝑉 = {𝑣 − 𝑢, 𝑤 − 𝑣, 𝑢 − 𝑤} . The expression to check whether the space vector is linearly dependent or independent is 𝑐1 (𝑣 − 𝑢) + 𝑐2 (𝑤 − 𝑣) + 𝑐3 (𝑢 − 𝑤) = (0,0,0) Further, solve the above expression. 𝑐1 𝑣 − 𝑐1 𝑢 + 𝑐2 𝑤 − 𝑐2 𝑣 + 𝑐3 𝑢 − 𝑐3 𝑤 = (0,0,0) (𝑐1 − 𝑐2 )𝑣 − (𝑐1 − 𝑐3 )𝑢 + (𝑐2 − 𝑐3 )𝑤 = (0,0,0) 𝑐1 − 𝑐2 = 0 𝑐1 − 𝑐3 = 0 Step 2 The above equation gives the value of the scalar multiples. 𝑐1 = 𝑐2 = 𝑐3 As the 𝑐1 , 𝑐2 , and 𝑐3 are the scalar multiples of the given space vector, and they are equal, so the space vector is linearly dependent. Thus, the given space vector is linearly dependent. Question #4 To simplify the expression, √19 . Answer: Concept used: To simplify an expression having square root of a number, firstly factorization of the number is done and then factors having even exponential powers are taken out from the square root function and their exponential powers are divided by two. 4 2 \Rightarrow √𝑚4 × 𝑛2 = 𝑚2 × 𝑛2 = 𝑚2 × 𝑛1 Calculations: As per the question, the expression is √19 . The factorization of a number is the product of numbers that equals the number. Now, 19 is a ' number. Hence, it has only two multiples i.e. 1 and 19. Therefore, 19 cannot be factorized anymore. Hence, the simplify value of √19 is √19 only. Question #5 Without dividing, determine which of the following represent terminating decimals. 1) 2) 72 81 21 28 a. Is b. Is 72 81 a terminating decimal? 21 28 A) Yes, because the only factors of the denominator, 81, are 2 and 5. B) No, because the only factors of the denominator, 81, are 2 and 5. C) No, the denominator of the simplified fraction contains a factor other than 2 or 5. D) Yes, the only factors of the denominator of the simplified fraction are 2 or 5. Answer: Step 1 Given the fraction a) 72 81 A rational number 𝑎 𝑏 in the simplest form can be written as a terminating decimal if an only if the ' factorization of the denominator contains no 's other than 2 or 5. Step 2 Here first simplify the given fraction 72 = 81 = 2×2×2×3×3 3×3×3×3 . 8 = 32 The denominator is 3 Step 3 Therefore, No the denominator of the simplified fraction contains a factor other than 2 or 5 Step 4 b) 21 28 Here first simplify the given fraction 3×7 = 2×2×7 3 = 23 Here the denominator is 2 Step 5 yes the only factors of the denominator of the simplified fraction is 2 Question #6 Fernando and Risa are factoring 5𝑥 2 − 𝑥 − 6 . They each made a chart below and selected a pair of integers. Which student is correct? Explain your reasoning. Fernando: Factors sum 7 −7 −1 Factors sum −3.10 1. − 6 −5 3. − 10 −1.6 5 2. − 3 −1 5. − 6 Risa: Factors sum −3.10 7 3. − 10 −7 5. − 6 −1 Who is correct? Answer: Step 1 Function is 5𝑥 2 − 𝑥 − 6 ⇒ 5𝑥 2 + 5𝑥 − 6𝑥 − 6 ⇒ 5𝑥(𝑥 + 5) − 6(𝑥 + 5) ⇒ (5𝑥 − 6)(𝑥 + 5) Hence factor are -5 and 6 5 Step 2 None of them are correct became corred factors are -5 and 6 5 with sum = −4.8 Question #7 A polynomial P is given. 𝑃(𝑥 ) = 𝑥 3 − 2𝑥 2 + 9𝑥 − 18 a) Factor P into linear and irreducible quadratic factor with real coefficients. 𝑃(𝑥 ) =? b) Factor P completely into linear factors with complex coefficients. 𝑃(𝑥 ) =? Answer: 𝑝(𝑥 ) = 𝑥 3 − 2𝑥 2 + 9𝑥 − 18 a) = 𝑥 2 (𝑥 − 2) + 9(𝑥 − 2) = (𝑥 − 2)(𝑥 2 + 9) Hence 𝑝(𝑥 ) = (𝑥 − 2)(𝑥 2 + 9) where (𝑥 − 2) is linear (𝑥 2 + 9) is irreducible quadratic factors. b) 𝑝(𝑥 ) = (𝑥 − 2)(𝑥 2 + 9) = (𝑥 − 2)(𝑥 2 + 32 ) = (𝑥 − 2)(𝑥 2 − 𝑖 2 32 )(𝑎𝑠𝑖 2 = −1) = (𝑥 − 2)(𝑥 2 − (3𝑖 )2 ) 𝑝(𝑥 ) = (𝑥 − 2)(𝑥 − 3𝑖 )(𝑥 + 3𝑖 ) Three linear factors are: (𝑥 − 2)(𝑥 − 3𝑖 ) and (𝑥 + 3𝑖 ) where (𝑥 − 3𝑖 ) and (𝑥 + 3𝑖 ) have complex coefficients. Question #8 Given 𝑃(𝑥 ) = 3𝑥 5 − 5𝑥 4 + 37𝑥 3 − 83𝑥 2 − 176𝑥 − 48 , and that 4i is a zero, write P in factored form (as a product of linear factors). Be sure to write the full equation, including 𝑃(𝑥 ) =? . Answer: Step 1 The given polynomial is: 𝑃(𝑥 ) = 3𝑥 5 − 5𝑥 4 + 37𝑥 3 − 83𝑥 2 − 176𝑥 − 48 Step 2 Now 𝑥 = 4𝑖 is a zero of the given polynomial, therefore 𝑥 = −4𝑖 is also zero of the given polynomial. Therefore: (𝑥 − 4𝑖 )(𝑥 + 4𝑖 ) divides the given polynomial. That is one of the factors of given polynomial is: 𝑥 2 + 16 . 𝑃(𝑥) 𝑥 2 +16 = 𝑃(𝑥) 𝑥 2 +16 3𝑥 5 −5𝑥 4 +37𝑥 3 −83𝑥 2 −176𝑥−48 𝑥 2 +16 = 3𝑥 3 − 5𝑥 2 − 11𝑥 − 3 𝑃(𝑥 ) = (𝑥 2 + 16)(3𝑥 3 − 5𝑥 2 − 11𝑥 − 3) 𝑃(𝑥 ) = (𝑥 2 + 16)(𝑥 + 1)(𝑥 − 3)(3𝑥 + 1) Question #9 A rectangular porch has dimensions of 𝑆(𝑥 + 12) and (𝑥 + 5) feet. If the area of the porch floor is 120 square feet, what are its length and width? Answer: Step 1 Given: A rectangular porch has dimensions of 𝑆 (𝑥 + 12) and (𝑥 + 5) feet. If the area of the porch floor is 120 square feet. To find: length and width of rectangular porch. Step 2 Solutions: Let the length of rectangular porch be (𝑥 + 12) feet. width of rectangular porch be (𝑥 + 5) feet. Area of rectangular porch = 120 sq. feet. Now according question, Area of rectangular body is given by ⇒ 𝐴𝑟𝑒𝑎 = 𝑙 × 𝑤 . ⇒ 120 = (𝑥 + 12)(𝑥 + 5) ⇒ (𝑥 + 12)(𝑥 + 5) = 120 ⇒ 𝑥 2 + 5𝑥 + 12𝑥 + 60 = 120 ⇒ 𝑥 2 + 5𝑥 + 12𝑥 = 120 − 60 ⇒ 𝑥 2 + 17𝑥 = 60 ⇒ 𝑥 2 + 17𝑥 − 60 = 0 Step 3 ⇒ 𝑥 2 + 20𝑥 − 3𝑥 − 60 = 0 ⇒ 𝑥(𝑥 + 20) − 3(𝑥 + 20) = 0 ⇒ (𝑥 + 20)(𝑥 − 3) = 0 ⇒ 𝑥 = +3, −20 We choose (+ue) value of 'x' i.e. 3 Length of Rectangular porch = (𝑥 + 12) feet = (3 + 12) feet = 15 feet width of rectangular porch = (𝑥 + 5) feet = (5 + 3) feet = 8 feet Question #10 On a black sheet of paper (lined or not), write any two different positive integers less than 10. In other words, two different single digit positive whole numbers. (Yes, that is redundant because whole numbers are positive.) Use those two numbers to create two factors, (𝑥 − 𝑎) and (𝑥 − 𝑏) , where a and b are your two numbers. (If you work together with others on this, which is ok for this assignment, be sure each of you chooses different numbers.) Multiply your factors to create a quadratic in standard form. Next write as a function: 𝑓(𝑥 ) = 𝑥 2 − 𝑏𝑥 + 𝑐 . Now create a quadratic equation by changing f(x) to zero: 0 = 𝑥 2 − 𝑏𝑥 + 𝑐 . Answer: Step 1 Let two different positive integers less than 10 are 2 and 4. Step 2 Create two factors (𝑥 − 2) and (𝑥 − 4) (𝑥 − 2)(𝑥 − 4) 𝑥 2 − 4𝑥 − 2𝑥 + 8 𝑥 2 − 6𝑥 + 8 Step 3 𝑓(𝑥 ) = 𝑥 2 − 6𝑥 + 8 Now create a quadratic by changing changing f(x) to zero. So we have 𝑥 2 − 6𝑥 + 8 = 0 Question #11 According to one source, the amount of data passing through mobile phone networks doubles each year. Explain why the amount of data passing through mobile phone networks is an exponential function of time. The amount of data grows by a factor of_________________ each year, this is an exponential function because the amount is growing by _____________________ multiples. Use 𝐷𝑜 for the initial amount of data and find a formula that gives the data D as an exponential function of the time t in years. If this trend continues, how long will it be before the amount of data is 100 times its initial value? Round to three decimal places. Answer: Step 1 Given: The amount of data passing through mobile phone networks doubles each year. Step 2 a) We have given that the amount of data passing through mobile phone networks doubles each year. That is if we assume the initial amount of data is 𝐴0 and A(t) is amount of data after t years, then after one year the amount of data becomes 𝐴(1) = 2𝐴0 . After 2 years the amount of data becomes 𝐴(2) = 2𝐴(1) = 2(2𝐴0 ) = 4𝐴0 . After 3 years the amount of data becomes 𝐴(3) = 2𝐴(2) = 2(4𝐴0 ) = 8𝐴0 . Continuing in this way after n years we get, 𝐴(𝑛) = 2𝐴(𝑛 − 1) . Therefore, as time increases by 1 year the amount of data is obtained by multiplying the previous amount by 2. That is the amount of data growing by constant multiples with time. Therefore, the amount of data passing through mobile phone networks is an exponential function of time. The exponential function is of the form 𝐴(𝑡) = 𝐴0 𝑒 𝑘𝑡 where A(t) is amount of data after t years, 𝐴0 is the initial amount of data and k is a growth factor. Now, we have to find the value of k. After 1 year the amount of data becomes 𝐴(1) = 2𝐴0 . By using the exponential function we get, ⇒ 2𝐴0 = 𝐴0 𝑒 𝑘(1) ⇒ 2 = 𝑒𝑘 ⇒ 𝑘 = 𝑙𝑛(2) ≈ 0.6931 Therefore, the amount of data grows by a factor of 𝑙𝑛(2) ≈ 0.6931 each year, this is an exponential function because the amount is growing by 2(twice) multiples. Step 3 b) To find: The exponential function using 𝐷0 as the initial amount of data and D as the amount after t years. Use 𝐷0 as the initial amount of data and find a formula that gives the data D as an exponential function of the time t in years. We get, 𝑡 𝐷(𝑡) = 𝐷0 𝑒 𝑙𝑛(2)𝑡 = 𝐷0 (𝑒 𝑙𝑛(2) ) = 𝐷0 (2)𝑡 Therefore, the exponential function is 𝐷(𝑡) = 2𝑡 𝐷0 . Step 4 c) To find: How long will it be before the amount of data is 100 times its initial value if the trend continues. Use 𝐷(𝑡) = 100𝐷0 in the exponential function from part (b) we get, 100𝐷0 = 2𝑡 𝐷0 Solve for t, ⇒ 100 = 2𝑡 Apply log on both sides, ⇒ log(100) = log(2𝑡 ) ⇒ 2 = 𝑡 log(2) 2 ⇒ 𝑡 = log(2) ≈ 6.644 Therefore, it will take 6.644 years to make the amount of data 100 times its initial value. Question #12 Consider the following. 𝑓(𝑥 ) = 𝑥 4 + 20𝑥 2 − 125 a) Write the polynomial as the product of factors that are irreducible over the rationals. 𝑓(𝑥 ) =? b) Write the polynomial as the product of linear and quadratic factors that are irreducible over the reals. 𝑓(𝑥 ) =? c) Write the polynomial in completely factored form. 𝑓(𝑥 ) =? Answer: Step 1 Given that 𝑓(𝑥 ) = 𝑥 4 + 20𝑥 2 − 125 Here write the polynomial as the product of factors that are irreducible over the rationals. Step 2 (a) 𝑓(𝑥 ) = 𝑥 4 + 20𝑥 2 − 125 = 𝑥 4 − 5𝑥 2 + 25𝑥 2 − 125 = 𝑥 2 (𝑥 2 − 5) + 25(𝑥 2 − 5) = (𝑥 2 − 5)(𝑥 2 + 25) Here both (𝑥 2 − 5) and (𝑥 2 + 25) are irreducible over the rationals. Therefore, f(x) = (x^{2} -5)(x^{2}+25) Step 3 (b) 𝑓(𝑥 ) = (𝑥 2 − 5)(𝑥 2 + 25) = (𝑥 + √5)(𝑥 − √5)(𝑥 2 + 25) Therefore 𝑓(𝑥 ) = (𝑥 + √5)(𝑥 − √5)(𝑥 2 + 25) is the product of irreducible factors over R. Step 4 (d) 𝑓(𝑥 ) = 𝑥 4 + 20𝑥 2 − 125 = (𝑥 2 − 5)(𝑥 2 + 25) = (𝑥 + √5)(𝑥 − √5)(𝑥 2 + 25) = (𝑥 + √5)(𝑥 − √5)(𝑥 + 5𝑖 )(𝑥 − 5𝑖 ) Therefore 𝑓(𝑥 ) = (𝑥 + √5)(𝑥 − √5)(𝑥 + 5𝑖 )(𝑥 − 5𝑖 ), which is in completely factored form. Question #13 To find: the othermonomial. Given: the GCF of 12𝑥 2 𝑦 3 and another monomial is 6𝑥𝑦 3 , and their LCM is 36𝑥 2 𝑦 4 . Answer: Calculation: As per problem, Compare the coefficients, The GCF of 12 and the other number is 6, so the number is a multiple of 6 Their LCM is 36, so the number is factor of 36 The multiples of 6 which are also factors of 36 are 6,12,18 and 36 Now, LCM and GCF of 12 and 6 is 6, so the number is not 6 And, the LCM and GCF of 12 and 12 is 12, so the number is not 12 However, LCM and GCF of 12 and 18 is 12 = 2 ⋅ 6 18 = 3 ⋅ 6 𝐿𝐶𝑀 = 2 ⋅ 3 ⋅ 6 = 36 Therefore, the coefficient of the monomial is 18. The first monomial has x^{2} as variable, and the GCF has x, so the other monomial has x. The first monomial has y^{3} as variable, but the LCM has y^{4} , so the other monomial has y^{4} . Hence, the other monomial is 18xy^{4} Question #14 Lumber that is used to frame walls of houses is frequently sold in multiples of 2 ft. If the length of a board is not exactly a multiple of 2 ft, there is often no charge for the additional length. For example, if a board measures at least 8 ft, but less than 10 ft, then the consumer is charged for only 8 ft. (a) Suppose that the cost of lumber is $0.70 every 2 ft. Find a formula for a function f(x)that computes the cost of a board x feet long for 10 ≤ 𝑥 ≤ 16 . (b) Determine the costs of boards with lengths of 8.6 ft and 15.1 ft. Answer: Step 1 To express f(x) explicitly as an expression and use it for computing the required values Step 2 To start with consider the typical case of x , with x greater than equal 10 but x strictly less than 12. The cost is the same for all x in this range = 5 times 0.70. This can be captured by the formula (using the floor function) First consider 10 ≤ 𝑥 < 12 𝑓(𝑥 ) = cos 𝑡 𝑓𝑜𝑟10𝑓𝑡 = 5 × 0.70 𝑥 = [2] × 0.70 , 𝑥 𝑥 𝑥 where [2] = floor ( 2) integral part ( 2) Step 3 The formula holds for any positive number x greater than or equal to 2, in particular for the given range 10 ≤ 𝑥 ≤ 16 𝑥 𝑓(𝑥 ) = [2] × 0.70 , 𝑥 𝑥 𝑥 where [2] = floor ( 2) = integral part ( 2) Step 4 More explicitly, the formula is worked out for the given range 𝑥 𝑓(𝑥 ) = [2] × 0.70 , = $3.50, 𝑖𝑓 10 ≤ 𝑥 < 12 = $4.20, 𝑖𝑓 12 ≤ 𝑥 < 14 = $4.90, 𝑖𝑓 14 ≤ 𝑥 < 16 = $5.60, 𝑖𝑓 𝑥 = 16 Step 5 ANSWER : (b) b) 𝑓(8.6) = 𝑓 (8) = 4 × 0.70 = 2.80 𝑓(15.1) = 𝑓(14) = 7 × 0.70 = 4.90 Question #15 For the functions 𝑓 (𝑥 ) = √𝑥 − 3 and 𝑔(𝑥) = √𝑥 + 2, find and determine the domain in each case: (a)gf (b) 𝑓 + 𝑔 Answer: 𝑓(𝑥 ) = √𝑥 − 3, 𝑔(𝑥 ) = √𝑥 + 2 a) 𝑔𝑓 = √𝑥 + 2√𝑥 − 3 = √(𝑥 + 2)(𝑥 − 3) Since, square root is not defined for negative values ⇒ (𝑥 + 2)(𝑥 − 3) ≥ 0 ⇒𝑥 ≥3 and 𝑥 ≤ −2 ⇒ Domain = (−∞, −2] ∪ [3, ∞) b) 𝑔 + 𝑓 = √𝑥 + 2 + √𝑥 − 3 √𝑥 + 2 is defined for 𝑥 + 2 ≥ 0 𝑥 ≥ −2(1) √𝑥 − 3 is defined for 𝑥 − 3 ≥ 0 𝑥 ≥ 3(2 ) Combining (1) and (2) for x remain positive ⇒ 𝑥 ≥ 3 ⇒ Domain: = [3, ∞) Question #16 Linear and Quadratic Factors. A polynomial P is given (a) Factor P into linear and irreducible quadratic factors with real coefficients. (b) Factor P completely into linear factors with complex coefficients. 𝑃(𝑥 ) = 𝑥 4 + 8𝑥 2 + 16 Answer: a) To find: The factor of the polynomial into linear and irreducible quadratic factor with real coefficients. Consider, 𝑃(𝑥 ) = 𝑥 4 + 8𝑥 2 + 16 ⇒ 𝑃(𝑥 ) = 𝑥 4 + 4𝑥 2 + 4𝑥 2 + 16 ⇒ 𝑃 (𝑥 ) = 𝑥 2 (𝑥 2 + 4) + 4( 𝑥 2 + 4) ⇒ 𝑃(𝑥 ) = (𝑥 2 + 4)(𝑥 2 + 4) Therefore, 𝑃 (𝑥 ) = (𝑥 2 + 4)(𝑥 2 + 4) . b) To find: The factor completely into linear factors with complex coefficients. Consider, 𝑃(𝑥 ) = 𝑥 4 + 8𝑥 2 + 16 ⇒ 𝑃(𝑥 ) = 𝑥 4 + 4𝑥 2 + 4𝑥 2 + 16 ⇒ 𝑃 (𝑥 ) = 𝑥 2 (𝑥 2 + 4) + 4( 𝑥 2 + 4) ⇒ 𝑃(𝑥 ) = (𝑥 2 + 4)(𝑥 2 + 4) ⇒ 𝑃(𝑥 ) = (𝑥 − 2𝑖 )(𝑥 − 2𝑖 )(𝑥 + 2𝑖 )(𝑥 − 2𝑖 ) Therefore, 𝑃 (𝑥 ) = (𝑥 − 2𝑖 )(𝑥 − 2𝑖 )(𝑥 + 2𝑖 )(𝑥 − 2𝑖 ). Question #17 A polynomial P is given (a) Factor P into linear and irreducible quadratic factors with real coefficients. (b) Factor P completely into linear factors with complex coefficients. 𝑃(𝑥 ) = 𝑥 6 − 64 Answer: a) To find: The factor of the polynomial into linear and irreducible quadratic factor with real coefficients. Consider, 𝑃(𝑥 ) = 𝑥 6 − 64 Using the Rational Zeros Theorem, possible rational zeros are ±1, ±2, ±4, ±8, ±16, ±32, ±64 . Substitute 𝑥 = 2 in 𝑃(𝑥 ) = 𝑥 6 − 64 , we get 𝑃(2) = 26 − 64 ⇒ 𝑃(2) = 64 − 64 ⇒ 𝑃 (2) = 0 Therefore, 𝑥 = 2 is a factor of 𝑃 (𝑥) = 𝑥 6 − 64 . Factorize the polynomial, 𝑃(𝑥 ) = 𝑥 6 − 64 ⇒ 𝑃(𝑥 ) = (𝑥 − 2)(𝑥 5 + 2𝑥 4 + 4𝑥 3 + 8𝑥 2 + 16𝑥 + 32) Substitute 𝑥 = −2𝑖𝑛(𝑥 − 2)(𝑥 5 + 2𝑥 4 + 4𝑥 3 + 8𝑥 2 + 16𝑥 + 32), we get ((−2)5 + 2(−2)4 + 4(−2)3 + 8(−2)2 + 16(−2) + 32) = −32 + 32 − 32 + 32 − 32 + 32 =0 Therefore, 𝑥 = −2 is a factor of (𝑥 5 + 2𝑥 4 + 4𝑥 3 + 8𝑥 2 + 16𝑥 + 32) . Factorize the polynomial, (𝑥 5 + 2𝑥 4 + 4𝑥 3 + 8𝑥 2 + 16𝑥 + 32) ⇒ (𝑥 5 + 2𝑥 4 + 4𝑥 3 + 8𝑥 2 + 16𝑥 + 32) = (𝑥 + 2)(𝑥 4 + 4𝑥 2 + 16) ⇒ 𝑃(𝑥 ) = (𝑥 − 2)(𝑥 + 2)(𝑥 4 + 4𝑥 2 + 16) ⇒ 𝑃(𝑥 ) = (𝑥 − 2)(𝑥 + 2)[(𝑥 2 + 2𝑥 + 4)(𝑥 2 − 2𝑥 + 4)] Therefore, 𝑃 (𝑥 ) = (𝑥 − 2)(𝑥 + 2)(𝑥 2 + 2𝑥 + 4)(𝑥 2 − 2𝑥 + 4) . b) To find: The factor completely into linear factors with complex coefficients. Consider, 𝑃(𝑥 ) = 𝑥 6 − 64 Using the Rational Zeros Theorem, possible rational zeros are \𝑝𝑚 1, ±2, ±4, ±8, ±16, ±32, ±64 . Substitute 𝑥 = 2𝑖𝑛𝑃(𝑥 ) = 𝑥 6 − 64, we get 𝑃(2) = 26 − 64 ⇒ 𝑃(2) = 64 − 64 ⇒ 𝑃 (2) = 0 Therefore, 𝑥 = 2 is a factor of 𝑃 (𝑥) = 𝑥 6 − 64 . Factorize the polynomial, 𝑃(𝑥 ) = 𝑥 6 − 64 ⇒ 𝑃(𝑥 ) = (𝑥 − 2)(𝑥 5 + 2𝑥 4 + 4𝑥 3 + 8𝑥 2 + 16𝑥 + 32) Substitute 𝑥 = −2𝑖𝑛(𝑥 − 2)(𝑥 5 + 2𝑥 4 + 4𝑥 3 + 8𝑥 2 + 16𝑥 + 32) , we get ((−2)5 + 2(−2)4 + 4(−2)3 + 8(−2)2 + 16(−2) + 32) = −32 + 32 − 32 + 32 − 32 + 32 =0 Therefore, 𝑥 = −2 is a factor of (𝑥 5 + 2𝑥 4 + 4𝑥 3 + 8𝑥 2 + 16𝑥 + 32) . Factorize the polynomial, (𝑥 5 + 2𝑥 4 + 4𝑥 3 + 8𝑥 2 + 16𝑥 + 32) ⇒ (𝑥 5 + 2𝑥 4 + 4𝑥 3 + 8𝑥 2 + 16𝑥 + 32) = (𝑥 + 2)(𝑥 4 + 4𝑥 2 + 16) ⇒ 𝑃(𝑥 ) = (𝑥 − 2)(𝑥 + 2)(𝑥 4 + 4𝑥 2 + 16) ⇒ 𝑃(𝑥 ) = (𝑥 − 2)(𝑥 + 2)[(𝑥 2 + 2𝑥 + 4)(𝑥 2 − 2𝑥 + 4)] ⇒ 𝑃 (𝑥 ) = (𝑥 − 2)(𝑥 + 2) (𝑥 − (1 + 𝑖√3)) (𝑥 − (1 − 𝑖√3)) (𝑥 − (−1 + 𝑖√3)) (𝑥 − (−1 − 𝑖√3)) Therefore, 𝑃 (𝑥 ) = (𝑥 − 2)(𝑥 + 2) (𝑥 − (1 + 𝑖√3)) (𝑥 − (1 − 𝑖√3)) (𝑥 − (−1 + 𝑖√3)) (𝑥 − (−1 − 𝑖√3)) Question #18 A polynomial P is given (a) Factor P into linear and irreducible quadratic factors with real coefficients. (b) Factor P completely into linear factors with complex coefficients. 𝑃(𝑥 ) = 𝑥 5 − 16𝑥 Answer: a) To find: The factorization of the polynomial P(x) in linear and irreducible quadratic factors with real coefficients. Given information: The polynomial P(x) is, 𝑃(𝑥 ) = 𝑥 5 − 16𝑥 Concept used: Linear and Quadratic Factors Theorem: Any polynomial that has real coefficients can be factored into the product of linear and irreducible quadratic factors. Calculation: The given polynomial P(x) is, 𝑃(𝑥 ) = 𝑥 5 − 16𝑥 Rewrite the above polynomial as, 𝑃(𝑥 ) = 𝑥(𝑥 4 − 16) = 𝑥((𝑥 2 )2 − 42 ) Use identity 𝑎2 − 𝑏2 = (𝑎 − 𝑏)(𝑎 + 𝑏) to factor the above equation as, 𝑃(𝑥 ) = 𝑥(𝑥 2 − 4)(𝑥 2 + 4) = 𝑥(𝑥 2 − 22 )(𝑥 2 + 4) = 𝑥(𝑥 − 2)(𝑥 + 2)(𝑥 2 + 4) The factors x, (𝑥 − 2) and (𝑥 + 2) are linear factors. The factor (𝑥 2 + 4) is irreducible, since it has no real zeros. Conclusion: Thus, the factored form of the polynomial P(x) that has linear and irreducible quadratic factors is 𝑥(𝑥 − 2)(𝑥 + 2)(𝑥 2 + 4) b) To find: The factors of the polynomial P(x) that has linear factors with complex coefficients. Given: The polynomial P(x) is, 𝑃(𝑥 ) = 𝑥 5 − 16𝑥 Calculation: From part (a) the factored form of the polynomial P(x) is, 𝑃(𝑥 ) = 𝑥(𝑥 − 2)(𝑥 + 2)(𝑥 2 + 4) Now, factor the remaining quadratic factor to obtain the complete factorization as, 𝑃(𝑥 ) = 𝑥(𝑥 − 2)(𝑥 + 2)(𝑥 2 + 4) = 𝑥(𝑥 − 2)(𝑥 + 2)(𝑥 2 − (2𝑖 )2 ) = 𝑥(𝑥 − 2)(𝑥 + 2)(𝑥 − 2𝑖 )(𝑥 + 2𝑖 ) The above factors are linear factors with complex coefficients. Conclusion: Thus, the factored form of the polynomial P(x) that has linear factors is 𝑥(𝑥 − 2)(𝑥 + 2)(𝑥 − 2𝑖 )(𝑥 + 2𝑖 ) . Question #19 let 𝑢 = [2, −1, −4], 𝑣 = [0,0,0] , and 𝑤 = [−6, −9,8]. we want to determine by inspection (with minimal compulation) if {𝑢, 𝑣, 𝑤} is linearly dependent or independent. choose the best answer: A. The set is linearly dependent because one of the vectors is a scalar multiple of another vector. B. The set is linearly independent because we only have two vectors and they are not scalar multiples of each other. C. The set is linearly dependent because the number of vectors in the set is greater than the dimension of the vector space. D. The set is linearly dependent because two of the vectors are the same. E. The set is linearly dependent because one of the vectors is the zero vector. F. We cannot easily tell if the set is linearly dependent or linearly independent. Answer: Step 1 Any set containing the zero vector is linearly dependent since in {𝑢, 𝑣, 𝑤} the vector v is zero vector. Step 2 So the set is linearly dependent because one of the vectors is the zero vector. The correct option is E. Question #20 A polynomial P is given. a) Factor P into linear and irreducible quadratic factor with real coefficients. b) Factor P completely into linear factors with complex coefficients. 𝑃(𝑥 ) = 𝑥 3 − 5𝑥 2 + 4𝑥 − 20 Answer: a)To find: The factorization of the polynomial P(x) in linear and irreducible quadratic factors with real coefficients. Given information: The polynomial P(x) is, 𝑃(𝑥 ) = 𝑥 3 − 5𝑥 2 + 4𝑥 − 20 Concept used: Linear and Quadratic Factors Theorem: Any polynomial that has real coefficients can be factored into the product of linear and irreducible quadratic factors. Calculation: The given polynomial P(x) is, 𝑃(𝑥 ) = 𝑥 3 − 5𝑥 2 + 4𝑥 − 20 Rewrite the above polynomial as, 𝑃 (𝑥 ) = 𝑥 2 (𝑥 − 5) + 4(𝑥 − 5) Now, factor the polynomial P(x) as, 𝑃(𝑥 ) = (𝑥 − 5)(𝑥 2 + 4) The factor (𝑥 − 2) is a linear factor. The factor (𝑥 2 + 4) is irreducible, since it has no real zeros. Conclusion: Thus, the factored form of the polynomial P(x) that has linear and irreducible quadratic factors is (𝑥 − 5)(𝑥 2 + 4) . b) To find: The factors of the polynomial P(x) that has linear factors with complex coefficients. The polynomial P(x) is, 𝑃(𝑥 ) = 𝑥 3 − 5𝑥 2 + 4𝑥 − 20 Calculation: From part (a) the factored form of the polynomial P(x) is, 𝑃(𝑥 ) = (𝑥 − 5)(𝑥 2 + 4) Now, factor the remaining quadratic factor to obtain the complete factorization as, 𝑃(𝑥 ) = (𝑥 − 5)(𝑥 2 + 4) = (𝑥 − 5)(𝑥 2 − (2𝑖 )2 ) = (𝑥 − 5)(𝑥 − 2𝑖 )(𝑥 + 2𝑖 ) The above factors are linear factors with complex coefficients. Conclusion: Thus, the factored form of the polynomial P(x) that has linear factors is (𝑥 − 5)(𝑥 − 2𝑖 )(𝑥 + 2𝑖 ) .