QA #4 Factors and Multipless

Questions and Answers Sheet 4 Factors and Multiples Question #1 Simplify. Assume that all variables result in nonzero denominators. Enter the expression in simplest form. The numerator and denominator must be in explanded form (i.e. not a product of factors). 6𝑞 𝑞−4 6 ⇒ 𝑞+1 − 𝑞 + 𝑞+1 = Answer: 6𝑞 ⇒ 𝑞+1 − ⇒ ⇒ 𝑞−4 𝑞 6 + 𝑞+1 ⇒ 6𝑞2 −(𝑞−4)(𝑞+1)+6𝑞 𝑞(𝑞+1) 6𝑞2 −(𝑞2 +𝑞−4𝑞−4)+6𝑞 𝑞(𝑞+1) 6𝑞2 −(𝑞2 −3𝑞−4)+6𝑞 𝑞(𝑞+1) 2 ⇒ 6𝑞 − 𝑞 2 + 3𝑞 + 4 + 6𝑞 𝑞 (𝑞 + 1) ⇒ ⇒ ⇒ ⇒ ⇒ 5𝑞2 +9𝑞+4 𝑞(𝑞+1) 5𝑞2 +5𝑞+4𝑞+4 𝑞(𝑞+1) 5𝑞(𝑞+1)+4(𝑞+1) 𝑞(𝑞+1) (5𝑞+4)(𝑞+1) 𝑞(𝑞+1) 5𝑞+4 𝑞 So, the simplified expression is: ⇒ 5𝑞+4 𝑞 Question #2 Write the exprissuon as a sum and/or difference of logarithms. 7√𝑥+3 log (𝑥 4(𝑥−5)2) 𝑥 > 5 Answer: 7√𝑥+3 Now log (𝑥 4(𝑥−5)2) 𝑥 > 5 𝑚 = log(7√𝑥 + 3) − log(𝑥 4 (𝑥 − 5)2 ) [∵ log ( 𝑛 ) = log 𝑚 − log 𝑛] = log(7) + log(√𝑥 + 3) − log(𝑥4) − log(𝑥 − 5)2 [log(𝑚𝑛) = log 𝑚 + log 𝑛] 1 = log(7) + 2 log(𝑥 + 3)1/2 − 4 log 𝑥 − 2 log(𝑥 − 5) [log 𝑥 𝑚 = 𝑚 log 𝑥 ] 1 = log(7) + 2 log(𝑥 + 3) − 4 log 𝑥 − 2 log(𝑥 − 5) 7√𝑥+3 1 Hence log (𝑥 4(𝑥−5)2) = log(7) + 2 log(𝑥 + 3) − 4 log 𝑥 − 2 log(𝑥 − 5) Question #3 Canning Tomato Products. Jaeger Foods produces tomato sauce and tomato paste, canned in small, −4 1 medium, large, and giant-sized cans. The matrix A 4 0 7 28 0 −8 −32 −4 0 0 −28 0 0 32 gives the size 1 0 0 7 0 0 −8 0 (in ounces) of each container. Ounces Small 6 Medium 10 Large 14 Giant 28 = A The matrix B tabulates one day's production of tomato sauce and tomato paste. Cans of sauce Cans of paste Small 2000 2500 Medium 3000 1500 =B Large 2500 1000 Giant 1000 500 a) Calculate the product AB. b) Interpret the entries in the product matrix AB. Answer: Step 1 Matrix, a set of numbers arranged in rows and columns so as to form a rectangular array. The numbers are called the elements, or entries, of the matrix. And by using the product of the matrix 4 × 1 and 2 × 4 will result the matrix of 2 × 1 .  2000 2500 3000 1500   AB =  6 10 14 28  AB =   2500 1000    1000 500  Step 2 Now multiply the row of the A matrix with the first column of the matrix B and keep on adding the multiples then multiply the row of the A matrix with the second column of the matrix B and keep on adding the multiples.  2000 2500 3000 1500   AB =  6 10 14 28  AB =   2500 1000    1000 500  = [(6 × 2000) + (10 × 3000) + (14 × 2500) + (28 × 1000)(6 × 2500) + (10 × 1500) + (14 × 1000) + (28 × 500)] = [12000 + 30000 + 35000 + 2800015000 + 15000 + 14000 + 14000] = [10500058000] Question #4 A polynomial and one of its factors is given. Factor the polynomial COMPLETELY given that one of its factors is 𝑥 + 4 . Your final answer should be in factored form. 𝑓(𝑥 ) = 𝑥 7 + 4𝑥 6 + 7𝑥 4 + 28𝑥 3 − 8𝑥 − 32 Answer: Step 1 We use the synthetic division to get the other factor for the missing terms, we will use 0 as the coefficient. −4 1 4 0 7 28 0 −8 −32 −4 0 0 −28 0 0 32 1 0 0 7 0 0 −8 0 So the quotient is 𝑥 6 + 7𝑥 3 − 8 Step 2 Then we try to factor 𝑥 6 + 7𝑥 3 − 8 𝑥 6 + 7𝑥 3 − 8 = (𝑥 3 + 8)(𝑥 3 − 1) = (𝑥^{3} + 2^{3})(𝑥 3 − 13 ) = (𝑥 + 2)(𝑥 2 − 2𝑥 + 4)(𝑥 − 1)(𝑥 2 + 𝑥 + 1) Answer: (𝑥 + 4)(𝑥 + 2)(𝑥 2 − 2𝑥 + 4)(𝑥 − 1)(𝑥 2 + 𝑥 + 1) Question #5 Write the following expression as a sum and/or difference of logarithms. Express powers as factors. log 𝑑 (𝑢8 𝑣 3 ) 𝑢 > 0, 𝑣 > 0 log 𝑑 (𝑢8 𝑣 3 ) =? (Simplify your answer.) Answer: Step 1 It is required to write the expression as a sum or difference of logarithms. The given expression is: log 𝑑 (𝑢8 𝑣 3 ) Step 2 Use property: log 𝑑 (𝑎 ⋅ 𝑏) = log 𝑑 (𝑎) + log 𝑑 (𝑏) Step 3 Now, apply the above property to simplify: log 𝑑 (𝑢8 𝑣 3 ) = log𝑑 (𝑢8 ) + log 𝑑 (𝑣 3 ) Step Now use the property: log 𝑑 (𝑥 𝑛 ) = 𝑛 log𝑑 (𝑥 ) Step 5 By applying the property expression reduces to: log 𝑑 (𝑢8 𝑣 3 ) = log𝑑 (𝑢8 ) + log 𝑑 (𝑣 3 ) log 𝑑 (𝑢8 𝑣 3 ) = 8 log𝑑 (𝑢) + 3 log 𝑑 (𝑣) Question #6 Simplify. 2 𝑥+4 9 1+ 𝑥−3 1+ Step 1: Find the LCM of the denominators of the fractions in the numerator and denominator. Step 2: Multiply the numerator and denominator of the complex fraction by the LCM. Step 3: Factor 𝑥−3 𝑥+4 Step 4: Divide out common factors. Answer: Given fraction is: 2 𝑥+4 9 1+ 𝑥−3 1+ To simplify the given fraction. Step1. Find the LCM of the denominators of the fractions in numerators and denominators 2 𝑥+4 9 1+ 𝑥−3 1+ = (𝑥+4)+2 (𝑥+4) (𝑥−3)+9 (𝑥−3) (𝑥+6) (𝑥+4) (𝑥+6) 𝑥−3 Step 2. Multiply the numerator and denominator of the complex fraction by the LCM. (𝑥+6) (𝑥+4) (𝑥+6) 𝑥−3 (𝑥+6)×(𝑥−3) = (𝑥+4)×(𝑥+6) Step 3. Factor (𝑥+6)×(𝑥−3) (𝑥+4)×(𝑥+6) Step 4. Divide out common factors (𝑥+6)×(𝑥−3) (𝑥+4)×(𝑥+6) 𝑥−3 = (𝑥+4) 𝑥−3 Thus, the simplified form is (𝑥+4). Question #7 Determine the lowest common multiple (LCM) and the highest common factor (HCF) of the following algebraic terms by using ' factors: 22𝑎5 𝑏3 𝑐 2 ; 2𝑎4 𝑏4 𝑐 3 ; 4𝑎2 𝑏2 𝑐 5 Answer: Step 1 Given , 𝑋 = 22𝑎5 𝑏3 𝑐 2 𝑌 = 2𝑎4 𝑏4 𝑐 3 𝑍 = 4𝑎2 𝑏2 𝑐 5 Step 2 22𝑎5 𝑏3 𝑐 2 ; 2𝑎4 𝑏4 𝑐 3 ; 4𝑎2 𝑏2 𝑐 5 Let 𝑥 = 2 × 11 × 𝑎5 × 𝑏3 × 𝑐 2 𝑦 = 2 × 𝑎4 × 𝑏4 × 𝑐 3 𝑧 = 2 × 2 × 𝑎2 × 𝑏2 × 𝑐 5 𝐿𝐶𝑀 = 2 × 11 × 2 × 𝑎5 × 𝑏4 × 𝑐 5 𝐿𝐶𝑀 = 44𝑎5 𝑏4 𝑐 5 𝐻𝐶𝐹 = 2 × 𝑎2 × 𝑏2 × 𝑐 2 𝐻𝐶𝐹 = 2𝑎2 𝑏2 𝑐 2 Question #8 Write the expression as a sum and/or difference of logarithms. Express powers as factors. ln 2𝑥 √1+8𝑥 ln 2𝑥 √1+8𝑥 (𝑥−3)13 (𝑥−3)13 ,𝑥 > 3 = (Simplify your answer.) Answer: Step 1 The given expression is 2𝑥 √1+8𝑥 ln (𝑥−3)13 ,𝑥 > 3 Step 2 Using the formula 𝑎 ln ( ) = ln 𝑎 − ln 𝑏 𝑏 ln(𝑎𝑏) = ln 𝑎 + ln 𝑏 ln 𝑎𝑛 = 𝑛 ln 𝑎 Step 3 On simplifying with the help of given formula 1 ln 2𝑥 √1+8𝑥 (𝑥−3)13 2𝑥(1+8𝑥)2 = ln (𝑥−3)13 1 2 = ln 2 𝑥(1 + 8𝑥 ) − ln(𝑥 − 3)13 1 = ln 2 𝑥 + ln(1 + 8𝑥 )2 − ln(𝑥 − 3)13 1 = ln 2 𝑥 + ln(1 + 8𝑥 )2 − ln(𝑥 − 3)13 1 = ln 2 + ln 𝑥 + 2 ln(1 + 8𝑥 ) − 13 ln(𝑥 − 3) ⇒ ln 2𝑥 √1+8𝑥 (𝑥−3)13 1 = ln 2 + ln 𝑥 + 2 ln(1 + 8𝑥 ) − 13 ln(𝑥 − 3) Question #9 To simplify the expression, √23 . 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 ⇒ √𝑚4 × 𝑛2 = 𝑚2 × 𝑛2 = 𝑚2 × 𝑛1 Calculations: As per the question, the expression is √23 . The factorization of a number is the product of numbers that equals the number. Now, 23 is a ' number. Hence, it has only two multiples i.e. 1 and 23. Therefore, 12 cannot be factorized anymore. Hence, the simplified value of √23 is √23 only. Question #10 Write the expression as a sum and/or difference of logarithms. Express powers as factors. 𝑥(𝑥+4) log [(𝑥+2)5 ] , 𝑥 > 0 𝑥(𝑥+4) log [(𝑥+2)5 ] =? (Simplify your answer.) Answer: Step 1 It is required to simplify the given expression: 𝑥(𝑥+4) log [(𝑥+2)5 ] Step 2 Apply these properties of logarithm to simplify: 𝑎 log (𝑏 ) = log(𝑎) − log(𝑏) log(𝑎 ⋅ 𝑏) = log(𝑎) + log(𝑏) log(𝑥 𝑛 ) = 𝑛 log(𝑥 ) Step 3 Now, by using these properties simplify the given expression: 𝑥(𝑥+4) log ((𝑥+2)5 ) = log(𝑥 (𝑥 + 4)) − log((𝑥 + 2)5 ) = log(𝑥 ) + log(𝑥 + 4) − 5 log(𝑥 + 2) Question #11 To calculate: The simplified value of the expression 5√32𝑥 16 𝑦 10 using radical notation. Answer: Formula used: From the product rule, 𝑛 𝑛 𝑛 √𝑎𝑏 = √𝑎 ⋅ √𝑏 Where a and b are positive number and n is larger than 1. Calculation: Consider the provided expression, 5 √32𝑥 16 𝑦 10 Expand the expression by identifying the largest exponents and multiples of 5, 5 5 √32𝑥 16 𝑦 10 = √25 ⋅ 𝑥 15 ⋅ 𝑥 ⋅ 𝑦 10 Now, separate the terms with fifth power factors by using the product rule, 5 5 5 5 5 √25 ⋅ 𝑥 15 ⋅ 𝑥 ⋅ 𝑦 10 = √25 ⋅ √𝑥 15 ⋅ √𝑥 ⋅ √𝑦 10 Simplify it further, 5 5 √25 ⋅ √𝑥 15 ⋅ 5√𝑥 ⋅ 5√𝑦 10 = 2 ⋅ 𝑥 3 ⋅ 5√𝑥 ⋅ 𝑦 2 = 2𝑥 3 𝑦 2 5√𝑥 Hence, the simplified value of expression is 2𝑥 3 𝑦 2 5√𝑥 Question #12 a) Using the remainder theorem, determine whether (𝑥 − 4) and (𝑥 − 1) are factors of the expression 𝑥 3 + 3𝑥 2 − 22𝑥 − 24 . b) Hence, by use of long division, find all remaining factors of the expression. Answer: Step 1 Given: 𝑥 3 + 3𝑥 2 − 22𝑥 − 24 a) (𝑥 − 4) Factor 𝑥 − 4 = 0 𝑥−4 (4)3 + 3(4)2 − 22(4) − 24 = 64 + 48 − 88 − 24 = 112 − 112 = 0 (𝑥 − 1) Factor 𝑥 − 1 = 0 𝑥=1 (1)3 + 3(1)2 − 22(1) − 24 4 1 3 −22 −24 0 4 28 24 = 1 + 3 − 22 − 24 1 7 6 0 = 4 − 46 = −40 (not a factor) Step 2 b) (𝑥 − 4) Factor of 𝑥 [3} + 3𝑥 2 − 22𝑛 − 24 𝑥=4 4 1 3 −22 −24 0 4 28 24 1 7 6 0 𝑥 2 + 7𝑥 + 6 = 𝑥 2 + 6𝑥 + 𝑥 + 6 = 𝑥 (𝑥 + 6) + 1(𝑥 + 6) = (𝑥 + 6)(𝑥 + 1) (𝑥 + 6)(𝑥 + 1) are the remaining factors. Question #13 Given 𝑃 (𝑥 ) = 𝑥 3 + 5𝑥 2 + 25𝑥 + 125 , find the zeros, real and non-real of 𝑃 . The zeros are 5. Write P in factored form (as a product of linear factors). Be sure to write the full equation, including 𝑃(𝑥 ) =. Answer: Step 1 Given 𝑃 (𝑥 ) = 𝑥 3 + 5𝑥 2 + 25𝑥 + 125 We have to write P in factored from Step 2 given: 𝑝(𝑥 ) = 𝑥 3 + 5𝑥 2 + 25𝑥 + 125 𝑝(𝑥 ) = (𝑥 3 + 5𝑥 2 ) + (25𝑥 + 125) 𝑝(𝑥 ) = 𝑥 2 (𝑥 + 5) + 25(𝑥 + 5) 𝑝(𝑥 ) = (𝑥 2 + 25)(𝑥 + 5) Question #14 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: 𝑥2 + 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 #15 Given 𝑃 (𝑥 ) = 3𝑥 5 − 𝑥 4 + 81𝑥 3 − 27𝑥 2 − 972𝑥 + 324 , and that 6i 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 𝑃(𝑥 ) = 3𝑥 5 − 𝑥 4 + 81𝑥 3 − 27𝑥 2 − 972𝑥 + 324 Given: 6i is a zero. ⇒ −6𝑖 is also a zero. ⇒ (𝑥 + 6𝑖 )(𝑥 − 6𝑖 ) in a factor ⇒ (𝑥 2 + 36) in a factor of P(x). Step 2 𝑃(𝑥 ) = 3𝑥 5 − 𝑥 4 + 81𝑥 3 − 27𝑥 2 − 972𝑥 + 324 = 𝑥 4 (3𝑥 − 1) + 27𝑥 2 (3𝑥 − 1) − 324(3𝑥 − 1) = (3𝑥 − 1)(𝑥 4 + 27𝑥 2 − 324) = (3𝑥 − 1)[𝑥 2 (𝑥 2 + 36) − 9(𝑥 2 + 36)] = (3𝑥 − 1)(𝑥 2 + 36)(𝑥 2 − 9) = (3𝑥 − 1)(𝑥 2 + 36)(𝑥 + 3)(𝑥 − 3) Factored form of P(x): 𝑃(𝑥 ) = (3𝑥 − 1)(𝑥 2 + 36)(𝑥 + 3)(𝑥 − 3) Question #16 Given 𝑃 (𝑥 ) = 3𝑥 5 − 4𝑥 4 + 9𝑥 3 − 12𝑥 2 − 12𝑥 + 16 , and that 2i 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: 𝑃(𝑥 ) = 3𝑥 5 − 4𝑥 4 + 9𝑥 3 − 12𝑥 2 − 12𝑥 + 16 Given that, 2i is a zero of P(x) ∴ −2𝑖 is also a zero of P(x) ∴ (𝑥 − 2𝑖 ), (𝑥 + 2𝑖 ) are two factors of P(x) Now (𝑥 − 2𝑖 )(𝑥 + 2𝑖 ) = 𝑥 2 − (2𝑖 )2 = 𝑥 2 − 4𝑖 2 𝑥2 + 4 ∴ 𝑃 (𝑥 ) = 3𝑥 5 − 4𝑥 4 + 9𝑥 3 − 12𝑥 2 − 12𝑥 + 16 = 3𝑥 3 (𝑥 2 + 4) − 4𝑥 2 (𝑥 2 + 4) − 3𝑥(𝑥 2 + 4) + 4(𝑥 2 + 4) = (𝑥 2 + 4)(3𝑥 3 − 4𝑥 2 − 3𝑥 + 4) = (𝑥 2 + 4)[3𝑥 2 (3𝑥 − 4) − 1(3𝑥 − 4)] = (𝑥 2 + 4)(3𝑥 − 4)(𝑥 2 − 1) = (𝑥 2 + 4)(3𝑥 − 4)(𝑥 + 1)(𝑥 − 1) = (𝑥 + 2𝑖 )(𝑥 − 2𝑖 )(3𝑥 − 4)(𝑥 + 1)(𝑥 − 1) For zeros, 𝑃 (𝑥 ) = 0 ⇒ 3𝑥 5 − 4𝑥 4 + 9𝑥 3 − 12𝑥 [2} − 12𝑥 + 16 = 0 ⇒ (𝑥 + 2𝑖 )(𝑥 − 2𝑖 )(3𝑥 − 4)(𝑥 + 1)(𝑥 − 1) = 0 ∴ 𝑍𝑒𝑟𝑜𝑠𝑜𝑓𝑃(𝑥 ) are 4 −2𝑖, 2𝑖, 3 , −1,1 . Question #17 Given 𝑃 (𝑥 ) = 3𝑥 5 − 8𝑥 4 + 35𝑥 3 − 98𝑥 2 − 208𝑥 + 480 , 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 Given: 𝑃 (𝑥) = 3𝑥 5 − 8𝑥 4 + 35𝑥 3 − 98𝑥 2 − 208𝑥 + 480 It is given that 4i is a zero of P(x) We know that for polynomials with real coefficients, the complex zeros always exists in conjugate pairs Hence, -4i is also a zero of P(x) Hence, (𝑥 − 4𝑖 ), (𝑥 + 4𝑖 ) are factors of P(x) So, (𝑥 − 4𝑖 )(𝑥 + 4𝑖 ) is a factor of P(x) ⇒ 𝑥 2 + 16 is a factor of P(x) Dividing P(x) by 𝑥 2 + 16 So, 3𝑥 5 −8𝑥 4 +35𝑥 3 −98𝑥 2 −208𝑥+480 𝑥 2 +16 = 3𝑥 3 − 8𝑥 2 − 13𝑥 + 30 𝑃(𝑥 ) = (3𝑥 3 − 8𝑥 2 − 13𝑥 + 30)(𝑥 2 + 16) Step 2 Consider 𝐹(𝑥 ) = 3𝑥 3 − 8𝑥 2 − 13𝑥 + 30 We observe that 𝐹(−2) = 3(−2)3 − 8(−2)2 − 13(−2) + 30 𝐹(−2) = −24 − 32 + 26 + 30 𝐹(−2) = 0 Similarly, 𝐹(3) = 0 So, -2,3 are zeros of F(x) and hence of P(x) too Let 𝛼 be the fifth zero of P(x) We know that the sum of zeros of 𝑃(𝑥 ) = 8/3 8 𝛼 + 4𝑖 − 4𝑖 + (−2) + (3) = 3 8 𝛼 = 3−1 5 𝛼=3 5 Hence, the zeros are −2,3,4𝑖, −4𝑖, 3 5 Hence, 𝑃(𝑥 ) = 3(𝑥 + 2)(𝑥 − 3)(𝑥 − 4𝑖 )(𝑥 + 4𝑖 ) (𝑥 − 3) Question #18 To solve the given equation for x. Given equation: log 3(𝑥 2 − 𝑥 − 6) − log 3(𝑥 − 3) = 1 Answer: Concept used: Quotient property of logarithm: log 𝑎 log 𝑎 𝑎=1. 𝑏 − log 𝑎 𝑏 𝑐 = log 𝑎 ( 𝑐 ) Calculation: log 3 ( 𝑥 2 −𝑥−6 𝑥−3 )=1 Next step is no factor the numerator 𝑥 2 − 𝑥 − 6. To factorize the above trinomial, first step is to find the two multiples of the constant term -6 so that their addition will result the coefficient of x which is -1. So, 6 = (−3) ⋅ (2) Addition of -3 and 2 will result -1. So, the factors form of: 𝑥 2 − 𝑥 − 6 = (𝑥 − 3)(𝑥 + 2) Hence, the equation will be: log 3 ( (𝑥−3)(𝑥+2) 𝑥−3 )=1 log 3(𝑥 + 2) = 1 Cancel out (𝑥 − 3) from both numerator and denominator. 𝑙𝑜𝑔3 (𝑥 + 2) = 𝑙𝑜𝑔3 3 Since, log 𝑎 𝑎 = 1. 𝑥+2=3 𝑥 + 2 − 2 = 3 − 2 By subtracting 2 from each sides of the equation. 𝑥 =1. Question #19 Carla used a calculator and found out that 31.5 × 21.68 = 682.92 . She is suspicious about her answer because it doesn't have three decimal places in it. Explain to her why the answer is correct by reasoning about the size of the factors, and also why her answer doesn't have 2+1=3 decimal places. Answer: Step 1 31.5 × 21.68 = 682.92 We know that 3 × 2 = 6 Also, 30 × 20 = 600 Thus solution should be > 600 . Step 2 Also, 8 (which is the last decimal digit of 21.68) × 5 (which is the last decimal digit of 31.5) = 40 . This 0 cancels the 3rd decimal digit in the answer. Thus, the solution has 2 decimal digits. Question #20 To solve the given equation for x. Given equation: log 5(𝑥 2 − 5𝑥 + 6) − log5 (𝑥 − 2) = 1 Answer: Concept used: Quotient property of logarithm: log 𝑎 log 𝑎 𝑎=1. Calculation: 𝑏 − log 𝑎 𝑏 𝑐 = log 𝑎 ( ) 𝑐 log 5 ( 𝑥 2 −5𝑥+6 𝑥−2 )=1 Next step is no factor the numerator 𝑥 2 − 5𝑥 + 6 . To factorize the above trinomial, first step is to find the two multiples of the constant term 6 so that their addition will result the coefficient of x which is -5. So, 6 = (−3) ⋅ (2) Addition of -3 and 2 will result -5. So, the factors form of: 𝑥 2 − 5𝑥 + 6 = (𝑥 − 3)(𝑥 − 2) Hence, the equation will be: log 5 ( (𝑥−3)(𝑥−2) 𝑥−2 )=1 log 5(𝑥 − 3) = 1 Cancel out (x-2) from both numerator and denominator. 𝑙𝑜𝑔5 (𝑥 − 3) = 𝑙𝑜𝑔5 5 Since, log𝑎 𝑎=1. 𝑥−3=5 𝑥 − 3 + 3 = 5 + 3 By adding 3 from each sides of the equation. 𝑥=8.