QA #8 Factors and Multiples

Questions and Answers Sheet 8 Factors and Multiples Question #1 How do you factor 𝑥 2 − 2𝑥 + 2 Answer: To factor our quadratic, we can complete the square and use the difference of squares identity with A=(x1) and B=i. 𝑥 2 − 2𝑥 + 2 = 𝑥 2 − 2𝑥 + 1 + 1 = (𝑥 − 1)2 + 1 = (𝑥 − 1)2 − 𝑖 2 = ((𝑥 − 1) − 𝑖)((𝑥 − 1) + 𝑖) = (𝑥 − 1 − 𝑖 )(𝑥 − 1 + 𝑖 ) Question #2 1 Find the reciprocal of − 5 Answer: 1 5 The reciprocal of − 5 is − 1 = −5 Question #3 How do you simplify 𝑛! (𝑛−2)! ? Answer: Explanation: We can write it as: 𝑛⋅(𝑛−1)(𝑛−2)(𝑛−3)! (𝑛−2)(𝑛−3)! = where you used the fact that 𝑛! = 𝑛(𝑛 − 1)! and so: 𝑛⋅(𝑛−1)(𝑛−2)(𝑛−3)! (𝑛−2)(𝑛−3)! = 𝑛(𝑛 − 1 ) Check it with 𝑛 = 4 𝑛! (𝑛−2)! 4! = 2! = 12 𝑛(𝑛 − 1) = 4 ⋅ 3 = 12 Question #4 What is the factorial of (𝑛 + 1) ? Answer: It is 𝑛! ⋅ (𝑛 + 1) Explanation: Since factorial n (or n!)) is the product of all numbers up to and including n, we only have to multiply by the next number. Question #5 How do you simplify 6!4! 5!8! ? Answer: The factorial of any integer n is represented by 𝑛! = 𝑛(𝑛 − 1)(𝑛 − 2) … (3)(2)(1) So, if we wish to calculate 6!4! 5!8! the simplest way would be to expand each factorial, and cancel the common terms in the numerator and the denominator. 4!6! 5!8! (4⋅3⋅2⋅1)(6⋅5⋅4⋅3⋅2⋅1) = (5⋅4⋅3⋅2⋅1)(8⋅7⋅6⋅5⋅4⋅3⋅2⋅1) If you look closely, all the terms in the numerator are cancelled out by terms in the denominator. Arranging them in this form also makes it easier to figure out which terms are getting cancelled. There are only three terms remaining in the denominator. 4! 6! 1 1 = = 5! 8! (5)(8 ⋅ 7) 280 Question #6 If (𝑛 − 6)! = 6(𝑛 − 5)! , then what is the value of n? Answer: This can be readily simplified to (𝑛−6)! (𝑛−5)! =6 Because the n's cancel each other out. (𝑛−6)! (𝑛−5)(𝑛−6)! 1 𝑛−5 =6 =6 1 = 6𝑛 − 30 31 = 6𝑛 𝑛= 31 6 However, since this is a factorial equation, the solutions need to be integers. Hence the equation has no solution. Hopefully this helps! Question #7 Find the product of √30 and √610 . Express it in standard form. Give the value of a followed by the value of b separeted by a comma. Answer: As per comments question is √30 ⋅ 6√10 = 6 ⋅ √3 = 6 ⋅ 10√3 = 60√3 = 𝑎 ⋅ √𝑏 In the form 𝑎 ⋅ √𝑏: 𝑎 = 60 and 𝑏 = 3 Question #8 How do you factor 𝑥 3 − 27 ? Answer: Both 𝑥 3 and 27 = 33 are perfect cubes. So we can use the difference of cubes identity: 𝑎3 − 𝑏3 = (𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏2 ) with 𝑎 = 𝑥 𝑏 = 3 as follows: 𝑥 − 27 = 𝑥 3 − 33 = (𝑥 − 3)(𝑥 2 + 𝑥(3) + 32 ) = (𝑥 − 3)(𝑥 2 + 3𝑥 + 9) This is as far as you can go with Real coefficients. If you allow Complex coefficients then you can factor this a little further: = (𝑥 − 3)(𝑥 − 3ω)(𝑥 − 3ω2 ) 1 where ω = − 2 + √3 𝑖 2 is the primitive Complex cube root of 1 Question #9 How do you factor 𝑥 4 − 1 ? Answer: We make use of the difference of squares 𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏) 𝑥 4 − 1 = (𝑥 2 + 1)(𝑥 2 − 1) we can use dos for the second bracket once more 𝑥 4 − 1 = (𝑥 2 + 1)(𝑥 + 1)(𝑥 − 1) - (1) for real numbers we can proceed no further, but if we use complex numbers note 𝑖 2 = −1 we see 𝑎2 + 𝑏2 = 𝑎2 − (𝑖𝑏)2 = (𝑎 + 𝑖𝑏)(𝑎 − 𝑖𝑏) → 𝑥 4 − 1 = (𝑥 + 𝑖𝑏)(𝑥 − 𝑖𝑏)(𝑥 + 1)(𝑥 − 1) Question #10 How do you solve 2 3 𝑥=6 Answer: We will multiply both sides by the reciprocal of the fraction on the left-hand side. This will eliminate x's coefficient and give us our answer: 2 3 3 𝑥× =6× 2 3 2 3 𝑥 =6×2 6 3 𝑥 = 1×2 𝑥= 18 2 𝑥=9 Question #11 What is the LCM of 36,8,72,12 ? Answer: So LCM means the lowest common multiple which in other words means what is the smallest number that is divisible by all the numbers you have present. ie what is the smallest number which can be divided by 36,8,72 and 12 without any remainders? So looking at the multiples of all the numbers you have multiplies of 36: 36,72,108, … multiplies of 8: 8,16,24,32,40,48,56,64,72,80,88,96, … multiplies of 72: 72,144,216, … multiplies of 12: 12,24,36,48,60,72,84,96,108,120, … From above, you can clearly see that 72 is the answer because it is the first common number that you can find in each of the multiples of 36,8,72 and 12 . Question #12 How do you factor 2𝑥 2 − 𝑥 − 3 ? Answer: 2𝑥 2 − 𝑥 − 3 = (2𝑥 − 3)(𝑥 + 1) since it is 2𝑥 2 in one of the brackets there must be a x with coefficient 2 (2𝑥 … )(𝑥 … ) next think how 3, 2, 1 (coefficients of 𝑥 2 , 𝑥 𝑎𝑛𝑑 𝑥 0 respectively) relate to each other (using addition and subtraction) 3−2=1 what makes 3 and 2 and 1? using multiplication 3⋅1= 3 2⋅1= 2 Now we have clearer idea on what to do (2𝑥 − 3)(𝑥 + 1) there you go If you are having difficulties to factorise (provided that the equation is factorisable), don't frustrate. You just need more practices. Question #13 White out explicitly and simplity as possible ∑𝑛=5 𝑛=0 𝑛2 𝑛! Answer: Step1 ∑𝑛=5 𝑛=0 𝑛2 𝑛! Step2 ∑𝑛=5 𝑛=0 = 02 0! 𝑛2 𝑛! 12 + 1! + 0 1 22 2! 4 + 9 32 3! + 16 42 4! + 52 5! 25 = 0 + 1 + 2 + 6 + 24 + 120 = 0 + 1 + 2 + 1,5 + 0,67 + 0,21 = 5,38 Question #14 What is the ' factorization of a 150? a) 2 × 2 × 25 . b) 2 × 3 × 5 × 5 c) 5 × 30 . d) 2 × 3 × 5 Answer: Step 1 2×3×5×5 = 6×5×5 = 6 × 25 = 150 Therefore, 150 = 2 × 3 × 5 × 5 Question #15 Find two whole numbers between the product of 5 and Answer: 6 8 and product of 2 and 9 10 ? 6 3\𝑐𝑎𝑛𝑐𝑒𝑙6 Product of 5 × 8 = 5 × 4\𝑐𝑎𝑛𝑐𝑒𝑙8 = 15 4 3 = 34 9 9 9 4 Product of 2 × 10 = 1 𝑐𝑎𝑛𝑐𝑒𝑙 2 × 5\𝑐𝑎𝑛𝑐𝑒𝑙10 = 5 = 1 5 The two whole numbers between which these products lie could be 1 and 4. Question #16 How do you write the ' factorization of −120𝑟2 𝑠𝑡 3 ? Answer: A minus requires a -1 factor 120 = 2 × 60 2 × 60 = 2 × 2 × 30 2 × 2 × 30 = 2 × 2 × 2 × 15 2 × 2 × 2 × 15 = 2 × 2 × 2 × 3 × 5 𝑠𝑡 3 = 𝑠 × 𝑡 × 𝑡 × 𝑡 or 𝑠𝑡 3 So −1 × 2 × 2 × 2 × 3 × 5 × 𝑠 × 𝑡 × 𝑡 × 𝑡 = −1 × 23 × 3 × 5 × 𝑠 × 𝑡 3 Question #17 How do you write the ' factorization of 28𝑥 2 𝑦? Answer: The ' factors of 28 are: 2 \times 2 \times 7 Since x are y are unknowns, I will express the factors of 𝑥 2 𝑦as: 𝑥 × 𝑥 × 𝑦 [With the assumption that x and y are '.] Hence the ' factorisation of 28𝑥 2 𝑦is: 2 × 2 × 7 × 𝑥 × 𝑥 × 𝑦 Question #18 How do you write the ' factorization of −7𝑐 2 ? Answer: −7 × 𝑐 × 𝑐 [Assuming c is '] Explanation: Expression = −7𝑐 2 We are asked to find the ' factors. A few things to do first: (i) ignore the negative sign (ii) Realise that 7 is ' (iii) Assume c is ' Then Expression = 7 × 𝑐 × 𝑐 × (−1) Hence: The ' factors of Expression are: −7 × 𝑐 × 𝑐 [Assuming c is '] Question #19 How do you write the ' factorization of −75𝑎𝑏2 ? Answer: Step 1 Prime numbers are only divisible evenly by 1 or by themselves. The smaller ' numbers are 2, 3, 5, 7, 11. Try dividing the given number by a small ' number and continue dividing until it can no longer be divided. 75 cannot be divided evenly by 2, so try dividing by 3. You get 25. This cannot be divided by 3, so try the next smallest ' number, 5. 25 is divisible by 5, with the result of 5, which is also divisible by 5. Similarly, the alphanumerics a and 𝑏2 each stand for numerals. Their 's are a and b. Finally, the original number was a negative, which means it must have been multiplied by -1. Question #20 What number has the ' factorization 23 × 52 × 7 ? How did you find the number? Answer: Step 1 ( x 2 , x and x 0 Given: 2 3 × 52 × 7 You find the number by multiplying all its factors together. 23 × 52 × 7 8  25 7 \/  ( 200 7) \ / 14 00 Factor 1400 into its ' factors and see if that brings back the same factors that were in the original question. 1400 / \ 14  100 /\ /\ ( 2 7 2  50 )   /\ 2 7 2  2  25    /\ 2 7 2 2 5 5 23 × 52 × 7