Questions and Answers #10 Polynomial Arithmetic

Questions and Answers Sheet 10 Polynomial Arithmetic Question #1 What is the derivative of the function 𝑓(𝑥) = 3𝑥−5 5𝑥+2 ? Answer: The quotient rule states that if 𝑦 = 𝑓(𝑥) 𝑔(𝑥) , then 𝑑𝑦 𝑑𝑥 = 𝑓 ′(𝑥)𝑔(𝑥)−𝑔′ (𝑥)𝑓(𝑥) 2 (𝑔(𝑥)) Now if we apply this to the given function 𝑓 ′ (𝑥) = 3(5𝑥+2)−5(3𝑥−5) (5𝑥+2)2 15𝑥+6−15𝑥+25 𝑓 ′ (𝑥) = 𝑓 ′ (𝑥) = (5𝑥+2)2 (5𝑥+2)2 31 Question #2 What is the inverse function of 𝑓(𝑥) = 5 − 4𝑥 3 ? Answer: Find the inverse function of 𝑦 = 5 − 4𝑥 3 Let 𝑥 = 5 − 4𝑦 3 , then solve for y. 4𝑦 3 = 5 − 𝑥 𝑦3 = 𝑦 1 3× 3 5−𝑥 4 =( 1 5−𝑥 3 ) 4 3 5−𝑥 𝑦= √ 4 Question #3 Does the (n-1)th derivative of an n-degree polynomial give the arithmetic mean of all the roots of the polynomial? Answer: Consider, 𝑎0 𝑥 𝑛 + 𝑎1 𝑥 𝑛−1 + ⋯ + 𝑎𝑛 = 0 𝑎 We have, sum of roots = − 1 𝑎0 Now if u take the (𝑛 − 1)𝑡ℎ derivative of the expression you have, 𝑛! 𝑎0 𝑥 + (𝑛 − 1)! 𝑎1 = 0 𝑎1 −𝑎1 1 ⇒𝑥=− = × 𝑛𝑎0 𝑎0 𝑥 This is the arithmetic mean of the roots we have the sum of the roots divided by n. Question #4 How do I find value of k so that zeroes of polynomial 𝑛3 + 12𝑛2 + 39𝑛 + 𝑘 are in arithmetic progression? Answer: Let’s assume roots/zeros of the polynomial are r-d, r and r+d From the given polynomial the sum of the roots is -12 According to our assumption the sum is 3r So, 3r=-12 or r=-4 Now 𝐹(𝑛) = 𝑛3 + 12𝑛2 + 39𝑛 + 𝑘 has a root -4, so F(-4)=0 This gives k=28 If you want to find the roots: From the given polynomial the product of the roots is −𝑘 or − 28 According to our assumption the product of the roots is 4𝑑 2 − 64 So, 4𝑑 2 − 64 = −28 or 𝑑 = ±3 Thus the roots are -1, -4 and -7 Question #5 Is it true that if a, b, c are in arithmetic progression, and if the polynomials (𝑏 − 𝑐)𝑥 2 + (𝑐 − 𝑎)𝑥 + (𝑎 − 𝑏) and 2(𝑐 + 𝑎)𝑥 2 + (𝑏 + 𝑐)𝑥 have a common root, then 𝑎2 , 𝑐 2 , 𝑏 2 must be un arithmetic progression also? Answer: This one’s a damn easy. Let’s sail in the seas of AP. It is stated that a,b,c are in AP. So 2b=a+c Now let’s factorize the first quadratic polynomial with a bit of skill: (𝑏 − 𝑐)𝑥 2 + (𝑐 − 𝑎)𝑥 + (𝑎 − 𝑏) = 0 (𝑏 − 𝑐)𝑥 2 − (𝑏 − 𝑐)𝑥 − (𝑎 − 𝑏)𝑥 + (𝑎 − 𝑏) = 0 (𝑏 − 𝑐)𝑥(𝑥 − 1) − (𝑎 − 𝑏)(𝑥 − 1) = 0 ((𝑏 − 𝑐)𝑥 − (𝑎 − 𝑏))(𝑥 − 1) = 0 𝑥= 𝑏−𝑐 ,1 𝑎−𝑏 𝑥= , 𝑎−𝑏 𝑥 = 1, 1 Now, let's do something with the second one: 2(𝑐 + 𝑎)𝑥 2 + (𝑏 + 𝑐)𝑥 = 0 (𝑥)(2(𝑐 + 𝑎)𝑥 + (𝑏 + 𝑐)) = 0 𝑎−𝑏 𝑥 = 0, − 𝑥 = 0, − 𝑥 = 0, − 𝑏+𝑐 2(𝑐+𝑎) 𝑏+𝑐 2×2𝑏 𝑏+𝑐 4𝑏 Now it is completely sure that: − 𝑏+𝑐 4𝑏 =1 -4b=b+c c=-5b as a+c=2b a-5b=2b a=7b So, 𝑐 2 − 𝑎2 = 25𝑏 2 − 49𝑏 2 = −24𝑏 2 𝑏 2 − 𝑐 2 = 𝑏 2 − 25𝑏 2 = −24𝑏 2 So, 𝑐 2 − 𝑎2 = 𝑏 2 − 𝑐 2 So, it is sure that 𝑎2 , 𝑐 2 , 𝑏 2 are in AP. Question #6 How do I prove that 2𝑏 2 − 3𝑎𝑏𝑐 + 𝑑𝑎2 = 0 if the roots of the polynomial 𝑓(𝑥) = 𝑎𝑥 3 + 3𝑏𝑥 2 + 3𝑐𝑥 + 𝑑 are in arithmetic progression? Answer: Let x-y, x, x+y be roots of the given equation. 1) 𝑥 − 𝑦 + 𝑥 + 𝑥 + 𝑦 = − 3𝑏 𝑎 ⇒𝑥=− 𝑏 𝑎 2) (𝑥 − 𝑦)𝑥 + 𝑥(𝑥 + 𝑦) + (𝑥 + 𝑦)(𝑥 − 𝑦) = ⇒ 3𝑥 2 − 𝑦 2 = ⇒ 𝑦2 = 3𝑐 3(𝑏 2 −𝑎𝑐) 3𝑐 𝑎 𝑎 𝑎2 3) (𝑥 − 𝑦)𝑥(𝑥 + 𝑦) = − ⇒ 𝑥(𝑥 2 − 𝑦 2 ) = − 𝑑 𝑎 𝑑 𝑎 𝑏2 2 −𝑏 𝑎2 − 3(𝑏 − 𝑎𝑐) 𝑑 ⇒ × =− 2 𝑎 𝑎 𝑎 ⇒ 2𝑏 3 − 3𝑎𝑏𝑐 + 𝑑𝑎2 = 0 Question #7 If x-y=3, then what is 𝑥 3 − 𝑦 3 − 9𝑥𝑦? Answer: Simply cube the equation So (𝑥 − 𝑦)3 = 𝑥 3 − 𝑦 3 − 3𝑥𝑦(𝑥 − 𝑦) = 33 = 27 But we know x-y=3. So substituting in the equation So 𝑥 3 − 𝑦 3 − 3𝑥𝑦(3) = 𝑥 3 − 𝑦 3 − 9𝑥𝑦 = 27 Question #8 Is the number 𝑛2 + 𝑛 + 41 a ' for all positive integers n? Answer: In general the claim is not true. 𝑛2 + 𝑛 + 42 = 𝑛(𝑛 + 1) + 41 can not be a ' number when 𝑛(𝑛 + 1) ≡ 0(𝑚𝑜𝑑 41) For example for n=40 n(n+1)+41 =40(41)+41 =41(40+1) = 412 is not a' number Similarly for n=41 n(n+1)+41 =41(42)+41 41(42+1) =41(43) is not a' number. Question #9 1 1 𝑥 𝑥𝑛 If 𝑥 + = 𝑎, then what is 𝑥 𝑛 + equal to? Answer: Assuming x is a complex number, let 𝑥 = 𝑒 𝑖𝜙 for some complex 𝜙 Euler’s identity then tells us that 𝑥 = cos 𝜙 + 𝑖 sin 𝜙 1 Similarly, = 𝑒 −𝑖𝜙 , which is equal to cos 𝜙 − 𝑖 sin 𝜙 𝑥 1 Now, 𝑥 + = (cos 𝜙 + 𝑖 sin 𝜙) + (cos 𝜙 − 𝑖 sin 𝜙) = 2 cos 𝜙, which is equal to a 𝑥 This means cos 𝜙 = Or, 𝜙 = cos −1 𝑎 𝑎 2 2 Again, 𝑥 𝑛 can be written as 𝑒 𝑖𝑛𝜙 = cos 𝑛 𝜙 + 𝑖 sin 𝑛 𝜙 and This means 𝑥 𝑛 + 1 𝑥𝑛 1 𝑥𝑛 as 𝑒 −𝑖𝑛𝜙 = cos 𝑛 𝜙 − 𝑖 sin 𝑛 = 2 cos 𝑛 𝜙, which is equal to 2 cos (𝑛 cos −1 𝑎 2 ) And there we have it: 𝑥𝑛 + 1 𝑥𝑛 𝑎 = 2 cos (𝑛 cos −1 ) 2 Question #10 How can I find a root of a polynomial, say, 𝑥 101 + 𝑥 7 − 1, without any approximation methods? Answer: If you mean that you wish to express the roots using only the coefficients of the polynomial, the four basic arithmetic operators and radicals, if the polynomial is a quintic or higher, you cannot in general. There are, of course, many specific cases when you can. For instance, the polynomial in your question factors into (𝑥 2 − 𝑥 + 1)𝑃99 (𝑥) where 𝑃99 (𝑥) is a lengthy 99-th power polynomial that (as far as I can tell) cannot be factorized any further. The quadratic equation 𝑥 2 − 𝑥 + 1 = 0 can be easily solved yielding the complex roots 𝑥 = 1±√−3 2 . But the one real root of your polynomial, at 𝑥 ≅ 0.980097 , cannot be found this way. Having said that, once you reduced a more complex problem to finding the roots of a rational polynomial, it is often viewed as good as being solved, with the roots having been found for all practical intents and purposes. That is because the values of the roots can always be calculated easily and efficiently. In a sense it's no different from finding the root of the equation 𝑥 2 − 2 = 0; we cannot do that using the four arithmetic operators only, so we invent the radical symbol which allows us to write 𝑥 = ±√2.