Questions and Answers #10 Polynomials

Questions and Answers Sheet 10 Polynomials Question #1 Add the given polynomials. 2x-4,-7x+2, and -4x+9 Answer: Only like terms are added or subtracted. Group the like terms of the expressions to add these three expressions. (2𝑥 − 4) + (−7𝑥 + 2) + (−4𝑥 + 9) = (2𝑥 − 7𝑥 − 4𝑥) + (−4 + 2 + 9) =-9x+7 Conclusion: The sum of the expressions is -9x+7. Question #2 Subtract the polynomials. (8𝑎2 − 7𝑎 + 9) − (𝑎2 − 8𝑎 + 5) Answer: Explanation of the solution Subtract the polynomials (8𝑎2 − 7𝑎 + 9) − (𝑎2 − 8𝑎 + 5) 8𝑎2 − 7𝑎 + 9 − 𝑎2 + 8𝑎 − 5 (8𝑎2 − 𝑎2 ) − 7𝑎 + 8𝑎 + 9 − 5 7𝑎2 + 𝑎 + 4 Question #3 Use the second degree Maclaurin polynomial for ln(1 + 𝑥) to approximate the value of ln(2) Answer: Maclaurin series for a function f(x) is given by 𝑓(𝑥) = 𝑓(0) + 𝑓 ′ (0)𝑥 + 𝑓 ′′ (0) 2! 𝑥2 + ⋯ Take 𝑓(𝑥) = ln(1 + 𝑥) 1 𝑓 ′ (𝑥) = = (1 + 𝑥)−1 1+𝑥 −1 𝑓 ′′ (𝑥) = −1(1 + 𝑥)−2 = (1+𝑥)2 then 𝑓(0) = ln(1 + 0) = ln(1) = 0 1 𝑓 ′ (0) = =1 𝑓 ′′ (0) = 1+0 −1 1+02 = −1 Substitute the values in A ln(1 + 𝑥) = 0 + 1 × 𝑥 + ln(1 + 𝑥) = 𝑥 − 𝑥2 (−1)×𝑥 2 2! 2! is the second degree Maclaurin polynomial for ln(1 + 𝑥) Substitute x=1 in ln(1 + 𝑥) ln(1 + 1) = 1 − (1)2 2! ln(2) = 0.5 The approximate value of ln(2) is 0.5 Question #4 What is the sum of 6𝑥 3 − 5𝑥 + 2 and −2𝑥 3 − 4𝑥 + 𝑥 − 8 ? Answer: When we add two polynomials, we add the sum of the coefficients of the same powers of x. Let 𝑓(𝑥) = 6𝑥 3 − 5𝑥 + 2 and 𝑔(𝑥) = −2𝑥 3 − 4𝑥 + 𝑋 − 0 so that the sum of two polynomials will be 𝑓(𝑥) + 𝑔(𝑥) = (6𝑥 3 − 5𝑥 + 2) + (−2𝑥 3 − 4𝑥 + 𝑥 − 0) = 6𝑥 3 − 5𝑥 + 2 − 2𝑥 3 − 4𝑥 + 𝑥 − 0 = (6 − 2)𝑥 3 + (−5 − 4 + 1)𝑥 + 2 − 0 = 4𝑥 3 + (−8)𝑥 + (−6) 𝑓(𝑥) + 𝑔(𝑥) = 4𝑥 3 − 8𝑥 − 6 Hence the sum of the given polynomials is 4𝑥 3 − 8𝑥 − 6 Question #5 Add the polynomials. (9𝑡 2 + 16𝑡 − 14) + (−2𝑡 2 − 13𝑡 + 10) Answer: Given data is add the polynomials (9𝑡 2 + 16𝑡 − 14) + (−2𝑡 2 − 13𝑡 + 10) Solution: (9𝑡 2 + 16𝑡 − 14) + (−2𝑡 2 − 13𝑡 + 10) 9𝑡 2 + 16𝑡 − 14 − 2𝑡 2 − 13𝑡 + 10 (9𝑡 2 − 2𝑡 2 ) + (16𝑡 − 13𝑡) − 14 + 10 Result: 7𝑡 2 + 3𝑡 − 4 Question #6 Find the roots of the quadratic polynomials: 𝑓(𝑥) = 4𝑥 2 − 3𝑥 − 1 Answer: Given: 𝑓(𝑥) = 4𝑥 2 − 3𝑥 − 1 To find roots Put f(x)=0 4𝑥 2 − 3𝑥 − 1 = 0 4𝑥 2 − 4𝑥 + 𝑥 − 1 = 0 4𝑥(𝑥 − 1) + 1(𝑥 − 1) = 0 (4𝑥 + 1)(𝑥 − 1) = 0 𝑥= −1 4 ,1 Question #7 Find the roots of the quadratic polynomials: 𝑓(𝑥) = 𝑥 2 − 2𝑥 − 1 Answer: We know that, for an standard equation 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, roots are calculated by using quadratic formula as 𝑥 = −𝑏±√𝑏 2 −4𝑎𝑐 2𝑎 We have the given equation as 𝑓(𝑥) = 𝑥 2 − 2𝑥 − 1 On comparing the equation 𝑓(𝑥) = 𝑥 2 − 2𝑥 − 1 with standard equation 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, we get the result as a=1,b=-2, and c=-1 Therefore, by using equation, we get the zeros as 𝑥= 𝑥= 𝑥= −(−2)±√(−2)2 −4(1)(−1) (2)(1) 2±√4+4 2 2±2√2 2 𝑥 = 1 ± √2 Hence, zeros of the quadratic 𝑓(𝑥) = 𝑥 2 − 2𝑥 − 1 are 𝑥 = 1 + √2 and 𝑥 = 1 − √2 Question #8 Multiply the polynomials. (6𝑥 2 + 6𝑥 + 1)(3𝑥 − 1) Answer: Solution: (6𝑥 2 + 6𝑥 + 1)(3𝑥 − 1) ⇒ 3𝑥[6𝑥 2 + 6𝑥 + 1] − 1[6𝑥 2 + 6𝑥 + 1] ⇒ 18𝑥 3 + 18𝑥 2 + 3𝑥 − 6𝑥 2 − 6𝑥 − 1 ⇒ 18𝑥 3 + 18𝑥 2 − 6𝑥 2 − 6𝑥 + 3𝑥 − 1 ⇒ 18𝑥 3 + 12𝑥 2 − 3𝑥 − 1 Hence, the result we get after the multiplication of polynomials is 18𝑥 3 + 12𝑥 2 − 3𝑥 − 1 Question #9 Maclaurin and Taylor polynomials: Find third-order Maclaurin or Taylor polynomial for the given function about the indicated point. tan−1 𝑥, 𝑥0 = 0 Answer: We’ll answer the first question since the exact one wasn’t specified. Please submit a new question specifying the one you’d like answered. We have 𝑓(𝑥) = tan−1 (𝑥) , 𝑥0 = 0 The general form of a taylor expansion centered at a of an function 𝑓(𝑥) = ∑∞ 𝑛=0 𝑛 𝑓 𝑛(𝑎) 𝑛! (𝑥 − 𝑎)𝑛 , Here 𝑓 represent the n th derivative of function. We have a=0 then taylor expression reduce to 𝑓(𝑥) = ∑∞ 𝑛=0 𝑓 𝑛(𝑎) 𝑛! Now, 𝑓(𝑥) = tan−1 (𝑥) 1 𝑓 1 (𝑥) = 2 1+𝑥 𝑓 1 (0) = 1 𝑥𝑛 2𝑥 𝑓 2 (𝑥) = − (1+𝑥 2 )2 𝑓 2 (0) = 0 2 2(1+𝑥 2 ) −2(1+𝑥 2 )(2𝑥)(2𝑥) 𝑓 3 (𝑥) = =− (1+𝑥 2 )4 2 2(1+𝑥 2 ) −8𝑥 2 (1+𝑥 2 ) (1+𝑥 2 )4 3 (0) 𝑓 = −2 Therefore, 𝑓 0 (0) tan−1 (𝑥) = 𝑓 1 (0) 1 𝑥0 + 𝑥 0! 1! (−2) 3 + 𝑥 +⋯ 3! 1 =0+ 𝑥± =𝑥− 1! 𝑥3 3 + 𝑓 2 (0) 2! 𝑥2 + 𝑓 3 (0) 3! 𝑥3 + ⋯ +⋯ Question #10 Let p(x) be a nonzero polynomial of degree less than 1992 having no nonconstant factor in common with 𝑥3 for polynomials f (x) and g(x). Find the smallest possible degree of f (x)? 𝑑1992 𝑝(𝑥) 𝑓(𝑥) ( )= 𝑑𝑥 1992 𝑥 3 − 𝑥 𝑔(𝑥) Answer: Given p(x) be a nonzero polynomial of degree less than 1992 having no non constant factor in common with 𝑥 3 − 𝑥 Let 𝑑 1992 ( 𝑝(𝑥) 𝑑𝑥 1992 𝑥 3 −𝑥 )= 𝑓(𝑥) 𝑔(𝑥) , where f(x) and g(x) are polynomials. To find the smallest possible degree of f(x): BY division algorithm we know that, If f(x) and g(x) are any two polynomials with 𝑔(𝑥) ≠ 0, then 𝑓(𝑥) = 𝑔(𝑥). 𝑞(𝑥) + 𝑟(𝑥), where r(x)=0 or deg 𝑟(𝑥) < 𝑑𝑒𝑔𝑓(𝑥), where q(x) is the quotient and r(x) is the remainder. Hence p(x) can be written as 𝑝(𝑥) = (𝑥 3 − 𝑥)𝑞(𝑥) + 𝑟(𝑥) Here, we can see that the degree of r(x) is 2 and the degree of q(x) is less than 1989 Then, 𝑑 1992 𝑑𝑥 1992 ( 𝑝(𝑥) 𝑥 3 −𝑥 )= 𝑑 1992 ( 𝑟(𝑥) 𝑑𝑥 1992 𝑥 3 −𝑥 ) 𝑟(𝑥) To find 𝑥 3 −𝑥 Using partial fractions, we have 𝑟(𝑥) 𝑥 3 −𝑥 = 𝐴 𝐵 𝑥−1 Hence, + + 𝑥 1992 𝐶 𝑥+1 d r ( x) A B C ( 3 ) = 1992!( ) + 1993 + 1992 1993 dx x −x ( x − 1) x ( x + 1)1993 = 1992! ( = 1992! ( 𝐴𝑥 1993 (𝑥+1)1993 +𝐵(𝑥−1)1993(𝑥+1)1993+𝐶(𝑥−1)1993 𝑥 1993 (𝑥−1)1993 𝑥 1993(𝑥+1)1993 𝐴𝑥 1993 (𝑥+1)1993 +𝐵(𝑥−1)1993(𝑥+1)1993+𝐶(𝑥−1)1993 𝑥 1993 3986 (𝑥 3 −𝑥)1993 3985 ) ) (𝐴 + 𝐵 + 𝐶)𝑥 + 1993(𝐴 − 𝐶)𝑥 + 1993(996𝐴 − 𝐵 + 996𝐶)𝑥 3984 + ⋯ We get A=C=1 and B=2, so the degree is as low as 3984. ⇒𝐴+𝐵+𝐶 =0 A-C=0 996A-B+996C=0 Solving these, we get A=B=C=0 , which is a contradiction. 𝑑 1992 ( 𝑝(𝑥) 𝑑𝑥 1992 𝑥 3 −𝑥 ) cannot be expressed in lower terms since, it will only increase the degree of the numerator. Hence, the smallest possible degree of f(x) is 3984.