Questions and Answers #12 Polynomials

Questions and Answers Sheet 12 Polynomials Question #1 Subtract the polynomials (10𝑥 3 − 5𝑥 2 + 3𝑥 − 4) − (6𝑥 3 + 4𝑥 2 − 3𝑥 − 1) Answer: We have to subtract the following Subtract the polynomials (10𝑥 3 − 5𝑥 2 + 3𝑥 − 4) − (6𝑥 3 + 4𝑥 2 − 3𝑥 − 1) Solution: (10𝑥 3 − 5𝑥 2 + 3𝑥 − 4) − (6𝑥 3 + 4𝑥 2 − 3𝑥 − 1) = 10𝑥 3 − 5𝑥 2 + 3𝑥 − 4 − 6𝑥 3 − 4𝑥 2 + 3𝑥 + 1 = 4𝑥 3 − 9𝑥 2 + 6𝑥 − 3 Hence, (10𝑥 3 − 5𝑥 2 + 3𝑥 − 4) − (6𝑥 3 + 4𝑥 2 − 3𝑥 − 1) = 4𝑥 3 − 9𝑥 2 + 6𝑥 − 3 Question #2 Applying the theorem on bounds for real zeros of polynomials, determine the smallest and largest integers that are upper and lower bounds, respectively, for the real solutions of the equation 2𝑥 4 − 9𝑥 3 − 8𝑥 − 10 − 0 Answer: The given polynomial is 2𝑥 4 − 9𝑥 3 − 8𝑥 − 10 = 0 2𝑥 4 − 9𝑥 3 − 8𝑥 − 10 = 0 divide by 2, 9 𝑥 4 − 𝑥 3 − 4𝑥 − 5 = 0 2 The coefficients are, 1,-4.5,-4,-5 remove minus sign 1,4.5,4,5 Remove 1, we have the list of values 4.5,4,5 Bound 1: largest value is 5+1=6 Bound 2: sum of all values is 4.5+4+5=13.5 smallest value is 6 Therefore the roots lies between -6 and 6. Question #3 Subtract the polynomials using the horizontal format. 2𝑎2 − 6𝑎 − 4 from −4𝑎2 + 6𝑎 + 10 Answer: We have to subtract: (−4𝑎2 + 6𝑎 + 10) − (2𝑎2 − 6𝑎 − 4) The subtraction for the polynomials can be given as: (−4𝑎2 + 6𝑎 + 10) − (2𝑎2 − 6𝑎 − 4) = (−4𝑎2 − 2𝑎2 ) + (6𝑎 + 6𝑎) + (10 + 4) = −6𝑎2 + 12𝑎 + 14 which is the required solution Question #4 Find the roots of the quadratic polynomials: a) 𝑓(𝑥) = 4𝑥 2 − 3𝑥 − 1 b) 𝑓(𝑥) = 𝑥 2 − 2𝑥 − 1 Answer: We set them equal to 0 and solve for x we first factor the left side 4𝑥 2 − 3𝑥 − 1 = 0 4𝑥 2 − 4𝑥 + 𝑥 − 1 = 0 4𝑥(𝑥 − 1) + 1(𝑥 − 1) = 0 (𝑥 − 1)(4𝑥 + 1) = 0 then we set rach factor equal to 0 and find x x-1=0 x=1 4x+1=0 4x=-1 1 𝑥=− 4 b) This expression can not be factored so we use the quadratic formula 𝑥 2 − 2𝑥 − 1 = 0 𝑥= 𝑥= 𝑥= −(−2)±√(−2)2 −4(1)(−1) 2(1) 2±√8 2 2±2√2 2 𝑥 = 1 ± √2 Answer: 𝑥 = 1 ± √2 Question #5 Show that 𝑓(𝑥) = 𝑥 2 + 3𝑥 −1 and 𝑔(𝑥) = 3𝑥 3 − 9𝑥 + 𝑥 −2 are rational functions that is, quotients of polynomials. Answer: To show: That the functions are rational function: Given: The functions are 𝑓(𝑥) = 𝑥 2 + 3𝑥 −1 and 𝑔(𝑥) = 3𝑥 3 − 9𝑥 + 𝑥 −2 Concept used: The function h(x) is rational function if h(x) is represented as 𝑝(𝑥) 𝑞(𝑥) Here, p(x) and q(x) are polynomial function. Verification: Check first function 𝑓(𝑥) = 𝑥 2 + 3𝑥 −1 1 𝑓(𝑥) = 𝑥 2 + 3 × = 𝑥 𝑥 3 +3 𝑥 Therefore, the function 𝑓(𝑥) = 𝑥 2 + 3𝑥 −1 can be represented as is rational function. Check first function 𝑔(𝑥) = 3𝑥 3 − 9𝑥 + 𝑥 −2 𝑔(𝑥) = 3𝑥 3 − 9𝑥 + 𝑥 −2 1 = 3𝑥 3 − 9𝑥 + 2 = 3𝑥 5 −9𝑥 3 +1 𝑥2 𝑥 𝑝(𝑥) 𝑞(𝑥) , then the function 𝑓(𝑥) = 𝑥 2 + 3𝑥 −1 Therefore, the function 𝑔(𝑥) = 3𝑥 3 − 9𝑥 + 𝑥 −2 can be represented 𝑝(𝑥) 𝑞(𝑥) , then the function 𝑔(𝑥) = 3𝑥 3 − 9𝑥 + 𝑥 −2 is rational function. Question #6 Find all the roots (real and imaginary) of the given polynomials. Describe multiplicity when applicable. Please show all work and simplify your answers. 𝑓(𝑥) = 𝑥 4 − 8𝑥 2 − 8𝑥 + 15 Answer: Given 𝑓(𝑥) = 𝑥 4 − 8𝑥 2 − 8𝑥 + 15 The all the roots of the above expression is X=1 X=3 X=-2+i X=-2-i = (𝑥 − 1)(𝑥 − 3)(𝑥 + 2 − 𝑖)(𝑥 + 2 + 𝑖) = (𝑥 2 − 3𝑥 − 𝑥 + 3)(𝑥 2 + 2𝑥 + 𝑥𝑖 + 2𝑥 + 4 + 2𝑖 − 𝑥𝑖 − 2𝑖 + 1) = (𝑥 2 − 4𝑥 + 3)(𝑥 2 + 4𝑥 + 5) = (𝑥 4 + 4𝑥 3 + 5𝑥 2 − 4𝑥 3 − 16𝑥 2 − 20𝑥 + 3𝑥 2 + 12𝑥 + 15) = 𝑥 4 − 8𝑥 2 − 8𝑥 + 15 Question #7 Let 𝑓(𝑥) = 𝑥 3 + 𝑥 2 + 𝑥 + 1 ∈ 𝑍2 [𝑥]. Write f(x) as a product of irreducible polynomials over 𝑍2 Answer: Given: 𝑓(𝑥) = 𝑥 3 + 𝑥 2 + 𝑥 + 1 The given polynomial can be factorized as a product two irreducible polynomials as, 𝑥 3 + 𝑥 2 + 𝑥 + 1 = 𝑥 2 (𝑥 + 1) + (𝑥 + 1) = (𝑥 + 1)(𝑥 2 + 1) Here, the polynomials (x+1) is in irreducible form which cannot be factorized further. (𝑥 2 + 1) can be factorized over Z as, 𝑥 2 + 1 = (𝑥 + 𝑖)(𝑥 − 𝑖) Here, both (x+i) and (x-i) are in irreducible form. Hence, the given polynomial can be expressed as a product of irreducible polynomial over Z as, 𝑥 3 + 𝑥 2 + 𝑥 + 1 = (𝑥 + 1)(𝑥 − 𝑖)(𝑥 + 𝑖) Question #8 List all of the polynomials of degrees 2 and 3 in 𝑍𝟚 [𝑥]. Find all of the irreducible polynomials of degrees 2 and 3 in 𝑍𝟚 [𝑥]. Answer: Let 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 ∈ 𝑍𝟚 [𝑥] be a polynomial of degree 3. then we must have a=1 for this polynomial to be irreducible we must also d=1 since otherwise we will have a polynomial 𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 = 𝑥(𝑥 2 + 𝑏𝑥 + 𝑐) that can be factored an therefore reducible. so the polynomial 𝑓(𝑥) = 𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 1, where b, 𝑐 ∈ 𝑍𝟚 [𝑥] so the polynomial has no zero in 𝑍𝟚 [𝑥] so, we may not have f(0)=0 of f(1)=0 now f(0)=1 no matter chosen f(1)=b+c hence we have to choose b and c so, that b+c=1 and the two possibilities b=1,c=0 or b=0,c=1. that there are two irreducible polynomials of degree 3 in 𝑍𝟚 [𝑥] 𝑓1 (𝑥) = 𝑥 3 + 𝑥 2 + 1 𝑓2 (𝑥) = 𝑥 3 + 𝑥 + 1 Question #9 Determine which functions are polynomial functions. For those that are, identify the degree. 1 𝑓(𝑥) = 𝑥 2 − 3𝑥 2 + 5 Answer: 1 Consider the polynomials, 𝑓(𝑥) = 𝑥 2 − 3𝑥 2 + 5 A polynomial is a function of the form, 𝑓(𝑥) = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + ⋯ + 𝑎𝑛 𝑥 𝑛 Where, the coefficient of function are real number, and the exponent (Power) of x are non negative integer. 1 In 𝑓(𝑥) = 𝑥 2 − 3𝑥 2 + 5 1 Power of 𝑥 = , 2 2 The exponent of “x” is not integer. Here the exponent of “x” is fraction. Therefore the function is not polynomial function. Question #10 5 2𝑥 𝑥 𝑥+4 Find values of x, if any, at which f is not continuous: 𝑓(𝑥) = + Answer: A function is continuous if the left-hand limit at x=c and the right-hand limit at x=c both are equal to the value of the function at x=c. lim− 𝑓(𝑥) = lim+ 𝑓(𝑥) = 𝑓(𝑐) 𝑥→𝑐 𝑥→𝑐 In other words, a function is continuous if its graph is a single and non-broken curve or line. If a function has a vertical asymptote at x=b, then the function is not defined for x=b. It means the function is discontinuous at x=b. To find the vertical asymptotes, we need to find the values of x for which the denominator value is 0. The given function is: 5 2𝑥 𝑓(𝑥) = + 𝑥 𝑥+4 Taking LCM, we get 𝑓(𝑥) = 𝑓(𝑥) = 5(𝑥+4)+2𝑥(𝑥) 𝑥(𝑥+4) 5𝑥+20+2𝑥 2 𝑥(𝑥+4) Here, the numerator and denominator both are polynomials, and the polynomials are always continuous. So, the given function of continuous for all values of x except the values for which the denominator is equal to 0. x(x+4)=0 x=0 and x+4=0 x=0 and x=-4 It means the given function is not defined for x=0 and x=-4. So, the vertical asymptotes of the given function are x=0 and x=-4. The function is discontinuous at the vertical asymptotes. So, the given function is discontinuous at x=0 and x=-4. Answer: The function f(x) is discontinuous or not continuous at x=0 and x=-4.