Polynomial ProblemsDate: 12/4/95 at 19:16:56 From: Anonymous Subject: Polynomial problems (using various theorems) - High School Level Hello. I am totally stuck on 5 or 6 questions. I am desperate! They are polynomial questions and I don't think they will take long, but there are certain loops in the questions that I can't seem to figure out. I used the integral root, factor and remainder theorems to figure these out. Here they are: 1. Let m, n and o be the 3 distinct roots of x^3 + ax + b=0. Compute (m-n)^2(n-o)^2(o-m)^2 in terms of a and b. 2. If m, n and 1 are non-zero roots of the equation x^3 - mx^2 + nx -1=0, then what is the sum of the roots? 3. Solve 2x^3 - 3x^2 + 1=0. 4. Find a and b so that the quadratic function f(x)=a^2x^4 + 3x^3 + b^2x^2 + 4abx + 4ab leaves a remainder of 10 on division by x+1 and a remainder of 20 on division by x. 5. For the polynomial f(x)=ax^3 + bx^2 + cx +d, the sum of the coefficients is equal to zero (a+b+c+d=0). Show that the polynomial is divisible by x-1. Any help would be greatly appreciated. Thank you a great deal, J. Howard Date: 5/28/96 at 19:57:26 From: Doctor Anthony Subject: Re: Polynomial problems (using various theorems) - High School Level (1) For this question we use the fact that the sum of the roots = 0 since the coefficient of x^2 = 0, and the product of the roots = -b. I shall use p rather than o as the third root to distinguish it from 0. So we have m+n+p =0 and mnp = -b Now (m-n)^2 = (m+n)^2 - 4mn = (-p)^2 - 4(-b/p) = p^2 + 4b/p = (p^3 + 4b)/p Now from the original equation p^3 = -ap - b So (m-n)^2 = (-ap+3b)/p also (n-p)^2 = (-am+3b)/m and (p-m)^2 = (-an+3b)/n Multiply these three, and remembering that mnp = -b, we get (-1/b)(-ap+3b)(-am+3b)(-an+3b) Multiplying this out and collecting terms we get (-1/b){-a^3b + 3a^2b(mn+np+pm) - 9ab^2(m+n+p) + 27b^3} = a^3 - 3a^2(mn+np+pm) + 9ab(m+n+p) - 27b^2 Finally substitute m+n+p = 0 mn+np+pm = a mnp = -b = a^3 - 3a^3 + 0 - 27b^2 = -2a^3 - 27b^2 (2) The sum of the roots of x^3-mx^2 + nx -1 = 0 is minus the coefficient of x^2. So in this case the sum of the roots is m. (3) Solve 2x^3-3x^2+1=0 By inspection x=1 is a solution. So divide out by x-1 and find the other roots by solving the quadratic. After dividing by x-1 we get a quotient 2x^2-x-1=0 which factorizes to (2x+1)(x-1)=0. This gives the other roots as x=1 (again) and x= -1/2. Three roots are x=1 (twice), x=-1/2 (4) For this question, using the Remainder theorem f(-1)= 10 and f(0) = 20 f(-1) = a^2-3+b^2-4ab+4ab = 10 so a^2 + b^2 = 13 f(0) = 4ab = 20 and so ab=5 b= 5/a (1) a^2 + 25/a^2 = 13 => a^4 + 25 = 13a^2 or a^4 - 13a^2 + 25 = 0 a^2 = {13+-sqrt(169-100)}/2 = (13+-8.3066)/2 = 10.6533 or 2.3466 a = 3.2639 or 1.53189 Values of b are found from equation (1) above. (5) Using the Remainder theorem we find f(1) to get remainder on division by x-1. But f(1) = a+b+c+d Now we are told that this sum is zero, and that means that x-1 is a factor of f(x). -Doctor Anthony, The Math Forum |
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