Constructing PolynomialsDate: 01/20/99 at 04:04:08 From: Yang Zhong Subject: Polynomials I have two questions to ask. 1. Construct the polynomial P(x) having the following properties: a) P(x) is of degree 4 b) P(x) = P(-x) c) 2P(x)+3 is divisible by (x+1) d) P(x)+1 is divisible by (x-2)^2 2. A polynomial P(x) leaves a remainder of 2 when divided by (x-1) and a remainder of 3 when divided by (x-2). The remainder when P(x) is divided by (x-1)(x-2) is ax + b. Find the values of a and b. We're also given that P(x) is a cubic polynomial with coefficient of x^3 equal to unity, and that -1 is a root if the equation P(x)=0. Show that the equation P(x) = 0 has no other real roots. Thanks! Date: 01/20/99 at 10:36:44 From: Doctor Rob Subject: Re: Polynomials Thanks for writing to Ask Dr. Math! 1. (b) and (d) imply that (x+2)^2 = (-x-2)^2 | P(-x)+1 = P(x)+1. In other words, both (x+2)^2 and (-x-2)^2 divide P(x)+1. Thus P(x) + 1 = C*(x-2)^2*(x+2)^2 (a) implies that C is a constant. (If P(x) were a higher power, C would have to be a polynomial). Then 2*P(x) + 3 = 2*C*(x-2)^2*(x+2)^2 + 1 and, by the Remainder Theorem and/or the Factor Theorem, (c) implies 0 = 2*P(-1) + 3 = 2*C*([-1]-2)^2*([-1]+2)^2 + 1 This equation will allow you to solve for C, and then use the first equation above to determine P(x) uniquely. 2. By hypothesis, P(x) = Q(x)*(x-1)*(x-2) + a*x + b and P(1) = 2 and P(2) = 3. Substituting x = 1 and x = 2 in the above equation will give you two linear equations in the two unknowns a and b which are easy to solve. Now since the degree of P(x) is 3, Q(x) must be a first-degree polynomial, and since the leading term has coefficient 1, Q(x) = x + c. Now substitute x = -1 into the equation P(x) = (x+c)*(x-1)*(x-2) + a*x + b which will give you a single equation in c that is easy to solve. Now factor P(x) = (x+1)*R(x), and show that the quadratic equation R(x) = 0 has negative discriminant, and so no real roots. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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