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Constructing Polynomials

Date: 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.


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
Associated Topics:
High School Basic Algebra
High School Functions

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