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Polynomial Problems


Date: 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

    
Associated Topics:
High School Polynomials

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