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Dividing Polynomials


Date: 03/06/97 at 20:59:18
From: Patricia binnig
Subject: Algebra

The polynomial p(x) with integer coefficients satisfies:

(1) if p(x) is divided by x^2-4x+3, the remainder is 65x-68
(2) if p(x) is divided by x^2+6x-7, the remainder is -5x+a

(a) Find the number a
(b) Suppose that p(x) is divided by x^2 + 4x -21. Find the remainder.

I got as far as factoring the two divisors in (1) and (2). I can see 
that they have the common factor (x-1), but from here I am not sure 
where to go with it. I noticed that the last equation in (b) has 
factors (x-3) and (x+7), which are the same factors that the two prior 
equations don't have in common. 

This was on a college entrance exam for Japanese high school students.


Date: 03/06/97 at 22:09:20
From: Doctor Charles
Subject: Re: Algebra

You seem to have found everything that I found out but haven't yet put 
all the information to good use. When you say that the remainder when 
p(x) is divided by q(x) is r(x), it means that we can find a 
polynomial (in this question we can never tell what it is), let's call 
it h(x), such that:

         p(x) = h(x) * q(x) + r(x)

In our case where m(x) and n(x) are two more polynomials which we 
can't tell anything more about, we have:

         p(x) = m(x) * (x^2 - 4x + 3) + 65x - 68         
         p(x) = n(x) * (x^2 + 6x - 7) - 5x + a

Because we can't tell anything about m(x) and n(x), it would be good 
to get some equations that don't make use of them.

This is quite easy as you have already factorized the two divisors so 
that you should have (x^2 - 4x + 3) = (x-1)(x-3). Now if we put x = 1 
or x = 3 in the top equation, the quadratic just evaluates to 0 so we 
get (respectively):

         p(1) = m(1) * 0 + 65 * 1 - 68 = 0 - 3 = -3
         p(3) = m(3) * 0 + 65 * 3 - 68 = 0 + 195 - 68 = 127

If we try the same trick with the second equation we get:

         p(1) = n(1) * 0 - 5 * 1 + a = a - 5
         p(-7)= n(-7) * 0 - 5 * (-7) + a = 35 + a

But now we have two equations for p(1) so we know:

         -3 = p(1) = a - 5

so       a = 5 - 3 = 2

We have p(-7) = 35 + 2 = 37

Now for the last part. We know that the remainder is a linear term 
that is something like bx + c. This is just because it is a factor 
whose degree is one less than the quadratic. (If we had a remainder 
like e*x^2 + f*x + g, we could take off another e * (x^2 + 4x - 21) to 
get something which looked like bx+c.)

So we have:

        p(x) = k(x) * (x^2 + 4x - 21) + b * x + c

k(x) is yet another polynomial about which we know nothing!

As you found out, (x^2 + 4x - 21) = (x + 7)(x - 3)

So putting in 3 and -7 like before we get:

        p(3) = k(3) * 0 + b * 3 + c
       p(-7) = k(-7) * 0 + b * (-7) + c

We already have values for p(3) and p(-7). They are 127 and 37. So:

         127 = 3b + c
          37 = -7b + c
 
Subtracting we get:

          90 = 10b

This means that b = 9. Putting that back in we get 37 = -63 + c so 
c = 100

Thus the remainder when p(x) is divided by (x^2 + 4x - 21) is 
9 x + 100.

I hope this helps 

-Doctor Charles,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Basic Algebra
High School Polynomials

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