Dividing PolynomialsDate: 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/ |
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