SOLUTION: a polynomial f(x) with rational coefficients leaves remainder 15 when divided by x-3 and remainder 2x+1 when divided by (x-1)^2. Find the remainder when f(X) is divided by (x-3)(x-

Algebra ->  Equations -> SOLUTION: a polynomial f(x) with rational coefficients leaves remainder 15 when divided by x-3 and remainder 2x+1 when divided by (x-1)^2. Find the remainder when f(X) is divided by (x-3)(x-      Log On


   



Question 1049425: a polynomial f(x) with rational coefficients leaves remainder 15 when divided by x-3 and remainder 2x+1 when divided by (x-1)^2. Find the remainder when f(X) is divided by (x-3)(x-1)^2.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Since divisor %28x-3%29%28x-1%29%5E2 is a polynomial of degree 3, the remainder we look for is a polynomial whose degree is at most 2.
Let us say that remainder is ax%5E2%2Bbx%2Bc , and we have to find coefficients a , b , and c .
So, f%28x%29=%28x-3%29%2A%28x-1%29%5E2%2AQ%28x%29%2Bax%5E2%2Bbx%2Bc
Since the remainder is 15 when divided by x-3 , f%283%29=15 , so
%283-3%29%2A%283-1%29%2AQ%283%29%2B9a%2B3b%2Bc=15
0%2A2%2AQ%28x%29%2B9a%2B3b%2Bc=15
0%2B9a%2B3b%2Bc=15
highlight%289a%2B3b%2Bc=15%29
When f%28x%29=%28x-1%29%2A%28x-1%29%5E3%2AQ%28x%29%2Bax%5E2%D7bx%D7c is divided by %28x-1%29%5E2 the remainder is the remainder of dividing ax%5E2%2Bbx%2Bc by %28x-1%29%5E2=x%5E2-2x%2B1 .
That remainder is %28b%2B2a%29x%2B%28c-a%29=2x-1 .
Since the polynomials 2x%2B1and %28b%2B2a%29x%2B%28c-a%29 are the same polynomial,
system%28b%2B2a=2%2C%22and%22%2Cc-a=1%29 <---> system%28b=2-2a%2Cc=a%2B1%29 .
Along with the equation 9a%2B3b%2Bc=15 highlighted above, we have the system of linear equations
system%289a%2B3b%2Bc=15%2Cb=2-2a%2Cc=a%2B1%29 .
Substituting into the top equation the expressions for b and c from the bottom two equations, we get
9a%2B3%28+2-2a++%29+%2B+%28+a%2B1++%29+=+15
9a%2B6-6a%2Ba%2B1=15
4a%2B7=15
4a=25-7
4a=8
a=8%2F4
highlight%28a=2%29 , and substituting 2 for a in the bottom two equations of the system, we get
b=2-2%2A2 b=2-4 highlight%28b=-2%29 , and
c=2%2B1 highlight%28c=3%29 .
So, it turns out that the remainder ax%5E2%2Bbx%2Bc that we were looking for is
highlight%282x%5E2-2x%2B3%29 .