document.write( "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. \n" ); document.write( "

Algebra.Com's Answer #665055 by KMST(5328) $\"\"$ $\"About$

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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.
\n" ); document.write( "Let us say that remainder is $\"ax%5E2%2Bbx%2Bc\"$ , and we have to find coefficients $\"a\"$ , $\"b\"$ , and $\"c\"$ .
\n" ); document.write( "So, $\"f%28x%29=%28x-3%29%2A%28x-1%29%5E2%2AQ%28x%29%2Bax%5E2%2Bbx%2Bc\"$
\n" ); document.write( "Since the remainder is $\"15\"$ when divided by $\"x-3\"$ , $\"f%283%29=15\"$ , so
\n" ); document.write( " $\"%283-3%29%2A%283-1%29%2AQ%283%29%2B9a%2B3b%2Bc=15\"$
\n" ); document.write( " $\"0%2A2%2AQ%28x%29%2B9a%2B3b%2Bc=15\"$
\n" ); document.write( " $\"0%2B9a%2B3b%2Bc=15\"$
\n" ); document.write( " $\"highlight%289a%2B3b%2Bc=15%29\"$
\n" ); document.write( "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 $\"%28x-1%29%5E2=x%5E2-2x%2B1\"$ .
\n" ); document.write( "That remainder is $\"%28b%2B2a%29x%2B%28c-a%29=2x-1\"$ .
\n" ); document.write( "Since the polynomials $\"2x%2B1\"$ and $\"%28b%2B2a%29x%2B%28c-a%29\"$ are the same polynomial,
\n" ); document.write( " $\"system%28b%2B2a=2%2C%22and%22%2Cc-a=1%29\"$ <---> $\"system%28b=2-2a%2Cc=a%2B1%29\"$ .
\n" ); document.write( "Along with the equation $\"9a%2B3b%2Bc=15\"$ highlighted above, we have the system of linear equations
\n" ); document.write( " $\"system%289a%2B3b%2Bc=15%2Cb=2-2a%2Cc=a%2B1%29\"$ .
\n" ); document.write( "Substituting into the top equation the expressions for $\"b\"$ and $\"c\"$ from the bottom two equations, we get
\n" ); document.write( " $\"9a%2B3%28+2-2a++%29+%2B+%28+a%2B1++%29+=+15\"$
\n" ); document.write( " $\"9a%2B6-6a%2Ba%2B1=15\"$
\n" ); document.write( " $\"4a%2B7=15\"$
\n" ); document.write( " $\"4a=25-7\"$
\n" ); document.write( " $\"4a=8\"$
\n" ); document.write( " $\"a=8%2F4\"$
\n" ); document.write( " $\"highlight%28a=2%29\"$ , and substituting $\"2\"$ for $\"a\"$ in the bottom two equations of the system, we get
\n" ); document.write( " $\"b=2-2%2A2\"$ $\"b=2-4\"$ $\"highlight%28b=-2%29\"$ , and
\n" ); document.write( " $\"c=2%2B1\"$ $\"highlight%28c=3%29\"$ .
\n" ); document.write( "So, it turns out that the remainder $\"ax%5E2%2Bbx%2Bc\"$ that we were looking for is
\n" ); document.write( " $\"highlight%282x%5E2-2x%2B3%29\"$ . \n" ); document.write( "

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