document.write( "Question 1183399: The remainder when ax^3 + bx^2 + 2x + 3 is divided by x-1 is twice that when it is divided by x+1. Show that b = 3a + 3 \n" ); document.write( "
Algebra.Com's Answer #813698 by robertb(5830)\"\" \"About 
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if \"f%28x%29+=+ax%5E3+%2B+bx%5E2+%2B+2x+%2B+3\", then by the remainder theorem, the remainder upon division of \"f%28x%29\" by \"x-1\" is equal to \"f%281%29+=+a%2Bb%2B2%2B3+=+a%2Bb%2B5\".\r
\n" ); document.write( "\n" ); document.write( "Similarly, the remainder upon division of \"f%28x%29\" by \"x%2B1\" is equal to \"f%28-1%29+=+-a%2Bb-2%2B3+=+-a%2Bb%2B1\".\r
\n" ); document.write( "\n" ); document.write( "From the given, \"f%281%29+=+2f%28-1%29\", hence\r
\n" ); document.write( "\n" ); document.write( "\"a%2Bb%2B5+=+2%28-a%2Bb%2B1%29+=+-2a%2B2b%2B2\" ==> \"highlight%283a+%2B+3+=+b%29\".
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