document.write( "Question 1157973: When ax³ + bx² + cx - 4 is divided by (x+2), the remainder is double that obtained when the expression is divided by (x+1). Show that c can have any value and find b in terms of a. \n" ); document.write( "

Algebra.Com's Answer #854525 by KMST(5377) $\"\"$ $\"About$

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$\"P%28x%29=ax%5E3%2Bbx%5E2%2Bcx-4\"$
\n" ); document.write( "If you divide $\"P%28x%29\"$ by $\"x%2B1\"$ , you obtain a quotient $\"Q%28x%29\"$ and remainder $\"r\"$ that is a constant.
\n" ); document.write( "You may remember that in then $\"P%28-1%29=r\"$
\n" ); document.write( "If you did not remember, you would understand that if the reminder is $\"r\"$ ,
\n" ); document.write( "it means that $\"P%28x%29=%28x%2B1%29Q%28x%29%2Br\"$ and that
\n" ); document.write( "for $\"x=-1\"$ , $\"P%28-1%29=%28-1%2B1%29Q%28-1%29%2Br=0%2AQ%28-1%29%2Br=0%2Br=r\"$ .
\n" ); document.write( "So $\"a%28-1%29x%5E3%2Bb%28-1%29%5E2%2Bc%28-1%29-4=r\"$ --> $\"highlight%28-a%2Bb-c-4=r%29\"$ .
\n" ); document.write( "Similarly the remainder, when dividing $\"P%28x%29\"$ by $\"x%2B2\"$ is
\n" ); document.write( " $\"a%28-2%29x%5E3%2Bb%28-2%29%5E2%2Bc%28-2%29-4=2r\"$ --> $\"highlight%28-8a%2B4b-2c-4=2r%29\"$ .
\n" ); document.write( "Then, $\"-8a%2B4b-2c-4=2%28-a%2Bb-c-4%29%29\"$ --> $\"-8a%2B4b-2c-4=-2a%2B2b-2c-8%29\"$ --> $\"+4b-2b-2c%2B2c=8a-2a-8%2B4\"$ --> $\"2b=6a-4\"$ --> $\"highlight%28b=3a-2%29\"$
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