document.write( "Question 1020769: Hii my beloved tutors, I really need help on this question, could you please help me?
\n" ); document.write( "Suppose $\"+a+\"$ is a real number, so that polynomial $\"+p%28x%29=+x%5E4%2B4x%2Ba+\"$ is divisible by $\"+%28x-c%29%5E2+\"$ for $\"+c+\"$ is a real number. What is the value of $\"+a+\"$ that satisfies?
\n" ); document.write( "Thanks \n" ); document.write( "

Algebra.Com's Answer #636609 by robertb(5830) $\"\"$ $\"About$

You can put this solution on YOUR website!
If $\"%28x-c%29%5E2\"$ is to divide $\"p%28x%29=+x%5E4%2B4x%2Ba+\"$ , then x-c should be able to divide the latter twice.\r
\n" ); document.write( "\n" ); document.write( "Using the remainder theorem, if c is to be root of p(x), then p(c) = 0, or $\"c%5E4+%2B+4c+%2Ba+=+0\"$ \r
\n" ); document.write( "\n" ); document.write( "Now divide $\"p%28x%29=+x%5E4%2B4x%2Ba+\"$ by x-c, and letting the remainder $\"c%5E4+%2B+4c+%2Ba\"$ equal to 0, we get the polynomial $\"x%5E3%2Bcx%5E2%2Bc%5E2x%2B%284%2Bc%5E3%29\"$ .
\n" ); document.write( "By applying the remainder theorem again on this resulting polynomial, we get $\"c%5E3%2Bc%5E3%2B+c%5E3+%2B4+%2B+c%5E3+=+4+%2B+4c%5E3+=+0\"$ \r
\n" ); document.write( "\n" ); document.write( "The last equation gives $\"highlight%28c+=+-1%29\"$ .
\n" ); document.write( "substitution of this value into $\"c%5E4+%2B+4c+%2Ba+=+0\"$ gives $\"highlight%28a+=+3%29\"$ . \n" ); document.write( "

\n" );