document.write( "Question 617533: if x^6+kx^5+x^4+kx^3+3x^2-k+2 is a multiple of (x-k) find the value of k \n" ); document.write( "

Algebra.Com's Answer #388402 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
If the polynomial is a multiple of (x-k), then (x-k) will divide evenly into the polynomial. (Just like \"24 is a multiple of 6\" means that \"6 will divide evenly into 24\".)

\n" ); document.write( "But (x-k) will not divide evenly into $\"x%5E6%2Bkx%5E5%2Bx%5E4%2Bkx%5E3%2B3x%5E2-k%2B2\"$ . So either there is an error in what you were given or there is an error in what you posted (or there is an error in my work). I'm guessing that the error may be in the area of the \"-k + 2\" at the end. Is that correct? There is not x^1 term in what you posted. Or should it be \"-x + 2\" or \"-kx + 2\"?

\n" ); document.write( "In case you are wondering how I am dividing by (x-k) to see if it divides evenly, I used synthetic division:
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document.write( "k ||   1   k   1           k           3              0                 -k+2\r\n" );
document.write( "====       k  2k^2    2k^3+k   2k^4+2k^2    2k^5+2k^3+3k  2k^6+2k^4+3k^2\r\n" );
document.write( "      ======================================================================\r\n" );
document.write( "       1  2k  2k^2+1  2k^3+2k  2k^4+2k^2+3  2k^5+2k^3+3k  2k^6+2k^4+3k^2-k+2\r\n" );
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\n" ); document.write( "The long expression in the lower right corner is the remainder. If (x-k) divides evenly then the remainder should be zero. So we set
\n" ); document.write( " $\"2k%5E6%2B2k%5E4%2B3k%5E2-k%2B2+=+0\"$
\n" ); document.write( "and try to solve it. But I cannot find any solutions to this. If you can then whatever solutions you find would be possible values for k.

\n" ); document.write( "If you find an error in what you posted, then maybe what I've shown you above will help you figure out how to find k on your own. If not, then re-post the corrected problem. \n" ); document.write( "

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