document.write( "Question 19916: Factor this equation \r
\n" ); document.write( "\n" ); document.write( "2c^3-5c^2+c+2
\n" ); document.write( "

Algebra.Com's Answer #9625 by longjonsilver(2297)\"\" \"About 
You can put this solution on YOUR website!
\"2c%5E3-5c%5E2%2Bc%2B2\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "By trial and error, put values into the equation. This will lead us to find one of the factors... start with easy numbers like +1, -1 then +2 and -2 etc until you find one.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "so, if c=1, we get that the polynomial is equal to zero, so c=1 is a solution (or root). If c=1, then (c-1)=0. Therefore, we can say that (c-1) is a factor.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "You now need to perform long division of this factor into the polynomial to find the other factor. Doing this, gives \"2c%5E2-3c-2\".\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "So, the factors are \"%28c-1%29%282c%5E2-3c-2%29\". Factoring the quadratic gives the final answer: \"%28c-1%29%282c%2B1%29%28c-2%29\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "jon.
\n" ); document.write( "
\n" ); document.write( "
\n" );