document.write( "Question 631736: completely factor the expression y^2+12y+35
\n" ); document.write( "a. (y-7)(y-5)
\n" ); document.write( "b. (y+7)(y-5)
\n" ); document.write( "c. (y+7)(y+5)
\n" ); document.write( "d. prime \n" ); document.write( "
Algebra.Com's Answer #397806 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
\r\n" );
document.write( "FOIL 'em all out and see:\r\n" );
document.write( "\r\n" );
document.write( "a. (y-7)(y-5) = y²-5y-7y+35 = y²-12y+35  No, the sign of -12y is wrong\r\n" );
document.write( "b. (y+7)(y-5) = y²-5y+7y+35 = y²+2y+35  No, the middle term is wrong\r\n" );
document.write( "c. (y+7)(y+5) = y²+5y+7y+35 = y²+12y+35  Yes, that's the same as the original.\r\n" );
document.write( "d. prime.  No for c is the correct factorization.\r\n" );
document.write( "\r\n" );
document.write( "Answer: c\r\n" );
document.write( "\r\n" );
document.write( "Edwin
\n" ); document.write( "
\n" );