document.write( "Question 300802: We learned how to factor trinomials awhile ago, by unfoiling like \"x%5E2%2B2x-3\" would be (x+3)(x-1), but how do you do it if there is a coefficient in front of the \"x%5E2\" like \"3b%5E2-17b%2B22\" i used to just factor it out when they were simple bye just making like \"2x%5E2%2B4x%2B6\" into \"2%28x%5E2%2B2x%2B3%29\" but it dosent work like that anymore because im getting ugly decimals. Please help. \n" ); document.write( "
Algebra.Com's Answer #215776 by Stitch(470)\"\" \"About 
You can put this solution on YOUR website!
Given: \"3B%5E2+-+17B+%2B+22\"
\n" ); document.write( "You still use the unfoil method however you need to include the coefficient.
\n" ); document.write( "Find the factors of 22: 1,2,11,22
\n" ); document.write( "You need to look at those factors and figure out how to get -17 while using the coefficient with one of the factors.
\n" ); document.write( "-17 = -11 - 3*(2)
\n" ); document.write( "\"%283B+-+11%29%2A%28B+-+2%29+=+3B%5E2+-+17B+%2B+22\"
\n" ); document.write( "Set that equal to 0 to solve for B
\n" ); document.write( "-----------------
\n" ); document.write( "\"3B+-+11+=+0\"
\n" ); document.write( "\"3B+=+11\"
\n" ); document.write( "\"highlight%28B+=+11%2F3%29\" or 3.6667
\n" ); document.write( "-----------------
\n" ); document.write( "\"B+-+2+=+0\"
\n" ); document.write( "\"highlight_green%28B+=+2%29\"
\n" ); document.write( "
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