document.write( "Question 36570This question is from textbook
\n" ); document.write( ": Hi
\n" ); document.write( "I am having a bit of trouble with this problem from my worksheet for pratice problems, any help would be greatly appreciated.\r
\n" ); document.write( "\n" ); document.write( "Divide and simplify: \r
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\n" ); document.write( "\n" ); document.write( " 3x^2 + 13x + 4/16 – x^2 \r
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\n" ); document.write( "\n" ); document.write( " ÷
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\n" ); document.write( "\n" ); document.write( " 3x^2 – 5x – 2/3x – 12\r
\n" ); document.write( "\n" ); document.write( "I want to invert the divisor and multiply
\n" ); document.write( "so the new problem is \r
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\n" ); document.write( "\n" ); document.write( "3x^2 + 13x + 4/16 – x^2 \r
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\n" ); document.write( "\n" ); document.write( "x(times)\r
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\n" ); document.write( "\n" ); document.write( "3x – 12/3x^2 – 5x – 2
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\n" ); document.write( "\n" ); document.write( " I cant seem to factor the 3x^2 + 13x + 4 =, is this possible, I have tried every combination of numbers but nothing is working.\r
\n" ); document.write( "\n" ); document.write( "I think the 3x – 12 can be factored down to 3(x – 4)\r
\n" ); document.write( "\n" ); document.write( "I think the 16 – x^2 should be switched to read –x^2 + 16 which can be factored down to –(x – 4)(x – 4)\r
\n" ); document.write( "\n" ); document.write( "I think the 3x^2 – 5x – 2 can be factored down to (3x + 1)(x – 2)\r
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\n" ); document.write( "\n" ); document.write( "And I am stuck at this point, because I really think that the 3x^2 + 13 + 4 can be factored down to simpler terms to come up with the final answer\r
\n" ); document.write( "\n" ); document.write( "Please help?\r
\n" ); document.write( "\n" ); document.write( "Thank you very much!
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Algebra.Com's Answer #22462 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
3x^2 + 13x + 4\r
\n" ); document.write( "\n" ); document.write( "=3x^2+12x+x+4
\n" ); document.write( "=3x(x+4)+(x+4)
\n" ); document.write( "=(x+4)(3x+1)
\n" ); document.write( "Cheers,
\n" ); document.write( "stan H.
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