document.write( "Question 12560: I am trying to learn about the quadratic equations. I know how to turn (3x-4)(2x+1)=0, but I don't know how to turn 6x^2 -5x -4=0 into (3x-4)(2x+1)=0.
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Algebra.Com's Answer #6373 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
Well, going from:\"%283x-4%29%282x%2B1%29\" to \"6x%5E2-5x-5\" is a whole lot easier than going the other way.\r
\n" ); document.write( "\n" ); document.write( "Here are some things you might try on factoring \"6x%5E2+-+5x+-+4\" for example:\r
\n" ); document.write( "\n" ); document.write( "1) Look at the first term \"6x%5E2\" and ask...how can I factor \"6x%5E2\"?
\n" ); document.write( "There aren't that many ways to factor it.
\n" ); document.write( "\"%286x%29%28x%29+=+6x%5E2\"
\n" ); document.write( "\"%282x%29%283x%29+=+6x%5E2\"
\n" ); document.write( "\"%28-6x%29%28-x%29+=+6x%5E2\"
\n" ); document.write( "\"%28-2x%29%28-3x%29+=+6x%5E2\" \r
\n" ); document.write( "\n" ); document.write( "Now look at the last term \"-4\" and go through the same process.
\n" ); document.write( "\"%28-1%29%284%29+=+-4%29\"
\n" ); document.write( "\"%281%29%28-4%29+=+-4\"
\n" ); document.write( "\"%282%29%28-2%29+=+-4\"
\n" ); document.write( "\"%28-2%29%282%29+=+-4\" This is really the same as the one above.\r
\n" ); document.write( "\n" ); document.write( "Now, keeping in mind that you want -5x as the middle term, you can try a pair of factors from the first group with a pair of factors from the second group.\r
\n" ); document.write( "\n" ); document.write( "For example: Try (6x - 1)(x + 4) and you see, almost right away that, while you get the first and last terms of your quadratic, you do not get -5x as the middle term. The same with (6x + 1)(x - 4) and (6x + 4)(x - 1).
\n" ); document.write( "Now try (3x + 1)(2x - 4) again, you don't get the middle term of -5x, so try (3x - 4)(2x + 1) Now you get a middle term of -5x from 3x - 8x = -5x.\r
\n" ); document.write( "\n" ); document.write( "So, these are the required factors. \r
\n" ); document.write( "\n" ); document.write( "It's a little harder when, as in this case, the coefficient of the x^2 term is greater than 1. Indeed, in some cases, you may not be able to factor the quadratic at all. But, you can always solve a quadratic equation by using the quadratic formula: \"x+=+%28-b+%2B-+sqrt%28b%5E2+-+4ac%29%29%2F2a\"\r
\n" ); document.write( "\n" ); document.write( "I hope this is some help. I have one math book on Intermediate Algebra that devotes some 36 pages to the topic of factoring polynomials.
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