document.write( "Question 23858: factor \r
\n" ); document.write( "\n" ); document.write( "9x^2+3x-2 \n" ); document.write( "
Algebra.Com's Answer #12678 by PeterC(8)\"\" \"About 
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A quadratic funtion \"ax%5E2%2Bbx%2Bc\" can be factored as \"%28a%29%28x-r1%29%28x-r2%29\",
\n" ); document.write( "where r1 and r2 are the roots (solutions) to the equation \"ax%5E2+%2B+bx+%2B+c+=+0\".\r
\n" ); document.write( "\n" ); document.write( "We can find r1 and r2, using the quadratic formula
\n" ); document.write( "\"%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29\".\r
\n" ); document.write( "\n" ); document.write( "Here, \r
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\n" ); document.write( "\n" ); document.write( "Therefore, our factors are: \"%289%29+%2A+%28x+-+1%2F3%29+%2A+%28x+%2B+2%2F3%29\"\r
\n" ); document.write( "\n" ); document.write( "What I have shown (creating a factored expression by finding the roots of a polynomial equation) is a brute-force method that will always work. However, you can often solve the problem by examination, by posing the question, \"What are two numbers whose sum is b/a and product is c/a?\" (It is a bit easier to do this if a = 1.) In fact, the usual reason for factoring a quadratic expression is that it gives us a shortcut to finding its roots (which are the negatives of those two numbers). If we can recognize the numbers that create our desired sum and product, we can avoid a somewhat laborious computation with the quadratic formula.\r
\n" ); document.write( "\n" ); document.write( "In this problem, \r
\n" ); document.write( "\n" ); document.write( "\"9x%5E2+%2B+3x+-+2+=+%283x-1%29%283x%2B2%29+=+9%28x-1%2F3%29%28x%2B2%2F3%29\" \n" ); document.write( "
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