document.write( "Question 211128: How does one factor a polynomial where only two of items have negative exponents?\r
\n" ); document.write( "\n" ); document.write( "3x^-2 -7x^-1 -6 = 0\r
\n" ); document.write( "\n" ); document.write( "1/3x^2 - 1/7x - 6 = 0\r
\n" ); document.write( "\n" ); document.write( "Foil would give (1/3x - 3) (x + 2) - but how to get that - 1/7x in the middle (I can get - 7/3x)?
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Algebra.Com's Answer #159550 by algebrapro18(249)\"\" \"About 
You can put this solution on YOUR website!
Simple, you don't factor polynomials with negative exponents. You put the negative exponents in the denominator and the get a common denominator and solve that way. You will end up having to factor but the exponents will be positive.
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\n" ); document.write( "\n" ); document.write( "3x^-2 -7x^-1 -6 = 0
\n" ); document.write( "\"3%2Fx%5E2+-+7%2Fx+-+6+=+0\"
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\n" ); document.write( "\n" ); document.write( "The common denominator on this problem is going to be x^2 so multiply the second part of the equation by x and the 6 by x^2. This will get us:
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\n" ); document.write( "\n" ); document.write( "\"%283+-+7x+-+6x%5E2%29%2Fx%5E2+=+0\"
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\n" ); document.write( "\n" ); document.write( "Now we multiply both sides by x^2 and that leaves us with the following polynomial:
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\n" ); document.write( "\n" ); document.write( "3 - 7x - 6x^2 = 0
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\n" ); document.write( "\n" ); document.write( "Now we solve this like any regular polynomial.
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\n" ); document.write( "\n" ); document.write( "3 - 7x - 6x^2 = 0
\n" ); document.write( "6x^2 + 7x - 3 = 0
\n" ); document.write( "(2x+3)(3x-1) = 0
\n" ); document.write( "2x+3 = 0 and 3x-1 = 0
\n" ); document.write( "x = -3/2 and x = 1/3
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\n" ); document.write( "\n" ); document.write( "I will leave the second equation for you to do but its solved in a similar manor.
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