document.write( "Question 211128: How does one factor a polynomial where only two of items have negative exponents?\r
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document.write( "3x^-2 -7x^-1 -6 = 0\r
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document.write( "1/3x^2 - 1/7x - 6 = 0\r
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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)? \n" );
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Algebra.Com's Answer #159550 by algebrapro18(249)![]() ![]() ![]() 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. \r \n" ); document.write( "\n" ); document.write( "3x^-2 -7x^-1 -6 = 0 \n" ); document.write( " \r \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: \r \n" ); document.write( "\n" ); document.write( " \r \n" ); document.write( "\n" ); document.write( "Now we multiply both sides by x^2 and that leaves us with the following polynomial: \r \n" ); document.write( "\n" ); document.write( "3 - 7x - 6x^2 = 0 \r \n" ); document.write( "\n" ); document.write( "Now we solve this like any regular polynomial. \r \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 \r \n" ); document.write( "\n" ); document.write( "I will leave the second equation for you to do but its solved in a similar manor. \n" ); document.write( " \n" ); document.write( " |