document.write( "Question 1009738: If I have the quadratic function, f(x)= (x^2)-x-12, and I want to find the x-intercept, I have to set y = 0, and factor like so: 0 = (x-4)(x+3).\r
\n" ); document.write( "\n" ); document.write( "The questions is: why can't I complete the square or factor like so: 0 = x(x-1)-12 (which leads to 0 = (x-1)(x-12))? Doing either of those alternative techniques results in different x-intercepts. \n" ); document.write( "
Algebra.Com's Answer #625216 by MathTherapy(10552)\"\" \"About 
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\n" ); document.write( "If I have the quadratic function, f(x)= (x^2)-x-12, and I want to find the x-intercept, I have to set y = 0, and factor like so: 0 = (x-4)(x+3).\r
\n" ); document.write( "\n" ); document.write( "The questions is: why can't I complete the square or factor like so: 0 = x(x-1)-12 (which leads to 0 = (x-1)(x-12))? Doing either of those alternative techniques results in different x-intercepts.
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You can use the \"complete the square\" method to solve for the x-intercepts/solutions to the equation/roots.
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document.write( "However, factoring \"f%28x%29+=+x%5E2+-+x+-+12\" CANNOT be done the following way you proposed: 
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document.write( "\"f%28x%29+=+x%5E2+-+x+-+12\"
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document.write( "\"0+=+x%28x+-+1%29+-+12\"
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document.write( "From this, you CANNOT obtain x - 1 and x - 12 as factors. That's not the proper way to factor a trinomial in
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document.write( "order to find its roots. The 1st method: \"0+=+%28x+-+4%29%28x+%2B+3%29\", though is indeed correct. \n" );
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