document.write( "Question 387689: In solving the equation (x + 1)(x – 2) = 4, Eric stated that the solution would be
\n" ); document.write( "x + 1 = 4 => x = 3
\n" ); document.write( "or
\n" ); document.write( "(x – 2) = 4 => x = 6
\n" ); document.write( "However, at least one of these solutions fails to work when substituted back into the original equation. Why is that? Please help Eric to understand better; solve the problem yourself, and explain your reasoning. \n" ); document.write( "

Algebra.Com's Answer #274065 by robertb(5830) $\"\"$ $\"About$

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For $\"%28x-a%29%28x-b%29+=+c\"$ , x-a = c or x - b = c works only if c = 0, and nowhere else. It is called the zero-factor property. To solve this equation, multiply out the left side and then write in standard form, i.e., with zero on one side.
\n" ); document.write( " $\"x%5E2-x+-+2+=+4\"$ ;
\n" ); document.write( " $\"x%5E2+-+x+-+6+=+0\"$ ;
\n" ); document.write( "(x - 3)(x+2) = 0;
\n" ); document.write( "x = 3, -2. These are the solutions to the equation. \n" ); document.write( "

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