document.write( "Question 1117462: Demonstrate that the roots of the following equation are rational\r
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Algebra.Com's Answer #732584 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "abc²x² + 3a²cx + b²cx - 6a² - ab + 2b² = 0 \r\n" );
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document.write( "We put the left side \r\n" );
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document.write( "abc²x² + 3a²cx + b²cx - 6a² - ab + 2b²\r\n" );
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document.write( "in quadratic form: Ax² + Bx + C\r\n" );
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document.write( "(abc²)x² + (3a²cx + b²x) - 6a²-ab+2b²\r\n" );
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document.write( "(abc²)x² + (3a²c+b²)x + (-6a²-ab+2b²) \r\n" );
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document.write( "The x² term is abc²x² which has most likely\r\n" );
document.write( "factorization (acx)•(bcx)\r\n" );
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document.write( "So I'll try to factor it like this:\r\n" );
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document.write( "[acx          ][bcx         ]\r\n" );
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document.write( "The constant term is\r\n" );
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document.write( "-6a² - ab + 2b²\r\n" );
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document.write( "That factors, take out a - sign\r\n" );
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document.write( "-(6a² + ab - 2b²) \r\n" );
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document.write( "and again as\r\n" );
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document.write( "-(2a - b)(3a + 2b)\r\n" );
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document.write( "So I'll try to factor the original left side as \r\n" );
document.write( "one of these possible factorizations:\r\n" );
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document.write( "(1).   [acx + (2a-b)][bcx - (3a+2b)]\r\n" );
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document.write( "(2).   [acx - (2a-b)][bcx + (3a+2b)]\r\n" );
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document.write( "(3).   [acx + (3a+2b)][bcx - (2a-b)]\r\n" );
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document.write( "(4).   [acx - (3a+2b)][bck + (2a-b)]\r\n" );
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document.write( "I could multiply those out, but it would be easier to choose\r\n" );
document.write( "arbitrary easy values for the letters and check their\r\n" );
document.write( "results with the results of the original left side.  Just\r\n" );
document.write( "choose one of the letters as 0 and the rest as 1, so hopefully\r\n" );
document.write( "only one of the four possible factorizations will be the same\r\n" );
document.write( "as when we substitute them in the original left side.  If two\r\n" );
document.write( "or more are the same as that value, we'll have to make up\r\n" );
document.write( "different values for the letters.\r\n" );
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document.write( "a = 0, b = c = x = 1\r\n" );
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document.write( "Substituting in possible factorization (1). \r\n" );
document.write( "[acx + (2a-b)][bcx - (3a+2b)]\r\n" );
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document.write( "[0•1•1 + (2•0-1)][1•1•1 - (3•0+2•1)] = [0-1][1-2] = [-1][-1] = 1\r\n" );
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document.write( "Substituting in possible factorization (2).\r\n" );
document.write( "[acx - (2a-b)][bcx + (3a+2b)]\r\n" );
document.write( "[0•1•1 - (2•0-1)][1•1•1 + (3•0+2•1)] = [0+1][1+2] = [1][3] = 3\r\n" );
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document.write( "Substituting in possible factorization (3).\r\n" );
document.write( "[acx + (3a+2b)][bcx - (2a-b)]\r\n" );
document.write( "[0•1•1 + (3•0+2•1)][1•1•1 - (2•0-1)] = [0+2][1+1] = [2][2] = 4\r\n" );
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document.write( "Substituting in  possible factorization (4).\r\n" );
document.write( "[acx - (3a+2b)][bcx + (2a-b)]\r\n" );
document.write( "[0•1•1 - (3•0+2•1)][1•1•1 + (2•0-1)] = [0-2][1-1] = [-2][0] = 0\r\n" );
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document.write( "That's good. They are all different.  Now we can tell which one \r\n" );
document.write( "is probably the correct factorization by substituting those same \r\n" );
document.write( "values in the original left side:\r\n" );
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document.write( "a = 0, b = c = x = 1\r\n" );
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document.write( "abc²x² + 3a²cx + b²cx - 6a² - ab + 2b²\r\n" );
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document.write( "0•1•1²•1² + 3•0²•1•1 + 1²•1•x - 6•0² - 0•1 + 2•1²\r\n" );
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document.write( "0 + 0 + 1 - 0 - 0 + 2 = 3\r\n" );
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document.write( "So only possible factorization (2) can be the correct one.\r\n" );
document.write( "But we really haven't shown that it is the correct one, but\r\n" );
document.write( "only if any are correct, it's possible factorization (2).\r\n" );
document.write( "So let's be sure:\r\n" );
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document.write( "(2).   [acx - (2a-b)][bcx + (3a+2b)]\r\n" );
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document.write( "So pick arbitrary values for the letters, say\r\n" );
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document.write( "a=3, b=4, c=2, x=1\r\n" );
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document.write( "Substituting in (2)\r\n" );
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document.write( "(2).   [acx - (2a-b)][bcx + (3a+2b)]\r\n" );
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document.write( "       [3•2•1 - (2•3-4)][4•2•1 + (3•3+2•4)]\r\n" );
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document.write( "       [6 - (6-4)][8 + (9+8)]\r\n" );
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document.write( "       [6 - 2][8 + 17]\r\n" );
document.write( "       \r\n" );
document.write( "       [4][25]\r\n" );
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document.write( "       100\r\n" );
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document.write( "Substituting in the original left side:\r\n" );
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document.write( "a=3, b=4, c=2, x=1\r\n" );
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document.write( "3•4•2²•1² + 3•3²•2•1 + 4²•2•1 - 6•3² - 3•4 + 2•4²\r\n" );
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document.write( "3•4•4•1 + 3•9•2•1 + 16•2•1 - 6•9 - 3•4 + 2•16\r\n" );
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document.write( "48 + 54 + 32 - 54 - 12 + 32 \r\n" );
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document.write( "100\r\n" );
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document.write( "Since we got the same 100, we are 99.999% sure that (2) is \r\n" );
document.write( "the correct factorization. We can simplify (2) slightly as\r\n" );
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document.write( "(2).   [acx - (2a-b)][bcx + (3a+2b)]\r\n" );
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document.write( "       [acx - 2a + b][bcx + 3a + 2b]\r\n" );
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document.write( "So the original equation:\r\n" );
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document.write( "abc²x² + 3a²cx + b²cx - 6a² - ab + 2b² = 0\r\n" );
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document.write( "becomes\r\n" );
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document.write( "[acx - 2a + b][bcx + 3a + 2b] = 0\r\n" );
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document.write( "Using the zero factor property:\r\n" );
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document.write( "acx - 2a + b = 0;  bcx + 3a + 2b = 0\r\n" );
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document.write( "acx = 2a - b \r\n" );
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document.write( "x = \"%282a-b%29%2F%28ac%29\"  <-- one root\r\n" );
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document.write( "bcx + 3a + 2b = 0\r\n" );
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document.write( "bcx = -3a - 2b \r\n" );
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document.write( "x = \"%28-3a-2b%29%2F%28bc%29\"  <-- other root\r\n" );
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document.write( "If a,b and c are rational and not 0, then both roots are\r\n" );
document.write( "rational also.\r\n" );
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document.write( "Edwin
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