document.write( "Question 46537: find the polynomial f(x) of degree three that has zeroes at 1,2 and 4 such that
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Algebra.Com's Answer #30828 by AnlytcPhil(1806)\"\" \"About 
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document.write( "Find the polynomial f(x) of degree three that has zeroes \r\n" );
document.write( "at 1, 2 and 4 such that f(0) = -16.\r\n" );
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document.write( "All polynomials of degree n that have zeros r1, r2, ..., rn \r\n" );
document.write( "are of the form\r\n" );
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document.write( "f(x) = k(x-r1)(x-r2)···(x-rn) \r\n" );
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document.write( "where k can be any non-zero number.\r\n" );
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document.write( "So any polynomial of degree 3 that has zeros 1, 2, and 4 is \r\n" );
document.write( "of the form\r\n" );
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document.write( "f(x) = k(x-1)(x-2)(x-4)\r\n" );
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document.write( "where k can be any non-zero number.\r\n" );
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document.write( "But we also want f(0) to equal -16. So it has to be true that \r\n" );
document.write( "if we substitute 0 for x and then -16 for f(0), they should be \r\n" );
document.write( "equal.  So,\r\n" );
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document.write( "f(x) = k(x-1)(x-2)(x-4)\r\n" );
document.write( "f(0) = k(0-1)(0-2)(0-4)\r\n" );
document.write( " -16 = -8k\r\n" );
document.write( "   2 = k\r\n" );
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document.write( "So substitute 2 for k in \r\n" );
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document.write( "f(x) = k(x-1)(x-2)(x-4)\r\n" );
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document.write( "to give\r\n" );
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document.write( "f(x) = 2(x-1)(x-2)(x-4)    \r\n" );
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document.write( "Multiply all that out and you'll get\r\n" );
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document.write( "f(x) = 2x³ - 14x² + 28x - 16 \r\n" );
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document.write( "That is choice (b).\r\n" );
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document.write( "Edwin
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