document.write( "Question 43913This question is from textbook Algebra and Trigonometry with Analytic Geometey
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Algebra.Com's Answer #28814 by AnlytcPhil(1806)\"\" \"About 
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Find the polynomial f(x) of degree three that has zeros at 1,2, and 4\r\n" );
document.write( "such that f(0)=-16.  My answer is f(x)=x^3-7x^2+14x-16,  could this \r\n" );
document.write( "be right.  thanks for checking my answer.\r\n" );
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document.write( "No, that couldn't be the right answer because\r\n" );
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document.write( "f(1) = (1)³ - 7(1)² + 14(1) - 16 = 1 - 7 + 14 - 16 = -8\r\n" );
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document.write( "But f(1) would have to be 0, not -8, if 1 were a zero.\r\n" );
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document.write( "1. If r is a zero of a polynomial then (x-r) is a factor of the\r\n" );
document.write( "   polynomial.\r\n" );
document.write( "2. If a polynomial is multiplied by a constant k, it still has the\r\n" );
document.write( "   same zeros.\r\n" );
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document.write( "By 1, since 1 is a zero, (x-1) must be a factor of the polynomial.\r\n" );
document.write( "By 1, since 2 is a zero, (x-2) must be a factor of the polynomial.\r\n" );
document.write( "By 1, since 4 is a zero, (x-4) must be a factor of the polynomial.\r\n" );
document.write( "By 2, we can assume the polynomial has factor k, so we have\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( "Since f(0) = -16\r\n" );
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document.write( "f(0) = -16 = k(0-1)(x-2)(0-4) =\r\n" );
document.write( "       k(-1)(-2)(-4) =\r\n" );
document.write( "       -8k\r\n" );
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document.write( "since -16 = -8k, then k = 2, so\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( "f(x) = 2(x-1)(x²-6x+8)\r\n" );
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document.write( "f(x) = 2(x³-6x²+8x-x²+6x-8)\r\n" );
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document.write( "f(x) = 2(x³-7x²+14x-8)\r\n" );
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document.write( "f(x) = 2x³-14x²+28x-16\r\n" );
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document.write( "Edwin\r\n" );
document.write( "AnlytcPhil@aol.com
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