document.write( "Question 1209706: There are integers b, c for which both roots of the polynomial x^2 - x - 3 are also roots of the polynomial x^3 - bx^2 - c. Determine the ordered pair (b,c).
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Algebra.Com's Answer #849884 by ikleyn(52943) $\"\"$ $\"About$

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document.write( "Let p and q be the roots of the polynomial x^2 - x - 3.\r\n" );
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document.write( "Due to Vieta's theorem, \r\n" );
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document.write( "    p + q = 1,        (1)\r\n" );
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document.write( "    pq = -3.          (2)\r\n" );
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document.write( "According to the problem, p and q are also the roots of the polynomial x^3 - bx^2 - c. \r\n" );
document.write( "Let r be the third root of this polynomial.\r\n" );
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document.write( "Then, due to Vieta's theorem for polynomial x^3 - bx^2 - c\r\n" );
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document.write( "    p + q + r = b,          (3)\r\n" );
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document.write( "    p*q + p*r + q*r = 0,    (4)   (coefficient at x in polynomial x^3 - bx^2 - c)\r\n" );
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document.write( "    p*q*r = c.              (5)\r\n" );
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document.write( "In (3), replace p+q by 1, based on (3).  In (5), replace p*q by -3, based on (2).\r\n" );
document.write( "Then from (3) an (4) you will have\r\n" );
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document.write( "    1 + r = b,              (6)\r\n" );
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document.write( "    -3r =  c,               (7) \r\n" );
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document.write( "In equation (4), replace p*q  by -3, based on (2).  Then equation (4) takes the form\r\n" );
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document.write( "    -3 + pr + qr = 0,  \r\n" );
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document.write( "or\r\n" );
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document.write( "    p*r + q*r  = 3,\r\n" );
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document.write( "    (p + q)*r  = 3.\r\n" );
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document.write( "In the last equation, replace (p + q) by 1,  based on (1).  You will get\r\n" );
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document.write( "    1*r = 3,  i.e.  r = 3.\r\n" );
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document.write( "Now from  (6)  b = 1 + r = 1 + 3 = 4;\r\n" );
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document.write( "    from  (7)  c = -3r = -3*3 = -9.\r\n" );
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document.write( "ANSWER.  In polynomial  x^3 - bx^2 - c, coefficients  are b = 4,  c = -9.\r\n" );
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