document.write( "Question 1172086: Find the equation of a degree 4 polynomial with zeros x=5, x=−6, and x=1+i. It passes through the point (1,84). Write the equation in general form.\r
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Algebra.Com's Answer #796999 by ikleyn(52805)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "The polynomial is ASSUMED to be with real coefficients - it is the condition you missed in your post (!)\r\n" );
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document.write( "Then the roots are 5, -6, 1+i and 1-i  (together with 1+i, its conjugate 1-i is the root, too - for such a polynomial).\r\n" );
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document.write( "Then the associate linear binomials are \r\n" );
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document.write( "    (x-5),  (x+6),  (x-(1+i))  and  (x-)1-i)).\r\n" );
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document.write( "We seek for the product of these binomials with the unknown real coefficient \"a\"\r\n" );
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document.write( "    f(x) = a*(x-5)*(x+6)*(x-(1+i))*(x-(1-i)).\r\n" );
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document.write( "The product of the last two binomials is\r\n" );
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document.write( "    (x-(1+i))*(x-(1-i)) = x^2 - 2x + (1+i)*(1-i) = x^2 - 2x +2.\r\n" );
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document.write( "Therefore, the polynomial f(x) is\r\n" );
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document.write( "     f(x) = a*(x-5)*(x+6)*(x^2 - 2x + 2)\r\n" );
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document.write( "To find \"a\", use the fact that  f(1) = 84.   It gives\r\n" );
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document.write( "      84 = a*(1-5)*(1+6)*(1 - 2 + 2) = a*(-4)*7*1 = -28a.\r\n" );
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document.write( "Hence,  a= 84/(-28) = -4.\r\n" );
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document.write( "So, your polynomial is\r\n" );
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document.write( "    f(x) = -4*(x-5)*(x+6)*(x^2 - 2x + 2).\r\n" );
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document.write( "It is your ANSWER.  \r\n" );
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document.write( "You may transform it further, if you want / (if you need).\r\n" );
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