document.write( "Question 1100423: Form the fifth-degree polynomial function with real coefficients,
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Algebra.Com's Answer #714922 by Edwin McCravy(20054) $\"\"$ $\"About$

You can put this solution on YOUR website!

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document.write( "I assumed you meant 1+1 instead of 11+i, since 11+1 made it so complicated.\r\n" );
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document.write( "-2 and 1 are zeros because the graph passes through (-2,0) and (1,0).\r\n" );
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document.write( "1+i is a zero, and since it has real coefficients,\r\n" );
document.write( "its conjugate 1-i is also a zero.\r\n" );
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document.write( "Therefore when the polynomial, f(x) is set equal to 0, we get\r\n" );
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document.write( "x = -2, x = 1, x = 1+i, x = 1-i\r\n" );
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document.write( "But there must be one more real zero, so that it will have degree 5.\r\n" );
document.write( "Let that fifth zero be \"a\", making the polynomial have 5 zeros\r\n" );
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document.write( "x = -2, x = 1, x = 1+i, x = 1-i, x = a\r\n" );
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document.write( "Getting 0 on each side we have\r\n" );
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document.write( "x+2 = 0, x-1 = 0, x-1-i = 0, x-1+i = 0, x-a = 0\r\n" );
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document.write( "Multiplying all left sides and right sides:\r\n" );
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document.write( "(x+2)(x-1)(x-1-i)(x-1+i)(x-a) = (0)(0)(0)(0)(0)\r\n" );
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document.write( "(x+2)(x-1)[(x-1)-i][(x-1)+i](x-a) = 0\r\n" );
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document.write( "So \r\n" );
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document.write( "f(x) = (x+2)(x-1)(x-1-i)(x-1+i)(x-a)\r\n" );
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document.write( "We are also given that the graph passes through (0,8), so\r\n" );
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document.write( "f(0) = (0+2)(0-1)(0-1-i)(0-1+i)(0-a) = 8\r\n" );
document.write( "             (2)(-1)(-1+i)(-1-i)(-a) = 8 \r\n" );
document.write( "                    (2a)(-1+i)(-1-i) = 8\r\n" );
document.write( "                            2a(1-i²) = 8\r\n" );
document.write( "                          2a[1-(-1)] = 8\r\n" );
document.write( "                             2a(1+1) = 8\r\n" );
document.write( "                               2a(2) = 8\r\n" );
document.write( "                                  4a = 8\r\n" );
document.write( "                                   a = 2\r\n" );
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document.write( "Therefore \r\n" );
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document.write( "f(x) = (x+2)(x-1)(x-1-i)(x-1+i)(x-2)              \r\n" );
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document.write( "Now you hav to go to all the troubl to muiltiply\r\n" );
document.write( "that out and collect terms.  It may help to group\r\n" );
document.write( "the first two terms in the factors with the i's as\r\n" );
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document.write( "f(x) = (x+2)(x-1)[(x-1)-i][(x-1)+i](x-2)\r\n" );
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document.write( "So I'll leave the multiplying up to you.\r\n" );
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document.write( "Edwin

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