document.write( "Question 141543: Verify that x = 2 is a root of multiplicity 3 of the equation x^4 - 4x^3 + 16x - 16 = 0. What is the other root? \n" ); document.write( "

Algebra.Com's Answer #103140 by solver91311(24713) $\"\"$ $\"About$

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Do the binomial expansion on $\"%28x-2%29%5E3\"$ . Using the result as the divisor and $\"x%5E4+-+4x%5E3+%2B+16x+-+16+=+0\"$ as the dividend, perform polynomial long division or synthetic division. Remember to insert $\"0x%5E2\"$ as a placeholder for the missing 2nd degree term. If x = 2 is a root of multiplicity 3, you will get no remainder and the quotient will be different from $\"x-2\"$ .\r
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\n" ); document.write( "\n" ); document.write( "If the quotient IS $\"x-2\"$ , then $\"%28x-2%29\"$ is a root with multiplicity 4.\r
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\n" ); document.write( "\n" ); document.write( "If there is a remainder, then $\"x-2\"$ is not a root at all.\r
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\n" ); document.write( "\n" ); document.write( "The quotient, $\"x-a\"$ where $\"a%3C%3E2\"$ , will be the 4th factor, and you can solve $\"x-a=0\"$ to get the 4th root. \n" ); document.write( "

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