document.write( "Question 440387: `Given that f(x) = $\"x%5E6%2Bx%5E5-6x%5E4%2B4x%5E3%2B7x%5E2-21x-18\"$ has -1 as a zero of multiplicity 2, 2 as a zero, and -3 as a zero, find all other zeros \n" ); document.write( "

Algebra.Com's Answer #304249 by richard1234(7193) $\"\"$ $\"About$

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We could use synthetic division or long division and divide f(x) by a quartic polynomial, but there is a much simpler way that will take far less time.\r
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\n" ); document.write( "\n" ); document.write( "Keep in mind that four of the zeros are known (-1, -1, 2, -3), so there are only two unknown zeros, which we will denote

and

. If you know Vieta's formulas*,\r
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\r
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\r
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\n" ); document.write( "\n" ); document.write( "This implies

and

. Applying Vieta's formulas again, if

are the roots of a quadratic polynomial, then these roots are equal to the roots of the polynomial

, which by the quadratic formula, we obtain\r
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\r
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upon simplification.\r
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\n" ); document.write( "*Vieta's formulas say that for any polynomial

, the sum of the roots is

and the product of the roots is

. Other cyclic sums involving the roots can be obtained, by multiplying n monomials and equating coefficients.\r
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\n" ); document.write( "\n" ); document.write( "http://mathworld.wolfram.com/VietasFormulas.html
\n" ); document.write( "http://en.wikipedia.org/wiki/Vi%C3%A8te's_formulas
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