document.write( "Question 628683: How do you find a 5th degree polynomial in factored form with roots: at -2, 1, 3 only? I have absolutely no clue, please help. This is kind of urgent. Thank you very much. \n" ); document.write( "
Algebra.Com's Answer #395853 by jsmallt9(3758)![]() ![]() ![]() You can put this solution on YOUR website! A 5th degree polynomial will have 5 roots according to the Fundamental Theorem of Algebra. So there are two \"missing\" roots. The explanation for these \"missing\" roots could be:
\n" ); document.write( "I am going to assume that explanation for the \"missing\" roots is the second one. \n" ); document.write( "If a polynomial has a root, let's call it r, then (x-r) will be a factor of that polynomial. So our factored polynomial will have factors of \n" ); document.write( "(x - (-2)) or (x + 2) \n" ); document.write( "(x - 1) and \n" ); document.write( "(x - 3) \n" ); document.write( "We still need two more factors. \n" ); document.write( "If a polynomial more than one factor of (x-r) then that root is a multiple root (also called a root of multiplicity n where n is the number of times (x-r) is a factor. So we can \"fill in\" for the \"missing\" roots, by using multiple copies of one or more of the above 3 factors. For example: \n" ); document.write( "P(x) = (x+2)(x+2)(x+2)(x-1)(x-3) \n" ); document.write( "P(x) = (x+2)(x+2)(x-1)(x-1)(x-3) \n" ); document.write( "P(x) = (x+2)(x-1)(x-1)(x-1)(x-3) \n" ); document.write( "P(x) = (x+2)(x-1)(x-3)(x-3)(x-3) \n" ); document.write( "etc. \n" ); document.write( "As long as there are 5 factors of any combination of the three the polynomial will fit the description you provided. FWIW, you could also throw in a constant factor and it would still fit: \n" ); document.write( "P(x) = 6(x+2)(x-1)(x-1)(x-3)(x-3) \n" ); document.write( "P(x) = (-1/2)(x+2)(x+2)(x-1)(x-3)(x-3) \n" ); document.write( "etc \n" ); document.write( " \n" ); document.write( " |