document.write( "Question 628931: i don't understand how to do a polynomial equation, particularly ones with exponents higher then 2 like x^3+9x^2+23x+15=0 \n" ); document.write( "

Algebra.Com's Answer #395931 by solver91311(24713) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "First note that you have a 3rd degree polynomial equation. Therefore you have three zeros. They could be 3 real and distinct zeros, 1 real and distinct zero plus 1 other real zero with a multiplicity of 2, or 1 real zero and a conjugate pair of complex zeros of the form

where

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\n" ); document.write( "\n" ); document.write( "Next, apply the Rational Roots Theorem. If a polynomial function of the form

has a rational zero, then that zero must be of the form

where

is an integer divisor of

and

is an integer divisor of

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\n" ); document.write( "\n" ); document.write( "For your problem, given that the lead coefficient is 1, the only possible value for

is 1. Given the constant term value of 15, the possible integer divisors are 1, 3, 5, and 15. Therefore the only possible rational zeros are

, and

. Note that the Rational Roots Theorem does NOT guarantee that any of the possibile rational zeros are actually zeros of the polynomial. It only guarantees that if a rational zero exists, it will indeed be a member of the set of possibles as defined by the Rational Roots Theorem.\r
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\n" ); document.write( "\n" ); document.write( "The next step is to test each of the 8 possible values to determine which, if any, are rational zeros of the given polynomial equation. You could, if you are a glutton for arithmetic punishment, substitute each of the values into the equation one at a time until you either find 1, 3, or none that make the function value be zero. But there is a better way called Synthetic Division. See Purple Math Synthetic Division. Be sure you read all four pages.\r
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\n" ); document.write( "\n" ); document.write( "Start running your values through the synthetic division process. If and when you find a value that shows a zero remainder, then the remaining 3 values in the bottom row will be the coefficients of the quotient polynomial. Since that polynomial has to be of degree one less than the degree three polynomial you started with, it must be a quadratic that you can solve by any means you like (Hint: it factors).\r
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\n" ); document.write( "\n" ); document.write( "Note: This process does not always work. But when it does work it works on any degree polynomial. If you are working with polynomials of the 4th degree or less, then, should the above process fail, you can always fall back on the general solutions to the cubic and quartic. However, once you look up the general cubic and then consider that the general quartic is several times worse, you may decide that being drawn and quartered is a slightly less painful alternative. If you have something that is 5th degree or greater and you cannot find any rational roots, then you are stuck with numeric approximation methods since no general solution exists for the quintic and above. \r
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\n" ); document.write( "\n" ); document.write( "John
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\n" ); document.write( "My calculator said it, I believe it, that settles it
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$\"The$

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