document.write( "Question 274623: I know to find roots, but am not sure why this problem wont turn out.\r
\n" ); document.write( "\n" ); document.write( "Find all roots of the equation: \r
\n" ); document.write( "\n" ); document.write( "x to the fourth plus three x to the third subtract x squared plus nine x subtract twelve equals zero. \n" ); document.write( "
Algebra.Com's Answer #200501 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
\"x%5E4+%2B+3x%5E3+-x%5E2+%2B9x+-12+=+0\"
\n" ); document.write( "To find the roots we need to factor this. The Greatest Common Factor (GCF) is 1 (which we rarely bother factoring out). This expression has too many terms for any of the commonly used patterns and it has too many terms for factoring trinomials. And I don't see how factoring by grouping would work. So we are left with factoring by trial and error of the possible rational roots.

\n" ); document.write( "Trial and error can be tedious, especially if there are a lot of possible rational roots. The possible rational roots are all the positive and negative numbers which can be formed using a factor of the constant term (the number without a variable at the end) in the numerator and a factor of the leading coefficient in the denominator.

\n" ); document.write( "Your polynomial has a constant term of -12 and a leading coefficient of 1. So the possible rational roots are: 1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 12, -12. This could take a while. But I will try to explain how to avoid wasting time.

\n" ); document.write( "1 is always a possible rational root (because it is a factor of every constant term and leading coefficient) and it is easy to check mentally. Just add all the coefficients (since 1 to any power is 1 the exponents don't really matter after all. If the sum is 0 then 1 is a root.

\n" ); document.write( "Your coefficients are: 1 3 -1 9 and -12. If you add these up you find that the sum is 0!! So we know that 1 is a root. And if 1 is a root, then x-1 is a factor of your polynomial. We can use synthetic division to divide by x-1 and find the other factor:
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document.write( "1 |   1  3  -1  9  -12\r\n" );
document.write( "---      1   4  3   12              \r\n" );
document.write( "     -----------------\r\n" );
document.write( "      1  4   3  12   0\r\n" );
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\n" ); document.write( "So \"x%5E4+%2B+3x%5E3+-x%5E2+%2B9x+-12+=+%28x-1%29%28x%5E3+%2B+4x%5E2+%2B+3x+%2B+12%29\"
\n" ); document.write( "We could continue to try other rational roots. If you do, then
  • Find rational roots of the second factor above.
  • Do not bother trying any positive roots. There are no other positive rational roots. I can say this because \"%28x%5E3+%2B+4x%5E2+%2B+3x+%2B+12%29\" has coefficients that are all positive. And if both the coefficients and x are positive, how can \"%28x%5E3+%2B+4x%5E2+%2B+3x+%2B+12%29\" ever be zero? Answer: It can't. We have to get some negatives from somewhere in oder for it to add up to zero.

\n" ); document.write( "However, instead of trying more rational roots, I'm going to use factoring by grouping to finish the factoring:
\n" ); document.write( "\"%28x-1%29%28x%5E3+%2B+4x%5E2+%2B+3x+%2B+12%29\"
\n" ); document.write( "\"%28x-1%29%28x%5E2%28x+%2B+4%29+%2B+3%28x+%2B+4%29%29\"
\n" ); document.write( "\"%28x-1%29%28x%2B4%29%28x%5E2+-+3%29\"
\n" ); document.write( "So
\n" ); document.write( "\"x%5E4+%2B+3x%5E3+-x%5E2+%2B9x+-12+=+%28x-1%29%28x%2B4%29%28x%5E2+%2B+3%29+=+0\"
\n" ); document.write( "From this we can see that the rational roots are 1 and -4. And there two imaginary roots: \"i%2Asqrt%283%29\" and \"-i%2Asqrt%283%29\". (If you don't see these roots, take each factor, set it equal to zero and solve.)
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