```Question 274623
{{{x^4 + 3x^3 -x^2 +9x -12 = 0}}}
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.<br>
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.<br>
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.<br>
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.<br>
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:
<pre>
1 |   1  3  -1  9  -12
---      1   4  3   12
-----------------
1  4   3  12   0
</pre>
So {{{x^4 + 3x^3 -x^2 +9x -12 = (x-1)(x^3 + 4x^2 + 3x + 12)}}}
We could continue to try other rational roots. If you do, then<ul><li>Find rational roots of the second factor above.</li><li>Do not bother trying any positive roots. There are no other positive rational roots. I can say this because {{{(x^3 + 4x^2 + 3x + 12)}}} has coefficients that are all positive. And if both the coefficients and x are positive, how can {{{(x^3 + 4x^2 + 3x + 12)}}} ever be zero? Answer: It can't. We have to get some negatives from somewhere in oder for it to add up to zero.</li></ul>
However, instead of trying more rational roots, I'm going to use factoring by grouping to finish the factoring:
{{{(x-1)(x^3 + 4x^2 + 3x + 12)}}}
{{{(x-1)(x^2(x + 4) + 3(x + 4))}}}
{{{(x-1)(x+4)(x^2 - 3)}}}
So
{{{x^4 + 3x^3 -x^2 +9x -12 = (x-1)(x+4)(x^2 + 3) = 0}}}
From this we can see that the rational roots are 1 and -4. And there two imaginary roots: {{{i*sqrt(3)}}} and {{{-i*sqrt(3)}}}. (If you don't see these roots, take each factor, set it equal to zero and solve.)
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