Start with the Rational Zeros Test, which states that all potential rational zeros of a function is in the form of P/Q, where P is a factor of the constant term, and Q is a factor of the leading coefficient. The potential rational zeros of
$\displaystyle f(x)=x^5+9x^4+13x^3-39x^2-56x-12$
would be
$\displaystyle \frac{factors\ of -12}{factors\ of 1} = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$
Now, start testing the possible roots using synthetic division. -12 doesn't work (needs to give a remainder of 0), but -6 does:
Code:
-6| 1 9 13 -39 -56 -12
-- -6 -18 30 54 12
------------------------
1 3 -5 -9 -2 0
Continue to try the potential zeros, but now on the quotient of the above division, which is $\displaystyle x^4 + 3x^3 - 5x^2 - 9x - 2.$ -4, -3, and -2 don't work, but -1 does:
Code:
-1| 1 3 -5 -9 -2
-- -1 -2 7 2
-------------------
1 2 -7 -2 0
You're now left with the polynomial $\displaystyle x^3 + 2x^2 - 7x - 2.$ Using this as the new dividend, 1 doesn't work, but 2 does:
Code:
2| 1 2 -7 -2
- 2 8 2
---------------
1 4 1 0
So far we've found 3 roots: -6, -1, and 2. The factorization so far is
$\displaystyle f(x)=x^5+9x^4+13x^3-39x^2-56x-12=(x + 6)(x + 1)(x - 2)(x^2 + 4x + 1).$ To find the remaining two roots, solve the quadratic $\displaystyle x^2 + 4x + 1 = 0$. This can't be factored evenly, so either complete the square or use the quadratic formula.
Edit: hit Submit too soon...
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