f(x)=x^5+9x^4+13x^3-39x^2-56x-12

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- May 31st 2009, 05:02 PMValainaHow do you find all real ratios..for this eq..[step by step]
f(x)=x^5+9x^4+13x^3-39x^2-56x-12

- May 31st 2009, 06:17 PMpickslides
- Jun 1st 2009, 05:14 AMyeongil
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

Code:`-1| 1 3 -5 -9 -2`

-- -1 -2 7 2

-------------------

1 2 -7 -2 0

Code:`2| 1 2 -7 -2`

- 2 8 2

---------------

1 4 1 0

$\displaystyle f(x)=x^5+9x^4+13x^3-39x^2-56x-12=(x + 6)(x + 1)(x - 2)(x^2 + 4x + 1).$ 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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