# How do you find all real ratios..for this eq..[step by step]

• May 31st 2009, 06:02 PM
Valaina
How 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, 07:17 PM
pickslides
Quote:

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

Do you mean all real roots? If so I would suggest the factor theorem.
• Jun 1st 2009, 06:14 AM
yeongil
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
$f(x)=x^5+9x^4+13x^3-39x^2-56x-12$
would be
$\frac{factors\ of -12}{factors\ of 1} = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$

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 $x^4 + 3x^3 - 5x^2 - 9x - 2.$ -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 $x^3 + 2x^2 - 7x - 2.$ 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
$f(x)=x^5+9x^4+13x^3-39x^2-56x-12=(x + 6)(x + 1)(x - 2)(x^2 + 4x + 1).$ To find the remaining two roots, solve the quadratic $x^2 + 4x + 1 = 0$. 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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