How do you find all rational roots for #x^5 - 2x^4 + 11x^3 - 22x^2 - 12x + 24 = 0#?

1 Answer
José F.
Apr 7, 2016

#+-1,2#

Explanation:

We will use the Rational Root Theorem:

If the rational number r/s is a root of a polynomial whose coefficients are integers, then the integer r is a factor of the constant term, and the integer s is a factor of the leading coefficient.

r=24 and s=1

So the rational roots must be factor of 24/1=24:

#+-1, +-2,+-3,+-4,+-6,+-8, +-12#

By trying these possible roots, we discover that #+-1,2# are roots of the polynom.

If we divide the polynom by #(x-1)(x+1)(x-2)# we will obtain #x^2+12# which has the roots #+-isqrt(12)#, which are not rational.

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