How do you use the rational root theorem to list all possible roots for #x^3+5x^2-2x-15=0#?

1 Answer
Nov 18, 2016

Answer:

The "possible" rational roots are #+-1, +-3, +-5, +-15#

The actual roots are:

#x_k = 1/3(2sqrt(31) cos(1/3 cos^(-1)((65sqrt(31)) / 1922)+(2kpi)/3)-5)#

#k = 0, 1, 2#

Explanation:

Given:

#f(x) = x^3+5x^2-2x-15#

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Rational root theorem

The rational root theorem will only tell you about possible rational roots - not irrational or Complex ones.

By the rational root theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-15# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational roots are:

#+-1, +-3, +-5, +-15#

However, note that all of these are odd, resulting in all of the terms of #f(x)# being odd, except #-2x#, which will be even. So for any of these values we have:

#f(x) = "odd" + "odd" + "even" + "odd" = "odd" != 0#

So #f(x)# has no rational zeros.

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Bonus

Let us see what else we can find out about #f(x)#'s zeros...

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Discriminant

The discriminant #Delta# of a cubic polynomial in the form #ax^3+bx^2+cx+d# is given by the formula:

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

In our example, #a=1#, #b=5#, #c=-2# and #d=-15#, so we find:

#Delta = 100+32+7500-6075+2700 = 4257#

Since #Delta > 0# this cubic has #3# Real zeros.

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Descartes's Rule of Signs

The signs of the coefficients of #f(x)# are in the pattern #+ + - -#. With one change of sign, by Descartes' rule of signs, we can deduce that #f(x)# has exactly one positive Real zero. So the other two Real zeros are negative.

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Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

#0=27f(x)=27x^3+135x^2-54x-405#

#=(3x+5)^3-93(3x+5)-65#

#=t^3-93t-65#

where #t=(3x+5)#

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Trigonometric substitution

Look for a substitution of the form:

#t = k cos theta#

where #k# is chosen such that the resulting cubic in #cos theta# contains:

#4 cos^3 theta - 3 cos theta = cos 3 theta#

Let #k = 2sqrt(31)#

Then:

#0 = t^3-93t-65#

#color(white)(0) = k^3cos^3 theta - 93k cos theta - 65#

#color(white)(0) = 2sqrt(31)(124 cos^3 theta - 93 cos theta) - 65#

#color(white)(0) = 62sqrt(31)(4 cos^3 theta - 3 cos theta) - 65#

#color(white)(0) = 62sqrt(31)cos 3theta - 65#

Hence:

#cos 3 theta = 65/(62 sqrt(31)) = (65sqrt(31)) / 1922#

So:

#3 theta = +-cos^(-1) ((65sqrt(31)) / 1922) + 2kpi#

#t = 2sqrt(31) cos(1/3 cos^(-1)((65sqrt(31)) / 1922)+(2kpi)/3)#

Then #x = 1/3(t-5)# and hence the zeros of our original cubic are:

#x_k = 1/3(2sqrt(31) cos(1/3 cos^(-1)((65sqrt(31)) / 1922)+(2kpi)/3)-5)#

#k = 0, 1, 2#