How do you use the rational roots theorem to find all possible zeros of #y=2x^3+2x^2-23x-9#?

1 Answer
Jul 7, 2016

Answer:

Use a trigonometric method to find Real zeros:

#x_n = 1/3(sqrt(142) cos(1/3 arccos((8sqrt(142))/2201) + (2npi)/3) - 1)#

for #n = 0, 1, 2#.

Explanation:

#f(x) = 2x^3+2x^2-23x-9#

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Rational roots 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 #-9# and #q# a divisor of the coefficient #2# of the leading term.

So the only possible rational zeros are:

#+-1/2, +-1, +-3/2, +-3, +-9/2, +-9#

None of these works, so #f(x)# has no rational 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=2#, #b=2#, #c=-23# and #d=-9#, so we find:

#Delta = 2116+97336+288-8748+14904 = 105896#

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

As a result, pure algebraic methods like Cardano's method will yield results containing irreducible cube roots of Complex numbers. So choose to use a trigonometric solution instead.

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Tschirnhaus transformation

First, simplify the cubic using a linear substitution:

#108 f(x) = 216x^3+216x^2-2484x-972#

#=(6x+2)^3-426(6x+2)-128#

#=t^3-426t-128#

where #t = 6x+2#

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Trigonometric solution

We want to solve:

#t^3-426t-128 = 0#

Consider the substitution:

#t = k cos theta#

Then:

#t^3-426t-128#

#= k^3 cos^3 theta - 426 k cos theta - 128#

#= k^3/4 (4cos^3 theta) - 142k (3cos theta) - 128#

Solve:

#k^3/4 = 142k#

to get:

#k = +-2sqrt(142)#

Let #k = 2sqrt(142)# to get:

#0 = t^3-426t-128#

#= 284 sqrt(142)(4 cos^3theta - 3cos theta)-128#

#= 284 sqrt(142)cos(3 theta)-128#

Hence:

#cos(3 theta) = 128/(248sqrt(142)) = (8sqrt(142))/2201#

Hence roots:

#t_n =2 sqrt(142) cos(1/3 arccos((8sqrt(142))/2201) + (2npi)/3)#

giving distinct values for #n = 0, 1, 2#

Then #x = 1/6(t-2)# so the zeros of the original cubic are:

#x_n = 1/3(sqrt(142) cos(1/3 arccos((8sqrt(142))/2201) + (2npi)/3) - 1)#

for #n = 0, 1, 2#.