What are the zeros of #7x^3+70x^2+84x+8=0# ?

2 Answers
Aug 15, 2017

Answer:

#x_n = 1/3(-10+16cos(1/3cos^(-1)(-13/14)+(2npi)/3))" "n = 0, 1, 2#

Numerical approximations:

#x_0 ~~ -0.10419#

#x_1 ~~ -8.62388#

#x_2 ~~ -1.27193#

Explanation:

Given:

#f(x) = 7x^3+70x^2+84x+8#

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=7#, #b=70#, #c=84# and #d=8#, so we find:

#Delta = 34574400-16595712-10976000-84672+5927040 = 12845056#

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

#color(white)()#
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=1323f(x)=9261x^3+92610x^2+111132x+10584#

#=(21x+70)^3-9408(21x+70)+326144#

#=t^3-9408t+326144#

where #t=(21x+70)#

#color(white)()#
Trigonometric substitution

Since all three roots of this cubic are real, it falls into the irreducible case (casus irreducibilis) of Cardano's method. Using Cardano's method will result in a solution expressed in terms of cube roots of complex numbers which cannot be simplified.

In such cases I prefer to use a trigonometric substitution, putting:

#t = k cos theta#

and choosing #k# so that the resulting equation in #cos theta# contains #4cos^3 theta - 3cos theta = cos 3 theta#.

We have:

#0 = k^3cos^3theta-9408kcostheta+326144#

Putting #k = 2*sqrt(9408/3) = 2sqrt(3136) = 112# we find:

#color(white)(0) = 1404928cos^3theta-1053696costheta+326144#

#color(white)(0) = 351232(4cos^3theta-3costheta)+326144#

#color(white)(0) = 25088(14cos3theta+13)#

So:

#cos3theta = -13/14#

So:

#3 theta = +-cos^(-1)(-13/14)+2npi#

So:

#theta = +-1/3cos^(-1)(-13/14)+(2npi)/3#

So:

#cos theta = cos(1/3cos^(-1)(-13/14)+(2npi)/3)#

So the distinct roots are:

#t_n = 112 cos(1/3cos^(-1)(-13/14)+(2npi)/3)" "n = 0,1,2#

Then #x = 1/21(t-70)#

So the distinct roots of the original cubic are:

#x_n = 1/3(-10+16cos(1/3cos^(-1)(-13/14)+(2npi)/3))" "n = 0, 1, 2#

Numerical approximations:

#x_0 ~~ -0.10419#

#x_1 ~~ -8.62388#

#x_2 ~~ -1.27193#

Aug 15, 2017

Answer:

There are no rational roots.

Explanation:

That George is correct in providing the solutions through the method that he did is evident from using the rules found in Precalc I. Descartes' Rule of Signs reveals that there are 3 or 1 negative real roots and no positive real roots.

The Bounding results from Synthetic Divisions reveal that all of them lie in the interval (-10, 0).

The Rational Root Theorem (or Rational Zeros Theorem) indicates that the only possible negative rational zeros are -8, -4, -2, -8/7, -1, -4/7, -2/7, and -1/7.

None of these are solutions; hence, there are no rational solutions.
Therefore the polynomial will not factor.

We must use a more advanced approach, as George has taken.
Are you sure that you have copied the problem down correctly?