How do I find the zeros of x^4-9x^2-4x+12 with Cauchy's Bound?

How do I find the zeros of x^4-9x^2-4x+12 with Cauchy's Bound?

1 Answer
Cem Sentin
Feb 16, 2018

#x_1=x_2=-2#, #x_3=1# and #x_4=3#

Explanation:

According to Cuchy's Bound, I took absolute value of leading coefficient, in this case |1|=1 and divide it into the largest of the remaining ones, in this case |12|=12. Hence #M=12/1=12#. So it's guaranteed that the all real zeros of this polynomial lie in #[-13, 13]#.

Due to sum of coefficients of it #1-9-4+12=0#, #x=1# must be zero of it or #(x-1)# must be multiplier of it.

Hence,

#x^4-9x^2-4x+12=0#

#x^4-x^3+x^3-x^2-8x^2+8x-12x+12=0#

#x^3*(x-1)+x^2*(x-1)-8x*(x-1)-12*(x-1)=0#

#(x-1)*(x^3+x^2-8x-12)=0#

After using Rational Root Theorem, #x=3# is root of it or #(x-3)# is another multiplier of it.

Hence,

#x^4-9x^2-4x+12=0#

#(x-1)*(x^3+x^2-8x-12)=0#

#(x-1)*(x^3-3x^2+4x^2-12x+4x-12)=0#

#(x-1)(x-3)(x^2+4x+4)=0#

#(x-1)(x-3)(x+2)^2=0#

Thus, zeros of this polynomial are #x_1=x_2=-2#, #x_3=1# and #x_4=3#

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