How do you find all the zeros of #f(x) = x^4 - 3.3x^3 + 2.3x^2 + 0.6x#?

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Answer 1 · 2016-05-05T16:13:17

Zeros of #f(x)# are #{0,2,3/2,-1/5}#

#f(x)=x^4-3.3x^3+2.3x^2+0.6x#

= #x/10(10x^3-33x^2+23x+6)#

Hence one of the zeros is #0#. Another zero could be a factor of #6#

As putting #x=2# in #(10x^3-33x^2+23x+6)#, we get

#10*2^3-33*2^2+23*2+6=80-132+46+6=0#

Hence, #2# is another zero and dividing #10x^3-33x^2+23x+6# by #(x-2)#, we get

#10x^3-33x^2+23x+6=(x-2)(10x^2-13x-3)#

or #(x-2)(10x^2-15x+2x-3)#

or #(x-2)(5x(2x-3)+1*(2x-3))#

= #(x-2)(5x+1)(2x-3)#

Hence other zeros are given by #2x-3=0# and #5x+1=0# i.e. are #3/2# and #-1/5#

Hence zeros of #f(x)# are #{0,2,3/2,-1/5}#