How do you solve #4x^3-12x^2-3x+9=0#?

1 Answer
Shwetank Mauria
May 6, 2018

#x=3,-sqrt3/2# or #sqrt3/2#

Explanation:

According to factor theorem , if #f(a)=0#, then #(x–a)# is a factor of the polynomial #f(x)#. Converse of this theorem is also true i.e. if #(x-a)# is a factor of the polynomial #f(x)#, then #f(a)=0#.

Here for #x=3#, #f(x)=4x^3-12x^2-3x+9#, we have

#f(3)=4*3^3-12*3^2-3*3+9=108-108-9+9=0#

Hence #x-3# is a factor of #4x^3-12x^2-3x+9# and hence

#4x^3-12x^2-3x+9=0#

#=>4x^2(x-3)-3(x-3)=0#

or #(x-3)(4x^2-3)=0#

i.e. #(x-3)(2x+sqrt3)(2x-sqrt3)=0#

Hence #x=3,-sqrt3/2# or #sqrt3/2#

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