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

1 Answer
Harish Chandra Rajpoot
Jun 19, 2018

The sum of all coefficients of given cubic equation: #x^3-3x^2-x+3=0# is zero hence #x=1# is a root of given cubic equation i.e. #(x-1)# is a factor of #x^3-3x^3-x+3#
Now, cubic algebraic polynomial: #x^3-3x^2-x+3# can be factorized as follows
#x^3-3x^3-x+3#
#=x^2(x-1)-2x(x-1)-3(x-1)#
#=(x-1)(x^2-2x-3)#
#=(x-1)(x^2-3x+x-3)#
#=(x-1)(x(x-3)+(x-3))#
#=(x-1)(x-3)(x+1)#
hence, the solution of given cubic equation will be given as
#(x-1)(x-3)(x+1)=0#
#x=-1, 1, 3#

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