How do you find all the real and complex roots and use Descartes Rule of Signs to analyze the zeros of #P(x) = x^3 - 4x^2 + x + 6#?

1 Answer
Harish Chandra Rajpoot
Jul 9, 2018

#x=-1, 2, 3#

Explanation:

The given cubic polynomial: #P(x)=x^3-4x^2+x+6#

we find that the sum of coefficients of odd powers of #x# is equal to sum of coefficients of even powers of variable #x# hence #x=-1# is a root of cubic polynomial #x^3-4x^2+x+6# i.e. #(x+1)# is a factor of #x^3-4x^2+x+6# & it can be factorized as follows

#x^3-4x^2+x+6#

#=x^2(x+1)-5x(x+1)+6(x+1)#

#=(x+1)(x^2-5x+6)#

#=(x+1)(x^2-2x-3x+6)#

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

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

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

hence, the roots of given cubic polynomial #P(x)# are given as follows

#P(x)=0#

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

#x=-1, 2,3#

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