How do you find the roots of #x^3-5x^2+2x+8=0#?

1 Answer
Dec 24, 2016

Answer:

Roots of #x^3-5x^2+2x+8=0# are #{-1,2,4}#.

Explanation:

As the coefficient of #x^3# in #x^3-5x^2+2x+8=0# is #1#,

One of the roots must be factor of #8/1=8# i.e. it can be #+-1. +-2, +-4 or +-8#.

As sum of coefficients (#1-5+2+8=6!=0#) is not zero, it is evident that #1# is not the root of #x^3-5x^2+2x+8=0#.

But #-1# is a root as #-1-5-2+8=0# and hence as per factor theorem #(x+1)# is a factor of #x^3-5x^2+2x+8=0#

Dividing #x^3-5x^2+2x+8# by #(x+1)# we get

#x^3-5x^2+2x+8=x^2(x+1)-6x(x+1)+8(x+1)=(x+1)(x^2-6x+8)#

now we can further factorize #x^2-6x+8# by splitting middle term

#x^2-6x+8=x^2-4x-2x+8=x(x-4)-2(x-4)=(x-2)(x-4)#

Hence, roots of #x^3-5x^2+2x+8=0# are #{-1,2,4}#.