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

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Answer 1 · 2016-09-29T06:30:17

The roots are:

#x=1" "# with multiplicity #2#

#x=-3" "# (with multiplicity #1#)

#f(x) = x^3+x^2-5x+3#

Note that the sum of the coefficients of #f(x)# is #0#. That is:

#1+1-5+3 = 0#

Hence #f(1) = 0# and #(x-1)# is a factor:

#x^3+x^2-5x+3 = (x-1)(x^2+2x-3)#
#color(white)(x^3+x^2-5x+3) = (x-1)(x-1)(x+3)#

Hence the roots of the given equation are:

#x=1# with multiplicity #2#

#x=-3#

graph{x^3+x^2-5x+3 [-10.545, 9.455, -22.2, 27.8]}