How do you find the zeros of # x^3 - 7x + 6 = 0#?

1 Answer
Jun 14, 2016

Answer:

#x^3-7x+6=(x-1)(x-2)(x+3)#

Explanation:

We use the property that if #alpha# is zero of #f(x)# i.e. #f(alpha)=0#, then #(x-alpha)# is a factor of #f(x)#.

As here #f(x)=x^3-7x+6# and #f(1)=1^3-7*1+6=1-7+6=0#

#(x-1)# is a factor of #x^3-7x+6#.

Now dividing #x^3-7x+6# by #(x-1)#, we get

#x^2+x-6#, whose discriminant is #1^2-4*1*(-6)=25=5^2#, hence factors can be found by splitting middle term. Hence,

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

Hence, #x^3-7x+6=(x-1)(x-2)(x+3)#