How do you find all real zeros of #x^5 - x^4 - 3x^3 + 5x^2 - 2x = 0#?

1 Answer
Apr 11, 2018

Answer:

#x^5-x^4-3x^3+5x^2-2x=x(x+2)(x-1)^3#

Explanation:

The first real zero is easy. It's zero. We can factor out an #x#.

#x^5-x^4-3x^3+5x^2-2x=x(x^4-x^3-3x^2+5x-2)#

Now we note that when #x=-2#,

#x^4-x^3-3x^2+5x-2#

#=(-2)^4-(-2)^3-3(-2)^2+5(-2)-2#

#=16-(-8)-3(4)-10-2=0#.

This means that #x+2# is a factor of #x^4-x^3-3x^2+5x-2#.

Let's factor #x+2# from #x^4-x^3-3x^2+5x-2#.

#x^4-x^3-3x^2+5x-2#

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

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

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

Now we see that when #x=1#

#x^3-3x^2+3x-1#

#=1^3-3(1)^2+3(1)-1=1-3+3-1=0#.

This means that #x-1# is a factor of #x^3-3x^2+3x-1#. Let's factor #x-1# from #x^3-3x^2+3x-1#.

#x^3-3x^2+3x-1#

#=x^3-x^2-2x^2+2x+x-1#

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

Finally we recognize that

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

So putting it all together, we have

#x^5-x^4-3x^3+5x^5-2x=x(x+2)(x-1)(x-1)(x-1)#

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