How do you find all real zeros of x5−x4−3x3+5x2−2x=0?

1 Answer
Mr. Mike
Apr 11, 2018

x5−x4−3x3+5x2−2x=x(x+2)(x−1)3

Explanation:

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

x5−x4−3x3+5x2−2x=x(x4−x3−3x2+5x−2)

Now we note that when x=−2,

x4−x3−3x2+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 x4−x3−3x2+5x−2.

Let's factor x+2 from x4−x3−3x2+5x−2.

x4−x3−3x2+5x−2

=x4+2x3−3x3−6x2+3x2+6x−x−2

=x3(x+2)−3x2(x+2)+3x(x+2)−(x+2)

=(x3−3x2+3x−1)(x+2)

Now we see that when x=1

x3−3x2+3x−1

=13−3(1)2+3(1)−1=1−3+3−1=0.

This means that x−1 is a factor of x3−3x2+3x−1. Let's factor x−1 from x3−3x2+3x−1.

x3−3x2+3x−1

=x3−x2−2x2+2x+x−1

=x2(x−1)−2x(x−1)+(x−1)=(x2−2x+1)(x−1)

Finally we recognize that

x2−2x+1=(x−1)(x−1)

So putting it all together, we have

x5−x4−3x3+5x5−2x=x(x+2)(x−1)(x−1)(x−1)

=x(x+2)(x−1)3.

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