Find Polynomial Roots: x4-2x3-25x2+50x

[theakdad]

I have to find all solutions for X when:
x⁴-2x³-25x²+50x

I have done it so,but I am not sure if this is ok:

x(x³-2x²-25x+50)

= x(x²(x-2)-25(x+2)
= x(x²-25)(x-2)
=x(x-5)(x+5)(x-2)

Now i see that root/zeroes are +5,-5 and 2. I know that this polynomial has another zero that is 0,but how do i know that? Because x is in the front? Or did i make a mistake and should such problems deal another way? Thank you for all replies!

 

[MarkFL]

I have to find all solutions for X when:
x⁴-2x³-25x²+50x

I have done it so,but I am not sure if this is ok:

x(x³-2x²-25x+50)

= x(x²(x-2)-25(x-2))
= x(x²-25)(x-2)
=x(x-5)(x+5)(x-2)

Now i see that root/zeroes are +5,-5 and 2. I know that this polynomial has another zero that is 0,but how do i know that? Because x is in the front? Or did i make a mistake and should such problems deal another way? Thank you for all replies!

You had a minor typo in your working which did not affect the outcome. I have highlighted it in red, but is is obvious from your subsequent steps that you intended to write this. The four roots of the polynomial are found by equating each of the four factors to zero, including the factor $x$ in front, and solving for $x$. So you are correct that the four roots are (in ascending order):

\(\displaystyle x=-5,\,0,\,2,\,5\)

Great job! :D

 

[theakdad]

You had a minor typo in your working which did not affect the outcome. I have highlighted it in red, but is is obvious from your subsequent steps that you intended to write this. The four roots of the polynomial are found by equating each of the four factors to zero, including the factor $x$ in front, and solving for $x$. So you are correct that the four roots are (in ascending order):

\(\displaystyle x=-5,\,0,\,2,\,5\)

Great job! :D

Thank you! Sorry,that was a type mismatch.
one question,when x stays alone in front of other terms,then this root is always zero?

 

[MarkFL]

Thank you! Sorry,that was a type mismatch.
one question,when x stays alone in front of other terms,then this root is always zero?

Yes, equating that factor to zero, we get:

\(\displaystyle x=0\)

It is already solved for $x$, and it tells us that $x=0$ is a root.

 

[Petrus]

Thank you! Sorry,that was a type mismatch.
one question,when x stays alone in front of other terms,then this root is always zero?

Another way to se it if we want to divide by x Then we get 2 case
case 1 \(\displaystyle x=0\)
put \(\displaystyle x=0\) to the equation and we see that \(\displaystyle 0=0\) hence 0 is a root
case 2 \(\displaystyle x \neq 0\)
we know can divide by x (we get third degree polynom) and now it's just to use rational root Theorem and long polynom division and Then it becomes second grade polynom which is easy to solve

Regards,
\(\displaystyle |\pi\rangle\)

 

[topsquark]

x⁴-2x³-25x²+50x

=x(x-5)(x+5)(x-2)

To be specific we have that if ab = 0 then either a = 0 or b = 0. In this case that means when
x(x - 5)(x + 5)(x - 2) = 0

we get solutions when:
x = 0, or x - 5 = 0, or x + 5 = 0, or x - 2 = 0

The solutions to these four equations are your roots.

-Dan