Solve algebraically for all values of x:
- 4x^4 + 36x^3 - 80x2 = 0

Given:
The equation is
[tex]-4x^4+36x^3-80x^2=0[/tex]
To find:
The value of x.
Solution:
We have,
[tex]-4x^4+36x^3-80x^2=0[/tex]
It can be written as
[tex]-4x^2(x^2-9x+20)=0[/tex]
Splitting the middle term, we get
[tex]-4x^2(x^2-5x-4x+20)=0[/tex]
[tex]-4x^2(x(x-5)-4(x-5))=0[/tex]
[tex]-4x^2(x-5)(x-4)=0[/tex]
Using zero product property, we get
[tex]x^2=0,x-5=0,x-4=0[/tex]
[tex]x=0,x=5,x=4[/tex]
Therefore, the values of x are 0, 4 and 5.