Question

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

Answer 1

Given:

The equation is

`-4x^4+36x^3-80x^2=0`

To find:

The value of x.

Solution:

We have,

`-4x^4+36x^3-80x^2=0`

It can be written as

`-4x^2(x^2-9x+20)=0`

Splitting the middle term, we get

`-4x^2(x^2-5x-4x+20)=0`

`-4x^2(x(x-5)-4(x-5))=0`

`-4x^2(x-5)(x-4)=0`

Using zero product property, we get

`x^2=0,x-5=0,x-4=0`

`x=0,x=5,x=4`

Therefore, the values of x are 0, 4 and 5.