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.