# What is the solution of x^3-4x^2+x-4 = 0

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### 2 Answers

The solution of x^3-4x^2+x-4 = 0 has to be determined.

x^3-4x^2+x-4 = 0

=> x^2(x - 4) + 1(x - 4) = 0

=> (x^2 + 1)(x - 4) = 0

x^2 + 1 = 0

=> x^2 = -1

=> x = `+-sqrt(-1)`

=> x = `+-i`

x - 4 = 0

=> x = 4

**The solution of the equation x^3-4x^2+x-4 = 0 is {-i, i, 4}**

QUESTION:-

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

SOLUTION:-

We can solve this equation by factorization;

`x^2(x-4)+1(x-4)=0`

Take x^2 and 1 common, so that the bracket values are same,

Then separate the two solutions;

` `

`(x^2+1)(x-4)=0`

` <br data-mce-bogus="1"> `

Hence the solutions are;

`x^2=-1,x=4`

The power of one side become the root on other side;

`x=sqrt(-1),x=4`

The square root of -1 is equal to (+-i);

Hence the solution set is {i,-i,4}

Hence Solved!