F(x)=x^3-4x^2+6x-4  how do I use the complex zeros to write f in factored form ?

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gsarora17 | (Level 2) Associate Educator

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f(x)= `x^3-4x^2+6x-4`

Now in order to find a factor of f(x) , the constant term in f(x) is -4. So the factors of -4 are `+-` 1,`+-` 2 and `+-4`

Then we have to find out the value of x for which f(x)=0

f(2) = `2^3-4*2^2+6*2-4`

` `

f(2) = 0 Therefore (x-2) is a factor of the f(x)

Now divide f(x) by (x-2)

`(x^3-4x^2+6x-4)/(x-2) = (x^2-2x+2)`

`:.` f(x)= `(x-2)(x^2-2x+2)`

In order to find the other zeros we have to set f(x)=0

`:. (x^2-2x+2) = 0`

This is a quadratic equation of the form `ax^2+bx+c=0`

`:.`  x= `((-b+-sqrt(b^2-4ac)))/(2*a)`

substituting the values of a=1 , b=-2 , c=2 

  x = `((2+-sqrt((-2)^2-4*1*2)))/(2*1)`

`x= (2+-sqrt(4-8))/2`

`x= (2+-sqrt(4i^2))/2`

`x= 1+-i`

 Fully factored form of f(x) is

f(x) = (x-2)(x-1-i)(x-1+i)

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