Question

What is the completely factored form of f(x)=x3−7x2+2x+4?

f(x)=(x−1)(x−(3+13−−√))(x−(3−13−−√))
f(x)=(x+1)(x−(3+13−−√))(x−(3−13−−√))
f(x)=(x+1)(x−(3+52−−√))(x−(3−52−−√))
f(x)=(x−1)(x−(3+52−−√))(x−(3−52−−√))

Answer 1

Answer:

Factored form is

f(x)= (x-1)(x-3+√(-5))(x-3-√(-5))

Step-by-step explanation:

Given that  

f(x)=x3−7x2+2x+4

To solve for x  we factorise the right side

f(x)=x3−7x2+2x+4

Let us substitute x for various values to check whether remainder is zero.

i.e. f(1) = 1-7+2+4  =0

     

Hence x-1 is a factor

Do synthetic division to find the quotient

1       1    -7    2    4

              1     -6   -4

       ----------------------

        1    -6    -4   0

i.e. we get remainder 0 and quotient as

x^2-6x-4

Use completion of squares method to solve this

x^2-6x-4

= (x-3)^2+5=0

x-3= ±√(-5)

Or x=1,3+√(-5,)  3-√(-5,)

Are the roots

Factored form is

f(x)= (x-1)(x-3+√(-5))(x-3-√(-5))

Answer 2

Answer:

f(x)=(x-1)(x-(3+√13))(x-(3-√13))