How do you factor x^3+4x^2-6x+1

Question

I like to factor the above expression. I already know one factor . I found it by divison; synthetic. But I would like to know how to factor it from the scratch? Can you help me?

Answer 1

Hi Mahnaz,

 

X^3 + 4X^2 -6X + 1

 

=X^3 -X^2 +5X^2 - 5X- X +1

 

=X^2 ( X-1) +5X (X-1) -1(X-1)

 

=(X-1) (X^2 +5X-1)

 

using Quadratic eqn (aX^2 + bX +c=0)formula, roots of Quadratic eq are

X =-b +sq root of b^2-4ac/2a    or X= -b - sq root of b^2-4ac/2a

 

For X^2 +5X -1

 

a=1, b=5, c=-1

 

X= -5 + sq root of (25+4)/2    or X= -5 - sq root of (25+4)/2

   = -5 + sq root of 29/2 or  X= -5 - sq root of 29/2

so factors of X^2 +5X -1 are (X - (-5+sq root of 29/2))(X - (-5-sq root of 29/2))

    

                                        = (X +5/2-sq root of 29/2) ( X +5/2 +sq root of 29/2)                                                           

 

X^3+4X^2-6X+1=(X-1)(X+5/2-sq root of 29/2)(X+5/2+sq root of 29/2)

Answer 2

Seeing the polynomial

 

+1 , -1 are the possible factors

 

For x=1

 

The polynomial =1+4-6+1=0

 

That means x= 1 is a factor.

 

Using synthetic division

 

The polynomial=

 

(x-1)(x^2+5x-1)