How do you factor #h(x)=x^3-3x^2-x+3#?

1 Answer
Sep 27, 2015

Answer:

Factor by grouping and using the difference of squares identity to find:

#x^3-3x^2-x+3 = (x-1)(x+1)(x-3)#

Explanation:

The difference of squares identity is: #a^2-b^2 = (a-b)(a+b)#

We use that with #a=x# and #b=1# after factoring by grouping:

#x^3-3x^2-x+3#

#=(x^3-3x^2)-(x-3)#

#=x^2(x-3)-1*(x-3)#

#=(x^2-1)(x-3)#

#=(x^2-1^2)(x-3)#

#=(x-1)(x+1)(x-3)#