How do you factor h(x)=x^3-3x^2-x+3h(x)=x3−3x2−x+3?

1 Answer
George C.
Sep 27, 2015

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

x^3-3x^2-x+3 = (x-1)(x+1)(x-3)x3−3x2−x+3=(x−1)(x+1)(x−3)

Explanation:

The difference of squares identity is: a^2-b^2 = (a-b)(a+b)a2−b2=(a−b)(a+b)

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

x^3-3x^2-x+3x3−3x2−x+3

=(x^3-3x^2)-(x-3)=(x3−3x2)−(x−3)

=x^2(x-3)-1*(x-3)=x2(x−3)−1⋅(x−3)

=(x^2-1)(x-3)=(x2−1)(x−3)

=(x^2-1^2)(x-3)=(x2−12)(x−3)

=(x-1)(x+1)(x-3)=(x−1)(x+1)(x−3)

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