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

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Answer 1 · 2015-12-19T14:09:12

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

$x^3-3x^2-x+3 =0 =(x-1)(x+1)(x-3)$

The difference of squares identity may be written:

$a^2-b^2 = (a-b)(a+b)$

After factoring by grouping use this identity with $a=x$ and $b=1$ ...

$x^3-3x^2-x+3$

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

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

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

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

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

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