How do you factor $g (x) = x^{3} - x^{2} - x + 1$ $g(x)=x^3-x^2-x+1$ ?

Question

1247 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-16T01:39:39

$g (x) = (x + 1) \cdot {(x - 1)}^{2}$ $color(green)(g(x) = (x + 1) * (x- 1)^2$

$g (x) = x^{3} - x^{2} - x + 1$ $g(x) = x^3 - x^2 - x + 1$

$g (x) = x^{2} (x - 1) - 1 (x - 1)$ $g(x) = x^2 (x - 1) - 1 (x-1)$ Taking common $x^{2}$ $x^2$ out.

$\Rightarrow (x^{2} - 1) \cdot (x - 1)$ $=> (x^2 - 1) * (x - 1)$ Separating common terms.

As per identity, $(x + 1) \cdot (x - 1) = x^{2} - 1$ $(x + 1) * (x - 1) = x^2 - 1$

Hence $g (x) \Rightarrow (x + 1) (x - 1) (x - 1)$ $g(x) => (x + 1) (x - 1) (x - 1)$

$g (x) = (x + 1) \cdot {(x - 1)}^{2}$ $color(green)(g(x) = (x + 1) * (x- 1)^2$

How do you factor g(x)=x3−x2−x+1g(x)=x^3-x^2-x+1?