given $f(x)=x^{3}-2x^{2}-x + 2$, the roots of $f(x)$ are

Answer:

$x = 1, - 1,2$

Explanation:

Step1: Factor by grouping

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

Step2: Factor out common factors

$x^{2}(x - 2)-(x - 2)=(x - 2)(x^{2}-1)$

Step3: Factor the difference - of - squares

Since $x^{2}-1=(x + 1)(x - 1)$ (using the formula $a^{2}-b^{2}=(a + b)(a - b)$ with $a=x$ and $b = 1$), then $f(x)=(x - 2)(x + 1)(x - 1)$

Step4: Find the roots

Set $f(x)=0$, so $(x - 2)(x + 1)(x - 1)=0$. By the zero - product property, if $ab = 0$, then $a = 0$ or $b = 0$. So $x-2=0$ gives $x = 2$, $x + 1=0$ gives $x=-1$, and $x - 1=0$ gives $x = 1$.