(x + 1) is a factor of the polynomial

Solution:

(c) $x^{3}-2 x^{2}-x-2$

Let:

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

By the factor theorem, (x + 1) will be a factor of f (x) if f (−1) = 0.
We have:

$f(-1)=(-1)^{3}-2 \times(-1)^{2}+(-1)+2$

$=-1-2-1+2$

$=-2 \neq 0$

Hence, $(x+1)$ is not a factor of $f(x)=x^{3}-2 x^{2}+x+2$.

Now,
Let:

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

By the factor theorem, (x + 1) will be a factor of f (x) if f (−-1) = 0.
We have:

$f(-1)=(-1)^{3}+2 \times(-1)^{2}+(-1)-2$

$=-1+2-1-2$

$=-2 \neq 0$

Hence, $(x+1)$ is not a factor of $f(x)=x^{3}+2 x^{2}+x-2$.

Now,
Let:

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

By the factor theorem, (x + 1) will be a factor of f (x) if f (−-1) = 0.
We have:

$f(-1)=(-1)^{3}+2 \times(-1)^{2}-(-1)-2$

$=-1+2+1-2$

$=0$

Hence, $(x+1)$ is a factor of $f(x)=x^{3}+2 x^{2}-x-2$.