(x + 1) is a factor of the polynomial
(a) $x^{3}-2 x^{2}+x+2$
(b) $x^{3}+2 x^{2}+x-2$
(c) $x^{3}-2 x^{2}-x-2$
(d) $x^{3}-2 x^{2}-x+2$
(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 $f(x)=x^{3}+2 x^{2}-x-2$.