Question

(x + 1) is a factor of the polynomial
(a) x³ − 2x² + x + 2
(b) x³ + 2x² + x − 2
(c) x³ − 2x² − x − 2
(d) x³ − 2x² − x + 2

Answer 1

(c) x³ − 2x² − x − 2

Let:
$f\left(x\right)=x^3-2x^2+x+2$
By the factor theorem, (x + 1) will be a factor of f (x) if f (−1) = 0.
We have:
$$\begin{gathered} f\left(-1\right)={\left(-1\right)}^3-2\times {\left(-1\right)}^2+\left(-1\right)+2 \\ =-1-2-1+2 \\ =-2\ne 0 \end{gathered}$$
Hence, (x + 1) is not a factor of $f\left(x\right)=x^3-2x^2+x+2$.

Now,
Let:
$f\left(x\right)=x^3+2x^2+x-2$
By the factor theorem, (x + 1) will be a factor of f (x) if f ($-$1) = 0.
We have:
$$\begin{gathered} f\left(-1\right)={\left(-1\right)}^3+2\times {\left(-1\right)}^2+\left(-1\right)-2 \\ =-1+2-1-2 \\ =-2\ne 0 \end{gathered}$$
Hence, (x + 1) is not a factor of $f\left(x\right)=x^3+2x^2+x-2$.

Now,
Let:
$f\left(x\right)=x^3+2x^2-x-2$
By the factor theorem, (x + 1) will be a factor of f (x) if f ($-$1) = 0.
We have:
$$\begin{gathered} f\left(-1\right)={\left(-1\right)}^3+2\times {\left(-1\right)}^2-\left(-1\right)-2 \\ =-1+2+1-2 \\ =0 \end{gathered}$$
Hence, (x + 1) is a factor of $f\left(x\right)=x^3+2x^2-x-2$.