Question

plz answer it fast ​

Answer 1

Hey mate here is the solution

Answer 2

$\huge \bold{Given}$

POLYNOMIAL :-

$\boxed{p(x) =2 {x}^{3} + {x}^{2} - 2x - 1}$

To check whether (x + 1 = 0) is a factor of p(x)

$\mathsf{x = -1}$

On placing the value of x if the equation leaves remainder 0 it means x+1 is a factor.

$\mathsf{p(0) =2 {x}^{3} + {x}^{2} - 2x - 1}$

$\mathsf{p(0) =2 {(-1)}^{3} + {(-1)}^{2} - (-1 \times 2)- 1}$

$\mathsf{p(0) = - 2 + 1 + 2 - 1}$

$\mathsf{p(0) = - 1 + 1}$

$\mathsf{p(0) = 0}$

Since, on placing x as (-1) gives remainder 0 .

Therefore,

(x+1 ) is a factor of $\mathsf{2 {x}^{3} + {x}^{2} - 2x - 1}$