Question

क्या x+1,p(x)=x^2-1 का एक गुणनखंड है?

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Answer 1

Answer:

$x + 1 = 0$

$x = - 1$

Putting the value of x in p(x) :

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

$\: \: {( - 1)}^{2} - 1$

$\: \: 1 - 1$

$\: \: 0$

As the result is coming zero so, x + 1 is a factor of p(x)

Ans. Yes, x + 1 is a factor of p(x)

Hope it helps.

Answer 2

$x + 1 = 0$

$x = \: \: - 1$

$put \: \: value \: of \: x \: in \: above$

$= \: \: {x}^{2} - 1$

$= \: ( - 1)^{2} - 1$

$= \: 1 - 1$

$= \: \: 0$