Question

Verify in the following whether g(x) is a factor of p(x)

Answer 1

$\red{\large\underline{\sf{Solution-i}}}$

Given that,

$\rm \longmapsto\: p(x) = {x}^{4} - {x}^{3} - {x}^{2} - x - 2$

and

$\rm \longmapsto\:g(x) = x - 2$

We know,

Factor theorem states that if g(x) = x - a is a factor of polynomial f(x), then remainder f(a) = 0.

So, using Factor theorem, Consider

$\rm \longmapsto\:p(2)$

$\rm \: = \: {2}^{4} - {2}^{3} - {2}^{2} - 2 - 2$

$\rm \: = \: 16 - 8 - 4 - 4$

$\rm \: = \: 16 - (8 + 4 + 4)$

$\rm \: = \: 16 - 16$

$\rm \: = \: 0$

$\rm\implies \:p(2) \: = \: 0$

So,

$\bf\implies \:g(x) \: is \: factor \: of \: p(x)$

$\red{\large\underline{\sf{Solution-ii}}}$

$\rm \longmapsto\: p(x) = 2{x}^{3} + {x}^{2} - 2x + 1$

and

$\rm :\longmapsto\: g(x) = x + 1$

Now, By using Factor Theorem, Consider

$\rm \longmapsto\:p( - 1)$

$\rm \: = \: 2 {( - 1)}^{3} + {( - 1)}^{2} - 2( - 1) + 1$

$\rm \: = \: - 2 + 1 + 2 + 1$

$\rm \: = \: 2$

$\rm\implies \:p( - 1) \: \ne \: 0$

So,

$\bf\implies \:g(x) \: is \: not \: a\: factor \: of \: p(x)$