[tex] \red{\large\underline{\sf{Solution-i}}}[/tex]
Given that,
[tex] \rm \longmapsto\: p(x) = {x}^{4} - {x}^{3} - {x}^{2} - x - 2[/tex]
and
[tex]\rm \longmapsto\:g(x) = x - 2[/tex]
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
[tex]\rm \longmapsto\:p(2)[/tex]
[tex]\rm \: = \: {2}^{4} - {2}^{3} - {2}^{2} - 2 - 2[/tex]
[tex]\rm \: = \: 16 - 8 - 4 - 4[/tex]
[tex]\rm \: = \: 16 - (8 + 4 + 4)[/tex]
[tex]\rm \: = \: 16 - 16[/tex]
[tex]\rm \: = \: 0[/tex]
[tex]\rm\implies \:p(2) \: = \: 0[/tex]
So,
[tex]\bf\implies \:g(x) \: is \: factor \: of \: p(x)[/tex]
[tex] \red{\large\underline{\sf{Solution-ii}}}[/tex]
[tex]\rm \longmapsto\: p(x) = 2{x}^{3} + {x}^{2} - 2x + 1[/tex]
and
[tex]\rm :\longmapsto\: g(x) = x + 1[/tex]
Now, By using Factor Theorem, Consider
[tex]\rm \longmapsto\:p( - 1)[/tex]
[tex]\rm \: = \: 2 {( - 1)}^{3} + {( - 1)}^{2} - 2( - 1) + 1[/tex]
[tex]\rm \: = \: - 2 + 1 + 2 + 1[/tex]
[tex]\rm \: = \: 2[/tex]
[tex]\rm\implies \:p( - 1) \: \ne \: 0[/tex]
So,
[tex]\bf\implies \:g(x) \: is \: not \: a\: factor \: of \: p(x)[/tex]
