Question

$\color{violet}\huge{qu}$$\color{red}\huge{es}$$\color{blue}\huge{ti}$$\color{orange}\huge{on}$
$answer \: neeed \: of \: both \: with \: \\ step \: by \: step$
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Answer 1

Given Question

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

$\sf \:(i) \: \: g(x) = x - 2, \: p(x) = {x}^{4} - {x}^{3} - {x}^{2} - x - 2$

$\sf \:(ii) \: \: g(x) = x + 1, \: p(x) = 2{x}^{3} + {x}^{2} - 2x + 1$

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

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,

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

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

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

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,

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

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More Identities to know :-

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)

Answer 2

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