Question

Refer the attached attachment
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Answer 1

Step-by-step explanation:

Tip:

Remainder Theorem:If a polynomial p(x) is divided by a linear polynomial g(x) whose zero is x = a, the remainder is given by r = p(a).

Given:

$i) \bold{ p(x) = {x}^{3} - 2 {x}^{2} - 4x - 1 }\\ \bold{g(x) = x + 1 }$

Solution: To find remainder put value of x from g(x) into p(x)

$x + 1 = 0 \\ \\ x = - 1 \\ \\ p( - 1) = { (- 1)}^{3} - 2 {( - 1)}^{2} - 4( - 1) - 1 \\ \\ = - 1 - 2 + 4 - 1 \\ \\ = - 4 + 4 \\ \\ \bold{p( - 1)= 0}$

Thus,

Remainder is zero.

$ii) \:\bold{ p(x) = {x}^{3} - 3{x}^{2} + 4x + 50} \\ \bold{g(x) = x - 3}$

$g(x) = 0 \\ \\ x - 3 = 0 \\ \\ x = 3 \\ \\ p(3) = {(3)}^{3} - 3{(3)}^{2} + 4(3) + 50 \\ \\ = 27 - 27 + 12 + 50 \\ \\ \bold{p(3)= 62}$

Remainder is 62.

$iii) \: \bold{p(x) = 4 {x}^{3} -12 {x}^{2} + 14x - 3} \\ \bold{g(x) =2 x - 1}$

Put g(x)=0 and from that put value of x in p(x)

$2x - 1 = 0 \\ \\ 2x = 1 \\ \\ x = \frac{1}{2}$

$\: p( \frac{1}{2} ) = 4 {( \frac{1}{2} )}^{3} -12 {( \frac{1}{2} )}^{2} + 14( \frac{1}{2} ) - 3 \\ \\ = \frac{1}{2} - 3 + 7 - 3 \\ \\ = \frac{1}{2} + 1 \\ \\ \bold{ p( \frac{1}{2} ) = \frac{3}{2}}$

Remainder is 3/2.

$iv) \: \bold{p(x) = {x}^{3} -6 {x}^{2} + 2x - 4 }\\ \bold{g(x) =1 - \frac{3}{2} x}$

$1 - \frac{3}{2} x = 0 \\ \\ \frac{3}{2}x = 1 \\ \\ x = \frac{2}{3}$

$p( \frac{2}{3} ) = {( \frac{2}{3})}^{3} -6 {( \frac{2}{3} )}^{2} + 2( \frac{2}{3} )- 4 \\ \\ = \frac{8}{27} - 6 \times \frac{4}{9} + \frac{4}{3} - 4 \\ \\ = \frac{8}{27} - \frac{8}{3} + \frac{4}{3} - 4 \\ \\ = \frac{8 - 72 + 36 -10 8}{27} \\ \\ = \frac{44 -1 80}{27} \\ \\ \bold{p( \frac{2}{3} )=\frac{-136}{27} }$

Remainder is -136/27

Hope it helps you.

Answer 2

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

Remainder Theorem :- This theorem states that if a polynomial f (x) is divided by linear polynomial g(x) = x - a, then remainder is f(a).

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

$\rm :\longmapsto\:p(x) = {x}^{3} - {2x}^{2} - 4x - 1$

and

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

So, when p(x) is divided by g(x), the remainder is

$\rm :\longmapsto\:p( - 1) = {( - 1)}^{3} - {2( - 1)}^{2} - 4( - 1) - 1$

$\rm :\longmapsto\:p( - 1) = - 1 - 2 + 4 - 1$

$\rm :\longmapsto\:p( - 1) = - 4 + 4$

$\rm \implies\:\boxed{ \tt{ \: \:p( - 1) = 0 \: }}$

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

$\rm :\longmapsto\:p(x) = {x}^{3} - {3x}^{2} + 4x + 50$

and

$\rm :\longmapsto\:g(x) = x - 3$

So, when p(x) is divided by g(x), the remainder is

$\rm :\longmapsto\:p(3) = {3}^{3} - {3(3)}^{2} + 4(3) + 50$

$\rm :\longmapsto\:p(3) = 27 - 27 + 12 + 50$

$\rm \implies\:\boxed{ \tt{ \: \:p(3) = 62 \: }}$

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

$\rm :\longmapsto\:p(x) = {4x}^{3} - {12x}^{2} + 14x - 3$

and

$\rm :\longmapsto\:g(x) = 2x - 1$

So, when p(x) is divided by g(x), the remainder is

$\rm :\longmapsto\:p\bigg[\dfrac{1}{2} \bigg] = 4 {\bigg[\dfrac{1}{2} \bigg]}^{3} - 12{\bigg[\dfrac{1}{2} \bigg]}^{2} + 14\bigg[\dfrac{1}{2} \bigg] - 3$

$\rm :\longmapsto\:p\bigg[\dfrac{1}{2} \bigg] = 4 {\bigg[\dfrac{1}{2} \bigg]}^{3} - 12{\bigg[\dfrac{1}{2} \bigg]}^{2} + 14\bigg[\dfrac{1}{2} \bigg] - 3$

$\rm :\longmapsto\:p\bigg[\dfrac{1}{2} \bigg] = 4 {\bigg[\dfrac{1}{8} \bigg]} - 12{\bigg[\dfrac{1}{4} \bigg]} + 7 - 3$

$\rm :\longmapsto\:p\bigg[\dfrac{1}{2} \bigg] = {\bigg[\dfrac{1}{2} \bigg]} - 3 + 7 - 3$

$\rm :\longmapsto\:p\bigg[\dfrac{1}{2} \bigg] = \dfrac{1}{2} + 1$

$\rm :\longmapsto\:p\bigg[\dfrac{1}{2} \bigg] = \dfrac{1 + 2}{2}$

$\rm \implies\:\boxed{ \tt{ \: p\bigg[\dfrac{1}{2} \bigg] = \dfrac{3}{2} \: }}$

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

$\rm :\longmapsto\:p(x) = {x}^{3} - {6x}^{2} + 2x - 4$

and

$\rm :\longmapsto\:g(x) = 1 - \dfrac{3}{2}x$

So, when p(x) is divided by g(x), the remainder is

$\rm :\longmapsto\:p\bigg[\dfrac{2}{3} \bigg] = {\bigg[\dfrac{2}{3} \bigg]}^{3} - {6\bigg[\dfrac{2}{3} \bigg]}^{2} + 2\bigg[\dfrac{2}{3} \bigg] - 4$

$\rm :\longmapsto\:p\bigg[\dfrac{2}{3} \bigg] = \dfrac{8}{27} - 6 \times \dfrac{4}{9} + \dfrac{4}{3} - 4$

$\rm :\longmapsto\:p\bigg[\dfrac{2}{3} \bigg] = \dfrac{8}{27} - \dfrac{8}{3} + \dfrac{4}{3} - 4$

$\rm :\longmapsto\:p\bigg[\dfrac{2}{3} \bigg] = \dfrac{8}{27} - \dfrac{4}{3} - 4$

$\rm :\longmapsto\:p\bigg[\dfrac{2}{3} \bigg] = \dfrac{8 - 36 - 108}{27}$

$\rm \implies\:\boxed{ \tt{ \: p\bigg[\dfrac{2}{3} \bigg] = \dfrac{ - 136}{27} \: }}$

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More 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)