Question

Without actual division . find the reminder

$1) \frac{ \large \: x⁶ - 3x⁵ + 2x² - 7 }{ \large \: x - 3}$
$2) \frac{ \large \: x³ + 5x² + 9x + 10}{ \large \: x - 2}$
$3) \frac{ \large9x³ - 7x² + 3x-5}{\large3x - 1}$
$4) {\frac {x {}^{21} + 5x {}^{11} - 1 }{x + 1}}$
$5) \large \frac{4x³ - 12 x² + 11x }{x + \frac{3}{2} }$
$6) \frac{2x {}^{4} -3x {}^{3} - 5x {}^{2} + 8x - 3}{x - \frac{3}{2} }$
$7) \frac{5x³ + 9x² + 2x + 10}{ \frac{x}{2} + 1 }$
Class 9 ​

Answer 1

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

$\rm :\longmapsto\:1) \: \: \frac{ \large \: x⁶ - 3x⁵ + 2x² - 7 }{ \large \: x - 3}$

Let assume that

$\rm :\longmapsto\:p(x) \: = \: {x}^{6} - 3 {x}^{5} + 2 {x}^{2} - 7$

We Know

Remainder Theorem states that when a polynomial p (x) is divided by linear polynomial x - a, then remainder is p(a).

So,

Remainder is

$\rm :\longmapsto\:p(3) \: = \: {(3)}^{6} - 3 {(3)}^{5} + 2 {(3)}^{2} - 7$

$\rm :\longmapsto\:p(3) = 729 - 729 + 18 - 7$

$\rm :\longmapsto\:p(3) = 11$

$\rm :\longmapsto\:2) \: \: \frac{ \large \: {x}^{3} + 5 {x}^{2} + 9x + 10}{ \large \: x - 2}$

Let assume that

$\rm :\longmapsto\:p(x) = {x}^{3} + 5 {x}^{2} + 9x + 10$

So,

Using Remainder Theorem, we get

$\rm :\longmapsto\:p(2) = {2}^{3} + 5 {(2)}^{2} + 9(2) + 10$

$\rm :\longmapsto\:p(2) = 8 + 20 + 18 + 10$

$\rm :\longmapsto\:p(2) = 56$

$\rm :\longmapsto\:3) \: \frac{ \large9 {x}^{3} - 7 {x}^{2} + 3x-5}{\large3x - 1}$

Let assume that

$\rm :\longmapsto\:p(x) = {9x}^{3} - 7 {x}^{2} + 3x - 5$

So, using Remainder Theorem, we get

$\rm :\longmapsto\:p\bigg[\dfrac{1}{3} \bigg] = 9{\bigg[\dfrac{1}{3} \bigg]}^{3} - 7 {\bigg[\dfrac{1}{3} \bigg]}^{2} + 3\bigg[\dfrac{1}{3} \bigg] - 5$

$\rm :\longmapsto\:p\bigg[\dfrac{1}{3} \bigg] = \dfrac{1}{3} - \dfrac{7}{9} +1 - 5$

$\rm :\longmapsto\:p\bigg[\dfrac{1}{3} \bigg] = \dfrac{1}{3} - \dfrac{7}{9} - 4$

$\rm :\longmapsto\:p\bigg[\dfrac{1}{3} \bigg] = \dfrac{3 - 7 + 39}{9}$

$\rm :\longmapsto\:p\bigg[\dfrac{1}{3} \bigg] = \dfrac{35}{9}$

$\rm :\longmapsto\: \:4) \: {\dfrac {x {}^{21} + 5x {}^{11} - 1 }{x + 1}}$

Let assume that

$\rm :\longmapsto\:p(x) = {x {}^{21} + 5x {}^{11} - 1}$

So, Using Remainder Theorem, we get

$\rm :\longmapsto\:p( - 1) = {{( - 1)}^{21} + 5{( - 1)}^{11} - 1}$

$\rm :\longmapsto\:p( - 1) = - 1 - 5 - 1$

$\rm :\longmapsto\:p( - 1) = - 7$

$\rm :\longmapsto\:5) \: \: \large \dfrac{4 {x}^{3} - 12 {x}^{2} + 11x }{x + \frac{3}{2} }$

Let assume that,

$\rm :\longmapsto\:p(x) = 4 {x}^{3} - 12 {x}^{2} + 11x$

So, Using Remainder Theorem, we get

$\rm :\longmapsto\:p\bigg[ - \dfrac{3}{2} \bigg] = 4 {\bigg[ - \dfrac{3}{2} \bigg]}^{3} - 12 {\bigg[ - \dfrac{3}{2} \bigg]}^{2} + 11\bigg[ - \dfrac{3}{2} \bigg]$

$\rm :\longmapsto\:p\bigg[ - \dfrac{3}{2} \bigg] = - \dfrac{27}{2} - 27 - \dfrac{33}{2}$

$\rm :\longmapsto\:p\bigg[ - \dfrac{3}{2} \bigg] = \dfrac{ - 27 - 54 - 33}{2}$

$\rm :\longmapsto\:p\bigg[ - \dfrac{3}{2} \bigg] = \dfrac{ - 114}{2}$

$\rm :\longmapsto\:p\bigg[ - \dfrac{3}{2} \bigg] = - \: 57$

$\rm :\longmapsto\:6) \: \dfrac{2x {}^{4} -3x {}^{3} - 5x {}^{2} + 8x - 3}{x - \frac{3}{2} }$

Let assume that

$\rm :\longmapsto\:p(x) = 2x {}^{4} -3x {}^{3} - 5x {}^{2} + 8x - 3$

So, using Remainder Theorem, we have

$\rm :\longmapsto\:p\bigg[\dfrac{3}{2} \bigg] = 2{\bigg[\dfrac{3}{2} \bigg]}^{4} -3{\bigg[\dfrac{3}{2} \bigg]}^{3} - 5{\bigg[\dfrac{3}{2} \bigg]}^{2} + 8\bigg[\dfrac{3}{2} \bigg] - 3$

$\rm :\longmapsto\:p\bigg[\dfrac{3}{2} \bigg] = \dfrac{81}{8} - \dfrac{81}{8} - \dfrac{45}{4} + 12 - 3$

$\rm :\longmapsto\:p\bigg[\dfrac{3}{2} \bigg] = - \dfrac{45}{4} +9$

$\rm :\longmapsto\:p\bigg[\dfrac{3}{2} \bigg] = \dfrac{ - 45 + 36}{4}$

$\rm :\longmapsto\:p\bigg[\dfrac{3}{2} \bigg] = - \: \dfrac{9}{4}$

$\rm :\longmapsto\:7) \: \dfrac{5 {x}^{3} + 9 {x}^{2} + 2x + 10}{ \dfrac{x}{2} + 1 }$

Let assume that

$\rm :\longmapsto\:p(x) = 5 {x}^{3} + 9 {x}^{2} + 2x + 10$

So, by using Remainder Theorem, we get

$\rm :\longmapsto\:p( - 2) = 5 {( - 2)}^{3} + 9 {( - 2)}^{2} + 2( - 2) + 10$

$\rm :\longmapsto\:p( - 2) = - 40 + 36 - 4 + 10$

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