Polynomials – Exercise 8.2 – Class X

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Polynomials – Exercise 8.2

  1. Divide p(x) by g(x) in each o the following cases and verify division algorithm

(i)p(x) = x2 + 4x + 4, g(x) = x + 2

(ii)p(x) = x2 – 9x + 9, g(x) = x – 3

(iii) p(x) = x3 + 4x2 -5x + 6, g(x) = x + 1

(iv) p(x) = x4 – 3x2 – 4, g(x) = x + 2

(v) p(x) = x3 – 1, g(x) = x – 1

(vi) p(x) = x4 – 4x2+ 12x + 9, g(x) = x2 + 2x – 3

  1. Find the divisor g(x) , when the polynomial p(x) = 4x3 + 2x2 – 10x +2 is divided by g(x) and the quotient and the remainder obtained are (2x2 +4x + 1) and 5 respectively.
  2. On dividng the polynomial p(x) = x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were (x – 2) and (-2x + 4) respectively. Find g(x)
  3. A polynomial p(x) id divided by g(x), the obtained quotient q(x) and the remainder are given in the table. Find the p(x) in each case.
Slp(x)g(x)q(x)r(x)
i?x – 2x2 – x + 14
ii?x + 32x2 + x + 53x +1
iii?2x + 1x3 + 3x2 – x +10
iv?x – 1x3 – x2 – x – 12x – 4
v?x2 + 2x + 1x4 – 2x2 + 5x – 74x + 12
  1. Find the quotient and remainder on dividing p(x) by g(x) in each of the fooling cases, without actual division

(i) p(x) = x2 + 7x + 10; g(x) = x – 2

(ii) p(x) = x3 +4x2 – 6x + 2; g(x) = x – 3

  1. What must be subtracted from (x3 + 5x2 + 5x + 8) so that the resulting polynomial exactly divisible by (x2 + 3x – 2)?
  2. What should be added to (x4 – 1) so that it is exactly divisible by (x2 + 2x + 1)?

Polynomials – Exercise 8.2 – Solutions:

  1. Divide p(x) by g(x) in each o the following cases and verify division algorithm

(i)p(x) = x2 + 4x + 4, g(x) = x + 2

Solution:

Given, we have to divide p(x) by g(x) i.e., we have to divide x2 + 4x + 4 by x + 2.

Polynomials – Exercise 8.2 – Class X

Verifying:

Quotient, q(x) = (x + 2)

Remainder, r(x) = 0

By division algorithm for polynomials, p(x) = [g(x) * q(x)] + r(x)

Here, p(x) = x2 + 4x + 4 ;

g(x) = x + 2 ;

q(x) = x + 2 ;

r(x) = 0

⸫ p(x) = [(x + 2) * (x + 2)] + 0

= [x2 + 2x + 2x + 4] + 0

= x2 + 4x + 4 = p(x)

⸫ division algorithm is verified.


(ii)p(x) = x2 – 9x + 9, g(x) = x – 3

Solution:

Given, we have to divide p(x) by g(x) i.e., we have to divide x2 – 9x + 9 by x – 3 .

Polynomials – Exercise 8.2 – Class X

Verifying:

Quotient, q(x) = (x – 6)

Remainder, r(x) = 27

By division algorithm for polynomials, p(x) = [g(x) * q(x)] + r(x)

Here, p(x) = x2 – 9x + 9 ;

g(x) = x – 3 ;

q(x) = x – 6 ;

r(x) = 27

⸫ p(x) = [(x – 3) * (x – 6)] + 27

= [x2 – 6x – 3x – 18] + 27

= x2 – 9x – 18 + 27

=x2 – 9x + 9 = p(x)

⸫ division algorithm is verified.


(iii) p(x) = x3 + 4x2 -5x + 6, g(x) = x + 1

Solution:

Given, we have to divide p(x) by g(x) i.e., we have to divide x3 + 4x2 -5x + 6 by x + 1.

Polynomials – Exercise 8.2 – Class X

Verifying:

Quotient, q(x) = (x2 + 2x – 7)

Remainder, r(x) = 13

By division algorithm for polynomials, p(x) = [g(x) * q(x)] + r(x)

Here, p(x) = x3 + 4x2 – 5x + 6 ;

g(x) = x + 1 ;

q(x) = x2 + 3x – 8  ;

r(x) = 14

⸫ p(x) = [(x2 + 3x – 8) * (x + 1)] + 14

= x3 + 3x2 – 8x + x2 + 3x – 8 + 14

= x3 + 4x2 – 5x + 6

= p(x)

⸫ division algorithm is verified.


(iv) p(x) = x4 – 3x2 – 4, g(x) = x + 2

Solution:

Given, we have to divide p(x) by g(x) i.e., we have to divide x4 – 3x2 – 4 by x + 2.

Polynomials – Exercise 8.2 – Class X

Verifying:

Quotient, q(x) = (x3 – 2x2 + x – 2)

Remainder, r(x) = 0

By division algorithm for polynomials, p(x) = [g(x) * q(x)] + r(x)

Here, p(x) = x4 – 3x2 – 4;

g(x) = x + 2 ;

q(x) = x3 – 2x2 + x – 2;

r(x) = 0

⸫ p(x) = [(x + 2) * (x3 – 2x2 + x – 2)] + 0

= [x4 – 2x3 + x2 – 2x + 2x3 – 4x2 + 2x – 4  ] + 0

= x4 + 0 – 3x2 + 0 – 4

x4 – 3x2 – 4  = p(x)

⸫ division algorithm is verified.


(v) p(x) = x3 – 1, g(x) = x – 1

Solution:

Given, we have to divide p(x) by g(x) i.e., we have to divide x3 – 1 by x – 1.

Polynomials – Exercise 8.2 – Class X

Verifying:

Quotient, q(x) = (x2 + x + 1)

Remainder, r(x) = 0

By division algorithm for polynomials, p(x) = [g(x) * q(x)] + r(x)

Here, p(x) = x3 – 1 ;

g(x) = x – 1  ;

q(x) = x2 + x + 1;

r(x) = 0

⸫ p(x) = [(x – 1) * (x2 + x + 1)] + 0

= [x3 + x2 + x – x2 – x – 1  ] + 0

= x3 + 0 – 0 – 1

= x3 – 1  = p(x)

⸫ division algorithm is verified.


(vi) p(x) = x4 – 4x2+ 12x + 9, g(x) = x2 + 2x – 3

Solution:

Given, we have to divide p(x) by g(x) i.e., we have to divide x4 – 4x2+ 12x + 9  by x2 + 2x – 3.

Polynomials – Exercise 8.2 – Class X

Verifying:

Quotient, q(x) = (x2 – 2x + 3)

Remainder, r(x) = 0

By division algorithm for polynomials, p(x) = [g(x) * q(x)] + r(x)

Here, p(x) = x4 – 4x2+ 12x + 9;

g(x) = x2 + 2x – 3;

q(x) = x2 – 2x + 3;

r(x) = 18

⸫ p(x) = [(x2 – 2x + 3) * (x2 + 2x – 3)] + 18

= [x4 +2x3 – 3x2 – 2x3  – 4x2 + 6x + 3x2 + 6x – 9 ] + 18

= x4 + 0 – 4x2 + 12x + 9

= x4 – 4x2 + 12x + 9

= p(x)

⸫ division algorithm is verified.


  1. Find the divisor g(x) , when the polynomial p(x) = 4x3 + 2x2 – 10x +2 is divided by g(x) and the quotient and the remainder obtained are (2x2 +4x + 1) and 5 respectively.

Solution:

By division algorithm for polynomials, p(x) = [g(x) * q(x)] + r(x)

p(x) = 4x3 + 2x2 – 10x +2

q(x) = 2x2 +4x + 1

r(x) = 5

We have to find g(x),

p(x) = [g(x) * q(x)] + r(x)

g(x) = p(x) – r(x)/q(x)

= (4x^3 + 2x^2 – 10x +2) – (5)/2x^2 +4x + 1

= 4x^3 + 2x^2 – 10x +2 – 5/2x^2 +4x + 1

= 4x^3 + 2x^2 – 10x – 3/2x^2 +4x + 1

Polynomials – Exercise 8.2 – Class X

g(x) = 2x – 2


  1. On dividing the polynomial p(x) = x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were (x – 2) and (-2x + 4) respectively. Find g(x)

Solution:

By division algorithm for polynomials, p(x) = [g(x) * q(x)] + r(x)

p(x) = x3 – 3x2 + x + 2

q(x) = (x – 2)

r(x) = (-2x + 4)

We have to find g(x),

p(x) = [g(x) * q(x)] + r(x)

g(x) = p(x) – r(x)/q(x)

= (x^3 – 3x^2 + x + 2)-(-2x + 4)/x – 2

= x^3 – 3x^2 + x + 2 +2x – 4/x – 2

= x^3 – 3x^2 + 3x – 2/x – 2

Polynomials – Exercise 8.2 – Class X

g(x) = x2  – x + 1


  1. A polynomial p(x) id divided by g(x), the obtained quotient q(x) and the remainder are given in the table. Find the p(x) in each case.
Slp(x)g(x)q(x)r(x)
i?x – 2x2 – x + 14
ii?x + 32x2 + x + 53x +1
iii?2x + 1x3 + 3x2 – x +10
iv?x – 1x3 – x2 – x – 12x – 4
v?x2 + 2x + 1x4 – 2x2 + 5x – 74x + 12

Solution:

(i) By division algorithm for polynomials, p(x) = [g(x) * q(x)] + r(x)

g(x) = x – 2 ;

q(x) = x2 – x + 1 ;

r(x) = 4 ;

p(x) = [(x – 2)*(x2 – x + 1)] + 4

= x3 – x2 + x – 2x2 + 2x – 2 + 4

= x3 – 3x2 + 3x + 2

(ii) By division algorithm for polynomials, p(x) = [g(x) * q(x)] + r(x)

g(x) = x + 3

q(x) = 2x2 + x + 5

r(x) = 3x + 1

p(x) = [(x + 3)*(2x2 + x + 5)] + (3x + 1)

= 2x3 + x2 + 5x + 6x2 + 3x + 15 + 3x + 1

= 2x3 + 7x2 + 11x + 16

(iii) By division algorithm for polynomials, p(x) = [g(x) * q(x)] + r(x)

g(x) = 2x + 1

q(x) = x3 + 3x2 – x +1

r(x) = 0

p(x) = (2x + 1)(x3 + 3x2 – x + 1) + 0

=  2x4 + 6x3 – 2x2 + 2x + x3 + 3x2 – x + 1

= 2x4 + 7x3 + x2 + x + 1

(iv) By division algorithm for polynomials, p(x) = [g(x) * q(x)] + r(x)

g(x) = x – 1

q(x) = x3 – x2 – x – 1

r(x) = 2x – 4

p(x) = (x3 – x2 – x – 1)*(x – 1) + (2x – 4)

= x4 – x3 – x2 – x – x3 + x2 + x + 1 + 2x – 4

= x4 – 2x3 + 2x – 3

(v) By division algorithm for polynomials, p(x) = [g(x) * q(x)] + r(x)

g(x) = x2 + 2x + 1

q(x) = x4 – 2x2 + 5x – 7

r(x) = 4x + 12

p(x) = (x2 + 2x + 1)( x4 – 2x2 + 5x – 7) + 4x + 12

= x6 – 2x4 + 5x3 – 7x2 +2x5 – 4x3 + 10x2 – 14x + x4 – 2x2 + 5x – 7 + 4x + 12

p(x) = x6 + 2x5 – x4 + x3 + x2 – 5x + 5


  1. Find the quotient and remainder on dividing p(x) by g(x) in each of the following cases, without actual division

(i) p(x) = x2 + 7x + 10; g(x) = x – 2

Solution:

p(x) = x2 + 7x + 10

⸫degree of p(x) = 2

g(x) = x – 2

⸫degree of g(x) = 1

⸫degree of quotient q(x) = 2 – 1 = 1 and degree of remainder r(x) is 0.

Let q(x) = ax + b (polynomial of degree 1) and remainder, r(x) = k

By division algorithm for polynomials, p(x) = [g(x) * q(x)] + r(x)

x2 + 7x + 10 = (x – 2)*(ax + b) + k

x2 + 7x + 10 = ax2 + bx – 2ax – 2b + k

x2 + 7x + 10 = ax2 + x (b – 2a) – 2b + k

Let us compare the coefficients of x2, x and k to get the values of a, b,  and k

⸫ a = 1  ; coefficients of x2 on both the sides

⸫ b – 2a = 7 ; coefficients of x on both the sides

⸫ 10 = -2b + k  ; constants on both the sides

We have to solve these equations to get the value of a, b and k

Since a = 1

b – 2a = 7

⇒ b = 7 + 2a = 7 + 2(1) = 9

10 = -2b + k

k = 10 + 2b = 10 + 9×2 = 10 + 18 = 28

Since q(x) = ax + b = x + 9

r(x) = 28

Therefore, quotient = x + 9 and remainder 28.


(ii) p(x) = x3 +4x2 – 6x + 2; g(x) = x – 3

Solution:

p(x) = x3 +4x2 – 6x + 2

⸫degree of p(x) = 3

g(x) = x –3

⸫degree of g(x) = 1

⸫degree of quotient q(x) = 3 – 1 = 2 and degree of remainder r(x) is 0.

Let q(x) = ax2 + bx + c (polynomial of degree 1) and remainder, r(x) = k

By division algorithm for polynomials, p(x) = [g(x) * q(x)] + r(x)

x3 +4x2 – 6x + 2= (x – 3)*(ax2 + bx + c) + k

x3 +4x2 – 6x + 2 = ax3 + bx2 + cx – 3ax2 – 3bx – 3c + k

x3 +4x2 – 6x + 2 = ax3 +x2(b – 3a)+x (c – 3b) – 3c + k

Let us compare the coefficients of x3, x2, x and k to get the values of a, b, c and k

⸫ a = 1  ; coefficients of x3 on both the sides

⸫ b – 3a = 4 ; coefficients of x2 on both the sides

⸫ -6 = c – 3b  ; coefficients of x on both the sides

⸫ 2 = -3c + k  ; constants on both the sides

Solve these equations to get the value of a, b and k

Since a = 1

b – 3a = 4

⇒ b = 4 + 3a = 4 + 3 = 7

b = 7

-6 = c – 3b

⇒ – 6 = c – 3(7)

⇒ – 6 = c – 21

⇒ c = -6 + 21 = 15

c = 15

2 = -3c + k

k = 2 + 3c = 2 + 3×15 = 2 + 45 = 47

Since q(x) = ax2 + bx + c = x2 + 7x + 15

r(x) = 47

Therefore, quotient = x2 + 7x + 15 and remainder 47.


  1. What must be subtracted from (x3 + 5x2 + 5x + 8) so that the resulting polynomial exactly divisible by (x2 + 3x – 2)?

Solution:

To  find what must be subtracted from (x3 + 5x2 + 5x + 8) so that the resulting polynomial exactly divisible by (x2 + 3x – 2), we need to  divide x3 + 5x2 + 5x + 8 by x2 + 3x – 2

Polynomials – Exercise 8.2 – Class X

On dividing x3 + 5x2 + 5x + 8 by x2 + 3x – 2, we get quotient q(x) = (x +2) and the remainder r(x) = (-x + 8).

Therefore, we must subtract (-x + 8) from (x3 + 5x2 + 5x + 8) so that the resulting polynomial exactly divisible by (x2 + 3x – 2).


  1. What should be added to (x4 – 1) so that it is exactly divisible by (x2 + 2x + 1)?

Solution:

To  find what should be added to (x4 – 1) so that it is exactly divisible by (x2 + 2x + 1), we need to  divide x4 – 1 from x2 + 2x + 1

Polynomials – Exercise 8.2 – Class X

On dividing x4 – 1 by x2 + 2x + 1, we get quotient q(x) = (x2 – 2x + 3) and the remainder r(x) = (-4x – 4).

Therefore, we must add (4x + 4) from (x4 – 1) so that the resulting polynomial exactly divisible by (x2 + 2x + 1).


Next  Exercise – Polynomials – Exercise 8.3 – Class X