Polynomials – Exercise 8.3 – Class X

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Polynomials – Exercise 8.3

1.In each of the following cases, use the remainder  theorem and find the remainder when p(x) is divided by g(x)

(i) p(x) = x3 + 3x2 – 5x + 8 ; g(x) = x – 3

(ii) p(x) = 4x3 – 10x2 + 12x – 3 ; g(x) = x + 1

(iii) p(x) = 2x4 – 5x2 + 15x – 6 ; g(x) = x – 2

(iv) p(x) = 4x3 – 12x2 + 14x – 3 ; g(x) = 2x – 1

(v) p(x) = 7x3 – x2 + 2x – 1 ; g(x) = 1 – 2x

  1. If the polynomials (2x3 + ax2 + 3x – 5) and (x3 + x2 – 4x – a) leave the same remainder when divided by (x – 1), find the value of a.
  2. The polynomials (2x3 – 5x2 + x + a) and (ax3 + 2x2 – 3)when divided by (x – 2) leave the remainders R1 and R2 respectively. Find the value of a in each of the following cases, if

(i) R1 = R1

(ii)2R1 + R2 = 0

(iii) R1 – 2R2 = 0


Polynomials – Exercise 8.3 – Solutions:

  1. In each of the following cases, use the remainder theorem and find the remainder when p(x) is divided by g(x)

(i) p(x) = x3 + 3x2 – 5x + 8 ; g(x) = x – 3

Solution:

p(x) = x3 + 3x2 – 5x + 8

g(x) = x – 3

Remainder theorem: If a polynomial p(x) is divided by (x – a), then the reminder is p(a).

Therefore, by the remainder theorem, the remainder of p(x) = x3 + 3x2 – 5x + 8 is p(3)

p(3) = (3)3 + 3(3)2 – 5(3) + 8

= 27 + 27 – 15 + 8

= 47

Therefore, the remainder r(x) = 47.


(ii) p(x) = 4x3 – 10x2 + 12x – 3 ; g(x) = x + 1

Solution:

p(x) = 4x3 – 10x2 + 12x – 3

g(x) = x + 1

Remainder theorem: If a polynomial p(x) is divided by (x – a), then the reminder is p(a).

Therefore, by the remainder theorem, the remainder of p(x) = 4x3 – 10x2 + 12x – 3  is p(-1)

p(-1) = 4(-1)3 – 10(-1)2 + 12(-1) – 3

= 4(-1) – 10(1) + 12 (-1) – 3

= -4 – 10 -12 – 3

= -29

Therefore, the remainder r(x) = -29.


 (iii) p(x) = 2x4 – 5x2 + 15x – 6 ; g(x) = x – 2

Solution:

p(x) = 2x4 – 5x2 + 15x – 6

g(x) = x – 2

Remainder theorem: If a polynomial p(x) is divided by (x – a), then the reminder is p(a).

Therefore, by the remainder theorem, the remainder of p(x) = 2x4 – 5x2 + 15x – 6  is p(2)

p(2) = (2)3 + 3(2)2 – 5(2) + 8

= 8 + 12 – 10 + 8

= 18

Therefore, the remainder r(x) = 18.


(iv) p(x) = 4x3 – 12x2 + 14x – 3 ; g(x) = 2x – 1

Solution:

p(x) = 4x3 – 12x2 + 14x – 3

g(x) = 2x – 1

Remainder theorem: If a polynomial p(x) is divided by (x – a), then the reminder is p(a).

Therefore, by the remainder theorem, the remainder of p(x) = 4x3 – 12x2 + 14x – 3   is p(1/2)

p(1/2) = 4(1/2)3 – 12(1/2)2 + 14(1/2) – 3

= 4x1/8 – 12 x 1/4 + 14x1/2 – 3

= 1/2 – 3 + 7 – 3

=3/2

Therefore, the remainder r(x) = 3/2


(v) p(x) = 7x3 – x2 + 2x – 1 ; g(x) = 1 – 2x

Solution:

p(x) = 7x3 – x2 + 2x – 1

g(x) = 1 – 2x

Remainder theorem: If a polynomial p(x) is divided by (x – a), then the reminder is p(a).

Therefore, by the remainder theorem, the remainder of p(x) = 7x3 – x2 + 2x – 1  is p(-1/2)

p(-1/2)= 7(-1/2)3 – (-1/2)2 + 2(-1/2) – 1

= -7x1/81/4 – 1 – 1

= -7 – 2 – 8 – 8/8

= –25/8

Therefore, the remainder r(x) = –25/8


  1. If the polynomials (2x3 + ax2 + 3x – 5) and (x3 + x2 – 4x – a) leave the same remainder when divided by (x – 1), find the value of a.

Solution:

Let p(x) = (2x3 + ax2 + 3x – 5) and g(x) = (x3 + x2 – 4x – a)

Remainder theorem: If a polynomial p(x) is divided by (x – a), then the reminder is p(a).

Therefore, by the remainder theorem, the remainder of p(x) = 2x3 + ax2 + 3x – 5 and g(x) = (x3 + x2 – 4x – a)  is p(1) and g(1)

By the given condition, p(1) = g(1)

p(1) = 2(1)3 + a(1)2 + 3(1) – 5 = 2 + a + 3 – 5 = a

g(1) = (1)3 + (1)2 – 4(1) – a = 1 + 1 – 4 – a = -2 – a

Since p(1) = g(1)

⇒ a = -2 – a

a + a = – 2

2a = -2

a = – 1


  1. The polynomials (2x3 – 5x2 + x + a) and (ax3 + 2x2 – 3)when divided by (x – 2) leave the remainders R1 and R2 respectively. Find the value of a in each of the following cases, if

(i) R1 = R1

(ii)2R1 + R2 = 0

(iii) R1 – 2R2 = 0

Solution:

Let p(x) = (2x3 – 5x2 + x + a) and g(x) = (ax3 + 2x2 – 3)

R1 is the remainder when p(x) = (2x3 – 5x2 + x + a)is divided by (x – 2).

⸫R1 = p(2)

p(2) = (2(2)3 – 5(2)2 + (2) + a) = 16 – 20 + 2 + a = -2 + a

⇒ R1 = a – 2

R2 is the remainder when g(x) = (ax3 + 2x2 – 3)is divided by (x – 2).

R2 = g(2) = a(2)3 + 2(2)2 – 3 = 8a + 8 – 3 = 8a + 5

⇒ R2 = 8a + 5

(i) R1 = R2

a – 2 = 8a + 5

a – 8a = 5 + 2

-7a = 7

a = -1

(ii)2R1 + R2 = 0

2(a – 2) + (8a + 5) = 0

2a – 4 + 8a + 5 = 0

10 a + 1 = 0

10 a = -1

a = –1/10

(iii) R1 – 2R2 = 0

(a – 2) – 2(8a + 5) = 0

a – 2 – 16a – 10 = 0

– 15a – 12 = 0

a = – 12/15 = –4/5


Next Exercise – Polynomials 8.4 – Exercise 8.4 – Class X