**Next exercise – Polynomials – Exercise 8.2 – Class X**

#### 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) = x^{3} + 3x^{2} – 5x + 8 ; g(x) = x – 3

(ii) p(x) = 4x^{3} – 10x^{2} + 12x – 3 ; g(x) = x + 1

(iii) p(x) = 2x^{4} – 5x^{2} + 15x – 6 ; g(x) = x – 2

(iv) p(x) = 4x^{3} – 12x^{2} + 14x – 3 ; g(x) = 2x – 1

(v) p(x) = 7x^{3} – x^{2} + 2x – 1 ; g(x) = 1 – 2x

- If the polynomials (2x
^{3}+ ax^{2}+ 3x – 5) and (x^{3}+ x^{2}– 4x – a) leave the same remainder when divided by (x – 1), find the value of a. - The polynomials (2x
^{3}– 5x^{2}+ x + a) and (ax^{3}+ 2x^{2}– 3)when divided by (x – 2) leave the remainders R_{1}and R_{2}respectively. Find the value of a in each of the following cases, if

(i) R_{1} = R_{1}

(ii)2R_{1} + R_{2} = 0

(iii) R_{1} – 2R_{2} = 0

#### Polynomials – Exercise 8.3 – Solutions:

**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) = x ^{3} + 3x^{2} – 5x + 8 ; g(x) = x – 3**

Solution:

p(x) = x^{3} + 3x^{2} – 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) = x^{3} + 3x^{2} – 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) = 4x ^{3} – 10x^{2} + 12x – 3 ; g(x) = x + 1**

Solution:

p(x) = 4x^{3} – 10x^{2} + 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) = 4x^{3} – 10x^{2} + 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) = 2x ^{4} – 5x^{2} + 15x – 6 ; g(x) = x – 2**

Solution:

p(x) = 2x^{4} – 5x^{2} + 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) = 2x^{4} – 5x^{2} + 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) = 4x ^{3} – 12x^{2} + 14x – 3 ; g(x) = 2x – 1**

Solution:

p(x) = 4x^{3} – 12x^{2} + 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) = 4x^{3} – 12x^{2} + 14x – 3 is p(^{1}/_{2})

p(^{1}/_{2}) = 4(^{1}/_{2})^{3} – 12(^{1}/_{2})^{2} + 14(^{1}/_{2}) – 3

= 4x^{1}/_{8} – 12 x ^{1}/_{4} + 14x^{1}/_{2} – 3

= ^{1}/_{2} – 3 + 7 – 3

=^{3}/_{2}

Therefore, the remainder r(x) = ^{3}/_{2}

**(v) p(x) = 7x ^{3} – x^{2} + 2x – 1 ; g(x) = 1 – 2x**

Solution:

p(x) = 7x^{3} – x^{2} + 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) = 7x^{3} – x^{2} + 2x – 1 is p(-^{1}/_{2})

p(-^{1}/_{2})= 7(-^{1}/_{2})^{3} – (-^{1}/_{2})^{2} + 2(-^{1}/_{2}) – 1

= -7x^{1}/_{8} – ^{1}/_{4} – 1 – 1

= ^{-7 – 2 – 8 – 8}/_{8}

= –^{25}/_{8}

Therefore, the remainder r(x) = –^{25}/_{8}

**If the polynomials (2x**^{3}+ ax^{2}+ 3x – 5) and (x^{3}+ x^{2}– 4x – a) leave the same remainder when divided by (x – 1), find the value of a.

Solution:

Let p(x) = (2x^{3} + ax^{2} + 3x – 5) and g(x) = (x^{3} + x^{2} – 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) = 2x^{3} + ax^{2} + 3x – 5 and g(x) = (x^{3} + x^{2} – 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

**The polynomials (2x**^{3}– 5x^{2}+ x + a) and (ax^{3}+ 2x^{2}– 3)when divided by (x – 2) leave the remainders R_{1}and R_{2}respectively. Find the value of a in each of the following cases, if

**(i) R _{1} = R_{1}**

**(ii)2R _{1} + R_{2} = 0**

**(iii) R _{1} – 2R_{2} = 0**

Solution:

Let p(x) = (2x^{3} – 5x^{2} + x + a) and g(x) = (ax^{3} + 2x^{2} – 3)

R_{1} is the remainder when p(x) = (2x^{3} – 5x^{2} + x + a)is divided by (x – 2).

⸫R_{1} = p(2)

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

⇒ R_{1} = a – 2

R_{2} is the remainder when g(x) = (ax^{3} + 2x^{2} – 3)is divided by (x – 2).

R_{2} = g(2) = a(2)^{3} + 2(2)^{2} – 3 = 8a + 8 – 3 = 8a + 5

⇒ R_{2} = 8a + 5

(i) R_{1} = R_{2}

a – 2 = 8a + 5

a – 8a = 5 + 2

-7a = 7

a = -1

(ii)2R_{1} + R_{2} = 0

2(a – 2) + (8a + 5) = 0

2a – 4 + 8a + 5 = 0

10 a + 1 = 0

10 a = -1

a = –^{1}/_{10}

(iii) R_{1} – 2R_{2} = 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**