**Download NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.3 – Polynomials**

**1.Find the remainder when x ^{3} + 3x^{2} + 3x + 1 is divided by**

** ****(i) x + 1
(ii) x – ½
(iii) x**

**(iv) x + π**

(v) 5 + 2x

(v) 5 + 2x

** ****Solution:**

According to remainder theorem if polynomial p(x) of greater than or equal to one is divided by the linear polynomial (x – a) , then the remainder is p(a) where a is any real number.

Here, p(x) = x^{3} + 3x^{2} + 3x + 1

**(i)** Given linear polynomial is (x + 1) , having zero equal to -1.

Putting x= -1 in p(x) , we get

p(-1) = (-1)^{3} + 3(-1)^{2} + 3(-1) + 1

= -1 + 3 – 3 + 1

= 0

So , by remainder theorem , remainder in this case is 0.

**Alternative Method – **We can also use long division method as follows :

Thus, we conclude that the remainder is zero in this case.

**(ii)** Given linear polynomial is (x – ½ ) , having zero equal to ½ .

Putting x= ½ in p(x) , we get

p(½ ) = (½ )^{3} + 3(½ )^{2} + 3(½ ) + 1

= ⅛ + ¾ + 3/2 + 1

= (1 + 6 + 12 + 8)/8

= 27/8

So , by remainder theorem , remainder in this case is 27/8 .

**Alternative Method – **We can also use long division method as follows :

Thus, we conclude that the remainder is (27/8) in this case.

**(iii)** Given linear polynomial is (x + 0) , having zero equal to 0.

Putting x= 0 in p(x) , we get

p(0) = (0)^{3} + 3(0)^{2} + 3(0) + 1

= 0 + 0 + 0 + 1

= 1

So , by remainder theorem , remainder in this case is 1.

**Alternative Method – **We can also use long division method as follows :

Thus, we conclude that the remainder is 1 in this case.

**(iv)**Given linear polynomial is (x + π ) , having zero equal to -π .

Putting x= -π in p(x) , we get

p(-π ) = (-π )^{3} + 3(-π )^{2} + 3(-π ) + 1

= – π^{3} + 3π^{2} – 3π + 1

So , by remainder theorem , remainder in this case is (- π^{3} + 3π^{2} – 3π + 1).

**Alternative Method – **We can also use long division method as follows :

Thus, we conclude that the remainder is (- π^{3} + 3π^{2} – 3π + 1) in this case.

**(v)** 5 + 2x

Given linear polynomial is (5 + 2x) , having zero equal to -5/2 .

Putting x= -5/2 in p(x) , we get

p(-5/2 ) = (-5/2)^{3} + 3(-5/2)^{2} + 3(-5/2) + 1

= -(125/8) + (75/4) -(15/2) + 1

= (-125 + 150 – 60 + 8)/8

= -(27/8)

So , by remainder theorem , remainder in this case is -(27/8).

**Alternative Method – **We can also use long division method as follows :

Thus, we conclude that the remainder is -(27/8) in this case.

**2. Find the remainder when x ^{3} – ax^{2} + 6x – a is divided by x – a.**

** ****Solution:**

Given

p(x) = x^{3} – ax^{2} + 6x – a

And the linear polynomial is (x – a) , having zero equal to a.

Putting x= a in p(x) , we get

p(a) = a^{3} – a(a)^{2} + 6a – a

= a^{3} – a^{3} + 6a – a

= 5a

According to remainder theorem if polynomial p(x) of greater than or equal to one is divided by the linear polynomial (x – a) , then the remainder is p(a) where a is any real number.

So , by remainder theorem , remainder in this case is 5a.

**3. Check whether 7 + 3x is a factor of 3x ^{3} + 7x.**

** ****Solution:**

We know that 7 + 3x is a factor of 3x^{3} + 7x , only if on division by 7 + 3x , 3x^{3} + 7x doesn’t leave any remainder .

And, According to remainder theorem if polynomial p(x) of greater than or equal to one is divided by the linear polynomial (x – a) , then the remainder is p(a) where a is any real number.

Here we have

p(x) = 3x^{3} + 7x

And the linear polynomial is (7 + 3x) , having zero equal to (-7/3).

Putting x= -7/3 in p(x) , we get

p(-7/3) = 3(-7/3)^{3} + 7(-7/3)

= (-343 – 147 ) / 9

= – 490/9

So , by remainder theorem , remainder in this case is -490/9.

Therefore, we conclude that 7 + 3x is not a factor of 3x^{3} + 7x because the remainder is not zero.

**Download NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.3 – Polynomials**

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