**Go To: Exercise 2.1 Exercise 2.2 ****Exercise 2.4**

**NCERT Solutions Maths Class 10 Exercise 2.3**

**1.
Divide the polynomial p(x) by the polynomial g(x)
and find the quotient and remainder in each of the following:**

**(i)
p(x) = x ^{3}**

**– 3x**

^{2 }+ 5x – 3, g(x) = x^{2}– 2**(ii)
p(x) = x ^{4}**

**– 3x**

^{2 }+ 4x + 5, g(x) = x^{2}+ 1 – x**(iii)
p(x) = x ^{4}**

**– 5x + 6, g(x) = 2 – x**

^{2}**Solution:
(i) p(x) = x ^{3}**

**– 3x**

^{2 }+ 5x – 3, g(x) = x^{2}– 2Note that we stop the division process when either the remainder is zero or its degree is less than the degree of the divisor.

Therefore, the quotient is x
– 3 and the remainder is 7x – 9.

**(ii)** **p(x) = x ^{4}**

**– 3x**

^{2 }+ 4x + 5, g(x) = x^{2}+ 1 – xNote that g(x) is not in standard form. First write it in standard form and then divide.

Then, g(x) = x^{2} – x + 1** **

We stop the division process when either the
remainder is zero or its degree is less than the degree of the divisor.

Therefore, the quotient is x

^{2 }+ x – 3 and the remainder is 8.

**(iii)** **p(x) = x ^{4}**

**– 5x + 6, g(x) = 2 – x**

^{2}Note that g(x) is not in standard form. First write it in standard form and then divide.

Then, g(x) = –x^{2} + 2** **

We stop the division process when either the
remainder is zero or its degree is less than the degree of the divisor.

Therefore, the quotient is –x

^{2}– 2 and the remainder is −5

*x*+ 10.

**2. Check whether the first polynomial is a factor of the second
polynomial by dividing the second polynomial by the first polynomial.**

**(i) t ^{2}**

**– 3, 2t**

^{4 }+ 3t^{3}– 2t^{2}– 9t – 12**(ii)
x ^{2 }+ 3x + 1, 3x^{4 }+ 5x^{3}**

**– 7x**

^{2 }+ 2x + 2**(iii) x ^{3}**

**– 3x + 1, x**

^{5}– 4x^{3 }+ x^{2 }+ 3x + 1**Solution:
(i) t ^{2}**

**– 3, 2t**

^{4 }+ 3t^{3}– 2t^{2}– 9t – 12Divide the second polynomial by the first polynomial. If the remainder is 0, then the first polynomial is a factor of the second polynomial.

Here, remainder = 0Hence, the first polynomial
is a factor of the second polynomial.

**(ii)
x ^{2 }+ 3x + 1, 3x^{4 }+ 5x^{3}**

**– 7x**

^{2 }+ 2x + 2Divide the second polynomial by the first polynomial. If the remainder is 0, then the first polynomial is a factor of the second polynomial.

Hence, the first polynomial
is a factor of the second polynomial.

**(iii) x ^{3}**

**– 3x + 1, x**

^{5}– 4x^{3 }+ x^{2 }+ 3x + 1Divide the second polynomial by the first polynomial. If the remainder is 0, then the first polynomial is a factor of the second polynomial.

Hence, the first polynomial
is not a factor of the second polynomial.

**3.
Obtain all other zeroes of (3x ^{4 }+ 6x^{3}**

**– 2x**

^{2}– 10x – 5) if two of its zeroes are**Solution: **Since two
zeroes of (3x^{4 }+ 6x^{3}
– 2x^{2} – 10x – 5)** **are

is a
factor of (3x^{4 }+ 6x^{3}
– 2x^{2} – 10x – 5).

⇒ (3x^{4 }+ 6x^{3}
– 2x^{2} – 10x – 5)** **= (x^{2}
– 5/3) (3x^{2 }+ 6x + 3)

= (x^{2 }– 5/3)3(x^{2 }+ 2x
+ 1)

= 3(x^{2
}– 5/3)(x^{2 }+ x + x + 1)

= 3(x^{2 }– 5/3)(*x *+ 1)(*x *+ 1)

^{4 }+ 6x^{3} –
2x^{2} – 10x – 5)** **are −1 and −1.

**4. On dividing (x ^{3 }**

**− 3x**

^{2 }+ x + 2)^{ }by a polynomial g(x), the quotient and remainder were (x − 2) and (−2x + 4), respectively. Find g(x).**Solution:
**Let p(x) = (x^{3
}− 3x^{2 }+ x + 2), q(x) = (x – 2) and r(x) = (–2x + 4)

According to Division
Algorithm of Polynomial, we have

p(x) = g(x).q(x) + r(x)

⇒ x^{3 }− 3x^{2 }+ x
+ 2** ^{ }**=

*g*(

*x*).(

*x*− 2) + (−2

*x*+ 4)

⇒ x^{3 }− 3x^{2 }+ x
+ 2 + 2x − 4 = *g*(*x*).(*x *− 2)

⇒
x^{3 }− 3x^{2 }+ 3x − 2 = *g*(*x*).(*x *− 2)

⇒ *g*(*x*) = (x^{3 }−
3x^{2 }+ 3x – 2)/(x – 2)

So, dividing (x^{3
}− 3x^{2 }+ 3x − 2) by (*x *− 2), we get

*g*(

*x*) = (x

^{3 }− 3x

^{2 }+ 3x – 2)/(x – 2) = x

^{2 }– x + 1

**5. Give examples of polynomials p(x), g(x), q(x) and r(x), which
satisfy the division algorithm and**

**(i)
deg p(x) = deg q(x) (ii)
deg q(x) = deg r(x) (iii) deg
r(x) = 0**

**Solution:
(i)** Let
p(x) = 3x^{2 }+ 3x + 6, g(x) = 3

Dividing
p(x) by g(x), we get

^{2 }+ x + 2

So, we can see in this
example that deg p(x) = deg q(x) = 2

**(ii)** Let p(x) = x^{3 }+ 5 and g(x) = x^{2 }–
1

Dividing p(x) by g(x), we get

q(x) = x

So, we can see in this
example that deg q(x) = deg r(x) = 1

**(iii)** Let p(x) = x^{2 }+ 5x − 3, g(x) = *x *+ 3

Dividing p(x) by g(x), we get

q(x) = x + 2

So, we can see in this
example that deg r(x) = 0