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*Class X,
Mathematics, NCERT (CBSE) Solutions*

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__Chapter 2, Polynomials (Division Algorithm for Polynomials __

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**Solutions of NCERT Math
Textbook Exercise 2.3**

(Page 36)

**Q 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}– 2(ii)

*p(x)*= x

^{4 }– 3x

^{2}+ 4x + 5,

*g(x)*= x

^{2}+ 1 – x

(iii) Do it yourself by taking hint from the above solutions.

**Q 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}+ x2 + 3x + 1__Solution__:

(i) Try to
solve this problem yourself after going through the solved examples given in
the NCERT mathematics textbook itself.

__Solution__: (ii) x

^{2 }+ 3x + 1, 3x

^{4}+ 5x

^{3}– 7x

^{2}+ 2x + 2

__Solution__: (iii) x

^{3}– 3x + 1, x

^{5}– 4x

^{3}+ x2 + 3x + 1

**Q 3: Obtain all other zeroes of 3x**

^{4}+ 6x^{3}– 2x^{2}– 10x – 5, if two of its zeroes are

__Solution__: Since two zeroes are

**Q 4: On dividing x**

^{3}– 3x^{2}+ x + 2 by a polynomial*g(x)*, the quotient and the remainder were x – 2 and -2x + 4, respectively. Find*g(x)*.__Solution__:

By the division
algorithm, we have

*f(x)*=

*g(x)*x

*q(x)*+

*r(x)*

Or,

*g(x)*x*q(x)*=*f(x)*–*r(x)*
Or,

*g(x)*(x – 2) = x^{3}– 3x^{2}+ x + 2 – (-2x + 4)
Or,

*g(x)*(x – 2) = x^{3}– 3x^{2}+ 3x – 2
Thus,

*g(x)*is a factor of x^{3}– 3x^{2}+ 3x – 2 other than the factor (x – 2). Hence, to get*g(x)*we will divide
(x

^{3}– 3x^{2}+ 3x – 2) by (x – 2),

**Q 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) deg

*p(x)*= deg*q(x)*
Let us assume the division of 6x

^{2}+ 2x + 2 by 2
Here,

*p(x)*= 6x^{2}+ 2x + 2*g(x)*= 2

*q(x)*= 3x

^{2}+ x + 1

*r(x)*= 0

Degree of

*p(x)*and*q(x)*is same i.e. 2.
Checking for division algorithm,

p(x) =

*g(x)*x*q(x)*+*r(x)*
Or, 6x

^{2}+ 2x + 2 = 2x (3x^{2}+ x + 1)
Hence, division algorithm is satisfied.

__Solution__: (ii) and (iii) to be loaded soon.

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