# Class10 NCRT Polynomials Exercise – 2.3 pdf || UP Board

__Polynomials__

__Exercise – 2.3__

__Polynomials__

__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**

**Solution****:**

Quotient = x – 2 and remainder = 7x – 9

**(****i****i****) ****p(x)
= x ^{4}-3x^{2}+4x+5, g(x) = x^{2}+1-x**

**Solution****:**

Quotient = x^{2} +
x - 3 and
remainder = 8

**(****ii****i****) ****p(x)
= x ^{4}-5x+6, g(x) = 2 – x^{2}**

**Solution****:**

Quotient = -x^{2} -2 and
remainder = -5x
+ 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**

**Solution****:**

Since, remainder is 0, therefore the polynomial
t^{2} – 3 is a
factor of the polynomial 2t^{4} + 3t^{3} – 2t^{2} – 9t
– 12.

**(****ii****) ****x ^{2}
+ 3x + 1, 3x^{4} + 5x^{3} – 7x^{2} +2x +2**

**Solution****:**

Since, remainder is 0, therefore the polynomial x^{2}
+ 3x + 1** **is a factor of the polynomial 3x^{4} + 5x^{3} –
7x^{2} +2x +2.

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

**Solution****:**

The
polynomial x^{3} -3x + 1** **is not a factor of the polynomial x^{5}
-4x^{3} +x^{2} +3x +1 because the remainder is 0 not 2.

**3. Obtain
all other zeroes of ****3x ^{4} +6x^{3} -2x^{2} -10x -5,
**

**if two of its zeroes are**

**√**

**(5/3)**

**and**

**-**

**√**

**(5/3)**

**.**

**Solution****:**

Let p(x) = 3x^{4} +6x^{3} -2x^{2} -10x -5

Therefore, x =√(5/3),
x =-√(5/3)

Or

3x^{2}
– 5 = 0

3x^{2}
– 5 is
a factor of p(x),

by division method,

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

Now, on factoring x^{2} + 2x +1,

x^{2} + 2x +1 = 0

x^{2} + x + x +1 = 0

x (x+1) +1(x+1) = 0

( x+1) (x+1) = 0

x = -1, -1

Hence two other zeroes are -1, -1.

**4. ****On
dividing**** ****x ^{3}
– 3x^{2} + x +2 **

**by a polynomial**

**g**

**(**

**x**

**)**

**,**

**the quotient and remainder were**

**x**

**– 2**

**and**

**-2**

**x**

**+4**

**, respectively. Find g**

**(**

**x**

**)**

**.**

**Solution****:**

Given that,

Dividend p(x) = x^{3} – 3x^{2} + x +2

Quotient q(x) = x – 2

Remainder r(x) = -2x +4

Divisor g(x) = ?

We know that,

p(x) = g(x)×
q(x) + r(x)

x^{3} – 3x^{2} + x +2 = g(x)×(x – 2) + (-2x +4)

by division method,

Hence, g(x) = 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)**

**Solution****:**

The degree of the dividend and the quotient can be equal only if the
divisor is a constant (to the degree 0) number.

Let p(x)
= 3x^{2} -6x +5

Let g(x)
= 3

Therefore, q(x) = x^{2} -2x +1 and r(x) = 2

**(****ii****) ****deg****
****q(x)****
= ****deg**** ****r(x)**

**Solution****:**

Let p(x)
= 2x^{2} -4x +3

Let g(x)
= x^{2} -2x +1

Therefore, q(x)
= 2 and r(x)
= 1

**(****iii****) ****deg****
****r(x)****
= 0**

**Solution****:**

Let p(x)
= 2x^{2} -4x +3

Let g(x)
= x^{2} -2x +1

Therefore, q(x) = 2 and r(x) = 1

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