﻿ Class 10 NCERT Math Solution
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TOPICS
Exercise - 2.3

Question-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) = x³ – 3x² + 5x – 3, g(x) = x² – 2
(ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x
(iii) p(x) = x⁴ – 5x + 6, g(x) = 2 – x²

Solution :-
```(i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2 q(x) = x - 3, r(x) = 7x - 9

Verification :
p(x) = g(x) x q(x) + r(x)
= (x² – 2)(x - 3) + 7x - 9
= x³ – 3x² - 2x + 6 + 7x – 9
= x³ – 3x² + 5x – 3
```
```(ii) p(x) = x⁴ – 3x² + 4x + 5,  g(x) = x² + 1 – x q(x) = x² + x - 3, r(x) = 8

Verification :
p(x) = g(x) x q(x) + r(x)
= (x² + 1 – x )(x² + x - 3) + 8
= x⁴ + x³ – 3x² + x² + x - 3 - x³ - x² + 3x + 8
= x⁴ – 3x² + 4x + 5
```
```(iii) p(x) = x⁴ – 5x + 6, g(x) = 2 – x² q(x) = -2 – x², r(x) = -5x + 10

Verification :
p(x) = g(x) x q(x) + r(x)
= (2 – x²)(-2 – x²) + (-5x + 10)
= -(4 - x⁴) - 5x + 10
= x⁴ – 5x + 6
```

Question-2 :-  Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial :
(i) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12
(ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2
(iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1

Solution :-
```(i) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12 The remainder is 0.
So, t² – 3 is a factor of 2t⁴ + 3t³ – 2t² – 9t – 12.
```
```(ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2 The remainder is 0.
So, x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2 .
```
```(iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1 The remainder is not 0.
So, x³ – 3x + 1 is not a factor of x⁵ – 4x³ + x² + 3x + 1 .
```

Question-3 :-  Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are √5/3 and -√5/3.

Solution :-
```   p(x) = 3x⁴ + 6x³ – 2x² – 10x – 5
Given that two zeroes are √5/3 and -√5/3.
So, (x - √5/3)(x + √5/3) = x² - 5/3           [a² - b² = (a + b)(a - b)] q(x) = 3x² + 6x + 3
Now, p(x) = 3x⁴ + 6x³ – 2x² – 10x – 5
= (x - √5/3)(x + √5/3)(3x² + 6x + 3)
= (x - √5/3)(x + √5/3) x 3 x (x² + 2x + 1)
= (x - √5/3)(x + √5/3) x 3 x (x + 1)²
= (x - √5/3)(x + √5/3) x 3 x (x + 1)(x + 1)
x + 1 = 0
x = -1
Therefore, all zeroes are √5/3, -√5/3, -1 and -1.
```

Question-4 :-  On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4, respectively. Find g(x).

Solution :-
```    p(x) = x³ – 3x² + x + 2
q(x) = x - 2
r(x) = -2x + 4
p(x) = g(x) x q(x) + r(x)
x³ – 3x² + x + 2 = g(x) x (x - 2) + (-2x + 4)
x³ – 3x² + x + 2 + 2x - 4 = g(x) x (x - 2)
g(x) = (x³ – 3x² + 3x - 2)/(x - 2) g(x) = x² - x + 1
```

Question-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)
We take an example in which deg p(x) = deg q(x) = 2 ```
```(ii) deg q(x) = deg r(x)
We take an example in which deg p(x) = deg r(x) = 2 ```
```(iii) deg r(x) = 0
We take an example in which deg deg r(x) = 0 ```
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