﻿ Class 10 NCERT Math Solution
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TOPICS
Unit-2(Examples)

Example-1 :-  Look at the graphs in Figures given below. Each is the graph of y = p(x), where p(x) is a polynomial. For each of the graphs, find the number of zeroes of p(x).

Solution :-
```(i) The number of zeroes is 1 as the graph intersects the x-axis at one point only.
(ii) The number of zeroes is 2 as the graph intersects the x-axis at two points.
(iii) The number of zeroes is 3 as the graph intersects the x-axis at three points.
(iv) The number of zeroes is 1 as the graph intersects the x-axis at one point only.
(v) The number of zeroes is 1 as the graph intersects the x-axis at one point only.
(vi) The number of zeroes is 4 as the graph intersects the x-axis at four points.
```

Example-2 :-  Find the zeroes of the quadratic polynomial x² + 7x + 10, and verify the relationship between the zeroes and the coefficients.

Solution :-
```    p(x) = x² + 7x + 10
= x² + 5x + 2x + 10
= x(x + 5) + 2(x + 5)
= (x + 2)(x + 5)
Zeroes of polynomial :
x + 2 = 0
x = -2
x + 5 = 0
x = -5

Relationship between the zeroes and the coefficients :
p(x) = x² + 7x + 10
a = 1, b = 7, c = 10
Sum of zeroes = α + β = -b/a = -7/1 = -7
Product of zeroes = αβ = c/a = 10/1 = 10
```

Example-3 :-  Find the zeroes of the polynomial x² – 3 and verify the relationship between the zeroes and the coefficients.

Solution :-
```    p(x) = x² – 3
= x² – (√3)²
= (x + √3)(x – √3)
Zeroes of polynomial :
x + √3 = 0
x = –√3
x – √3 = 0
x = √3

Relationship between the zeroes and the coefficients :
p(x) = x² – 3
a = 1, b = 0, c = -3
Sum of zeroes = α + β = -b/a = 0/1 = 0
Product of zeroes = αβ = c/a = -3/1 = -3
```

Example-4 :-  Find a quadratic polynomial, the sum and product of whose zeroes are –3 and 2, respectively.

Solution :-
```
Given that :
Sum of zeroes = α + β = -b/a = -3/1
Product of zeroes = αβ = c/a = 2/1
Therefore, a = 1, b = -3, c = 2.
In general form, p(x) = ax² + bx + c.
So, quadratic polynomial is x² - 3x + 2.
```

Example-5 :-  : Verify that 3, –1, -1/3 are the zeroes of the cubic polynomial p(x) = 3x³ – 5x² – 11x – 3, and then verify the relationship between the zeroes and the coefficients.

Solution :-
```   p(x) = 3x³ – 5x² – 11x – 3
p(3) = 3 × 3³ – (5 × 3²) – (11 × 3) – 3
= 81 – 45 – 33 – 3 = 0,
p(–1) = 3 × (–1)³ – 5 × (–1)² – 11 × (–1) – 3
= –3 – 5 + 11 – 3 = 0,
p(-1/3) = 3 x (-1/3)³ - 5 x (-1/3)² - 11 x (-1/3) - 3
= 3 x (-1/27) - 5 x (1/9) + 11/3 - 3
= -1/9 - 5/9 + 11/3 - 3
= (-1 - 5 + 33 - 27)/9
= 0/9 = 0
So, 3, –1, -1/3 are the zeroes of the cubic polynomial p(x) = 3x³ – 5x² – 11x – 3.

In general, cubic polynomial is ax³ + bx² + cx + d
p(x) = 3x³ – 5x² – 11x – 3
By compairing, a = 3, b = -5, c = -11, d = -3.
Relationship between the zeroes and the coefficients :
Given that α = 3, β = -1, γ = -1/3
α + β + γ = 3 - 1 - 1/3
= (9 - 3 - 1)/3 = 5/3
-b/a = -(-5)/3 = 5/3
αβ + βγ + γα = 3 x (-1) + (-1) x (-1/3) + (-1/3) x 3
= -3 + 1/3 - 1
= (-9 + 1 - 3)/3
= -11/3
c/a = -11/3
αβγ = 3 x (-1) x (-1/3) = 1
-d/a = -(-3)/3 = 1
```

Example-6 :-  Divide 2x² + 3x + 1 by x + 2.

Solution :-
```    p(x) = 2x² + 3x + 1
g(x) = x + 2

q(x) = 2x - 1, r(x) = 3

Verification :
p(x) = g(x) x q(x) + r(x)
= (x + 2)(2x - 1) + 3
= 2x² - x + 4x - 2 + 3
= 2x² + 3x + 1
```

Example-7 :-  Divide 3x³ + x² + 2x + 5 by 1 + 2x + x².

Solution :-
```
p(x) = 3x³ + x² + 2x + 5
g(x) = 1 + 2x + x²

q(x) = 3x - 5, r(x) = 9x + 10

Verification :
p(x) = g(x) x q(x) + r(x)
= (1 + 2x + x²) x (3x - 5) + (9x + 10)
= 3x - 5 + 6x² - 10x + 3x³ - 5x² + 9x + 10
= 3x³ + (6x² - 5x²) + (3x + 9x - 10x) + (10 - 5)
= 3x³ + x² + 2x + 5
```

Example-8 :-  : Divide 3x² – x³ – 3x + 5 by x – 1 – x², and verify the division algorithm.

Solution :-
```    p(x) = 3x² – x³ – 3x + 5
g(x) = x – 1 – x²

q(x) = x - 2, r(x) = 3

Verification :
p(x) = g(x) x q(x) + r(x)
= (x – 1 – x²)(x - 2) + 3
= x² - 2x - x³ + 2 - x + 2x² + 3
= 3x² – x³ – 3x + 5
```

Example-6 :-  Find all the zeroes of 2x⁴ – 3x³ – 3x² + 6x – 2, if you know that two of its zeroes are √2 and -√2.

Solution :-
```    p(x) = 2x⁴ – 3x³ – 3x² + 6x – 2
Given that two zeroes are √2 and -√2.
So, (x - √2)(x + √2) = x² - 2           [a² - b² = (a + b)(a - b)]

q(x) = 2x² - 3x + 1
Now, p(x) = 2x⁴ – 3x³ – 3x² + 6x – 2
= (x - √2)(x + √2)(2x² - 3x + 1)
= (x - √2)(x + √2)[2x² - 2x - x + 1]
= (x - √2)(x + √2)[2x(x - 1) - 1(x - 1)]
= (x - √2)(x + √2)(x - 1)(2x - 1)
x - 1 = 0
x = 1
2x - 1 = 0
2x = 1
x = 1/2
Therefore, all zeroes are √2, -√2, 1 and 1/2.
```
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