﻿ Class 10 NCERT Math Solution
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TOPICS
Exercise - 2.4

Question-1 :-  Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
(i) 2x³ + x² - 5x + 2; 1/2, 1, -2
(ii) x³ – 4x² + 5x - 2; 2, 1, 1

Solution :-
```(i) p(x) = 2x³ + x² - 5x + 2; 1/2, 1, -2
p(1/2) = 2 x (1/2)³ + (1/2)² - 5 x (1/2) + 2
= 2 x 1/8 + 1/4 - 5/2 + 2
= 1/4 + 1/4 - 5/2 + 2
= (1 + 1 - 10 + 8)/4
= 0/4 = 0
p(1) = 2 x (1)³ + (1)² - 5 x 1 + 2
= 2 + 1 - 5 + 2
= 5 - 5 = 0
p(-2) = 2 x (-2)³ + (-2)² - 5 x (-2) + 2
= 2 x (-8) + 4 + 10 + 2
= -16 + 6 + 10
= -16 + 16 = 0
So, 1/2, 1, -2 are the zeroes of the cubic polynomial p(x) = 2x³ + x² - 5x + 2.

In general, cubic polynomial is ax³ + bx² + cx + d
p(x) = 2x³ + x² - 5x + 2
By compairing, a = 2, b = 1, c = -5, d = 2.
Relationship between the zeroes and the coefficients :
Given that α = 1/2, β = 1, γ = -2
α + β + γ = 1/2 + 1 - 2
= 3/2 - 2
= -1/2
-b/a = -(1)/2 = -1/2
αβ + βγ + γα = 1/2 x 1 + 1 x (-2) + (-2) x 1/2
= 1/2 - 2  - 1
= (1 - 4 - 2)/2
= -5/2
c/a = -5/2
αβγ = 1/2 x 1 x (-2) = -1
-d/a = -(2)/2 = -1
```
```(ii) p(x) = x³ – 4x² + 5x - 2; 2, 1, 1
p(2) = (2)³ – 4 x (2)² + 5 x 2 - 2
= 8 - 16 + 10 - 2
= 18 - 18 = 0
p(1) = (1)³ – 4 x (1)² + 5 x 1 - 2
= 1 - 4 + 5 - 2
= 6 - 6 = 0
So, 2, 1, 1 are the zeroes of the cubic polynomial p(x) = x³ – 4x² + 5x - 2.

In general, cubic polynomial is ax³ + bx² + cx + d
p(x) = x³ – 4x² + 5x - 2
By compairing, a = 1, b = -4, c = 5, d = -2.
Relationship between the zeroes and the coefficients :
Given that α = 2, β = 1, γ = 1
α + β + γ = 2 + 1 + 1 = 4
-b/a = -(-4)/1 = 4
αβ + βγ + γα = 2 x 1 + 1 x 1 + 1 x 2
= 2 + 1 + 2 = 5
c/a = 5/1 = 5
αβγ = 2 x 1 x 1 = 2
-d/a = -(-2)/1 = 2
```

Question-2 :-  Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.

Solution :-
```   In general, the cubic polynomial is ax³ + bx² + cx + d.
Now, Given that α = 2, β = -7, γ = -14
α + β + γ = 2 = -b/a = -(-2)/1
αβ + βγ + γα = -7 = c/a = -(7)/1
αβγ = -14 = -d/a = -(14)/1
All coefficients a = 1, b = -2, c = 7, d = 14.
Therefore, cubic polynomial is x³ - 2x² + 7x + 14.
```

Question-3 :-  Obtain all other zeroes of If the zeroes of the polynomial x³ – 3x² + x + 1 are a – b, a, a + b, find a and b.

Solution :-
```   p(x) = x³ – 3x² + x + 1
Zeroes of polynomial are a - b, a, a + b.
In general, the cubic polynomial is px³ + qx² + rx + s.
So, p = 1, q = -3, r = 1, s = 1
Sum of zeroes = a - b + a + a + b = 3a
-q/p = -(-3)/1 = 3
3a = 3
a = 1
Product of zeroes = (a - b) x a x (a + b)
= a x (a² - b²)
= 1 x (1 - b²)
= 1 - b²
-s/p = -(1)/1 = -1
1 - b² = -1
- b² = -1 - 1
- b² = -2
b² = 2
b  = ±√2
All other zeroes are 1, ±√2.
```

Question-4 :-  If two zeroes of the polynomial x⁴ – 6x³ – 26x² + 138x – 35 are 2 ± √3 , find other zeroes.

Solution :-
```    p(x) = x⁴ – 6x³ – 26x² + 138x – 35
Given that two zeroes are 2 + √3 and 2 - √3.
Now, g(x) = (x - 2 - √3)(x - 2 + √3)
= x² + 4 - 4x – 3
= x² - 4x + 1

q(x) = x² - 2x - 35
= x² - 7x + 5x -35
= x(x - 7) + 5(x - 7)
= (x + 5)(x - 7)
x + 5 = 0
x = -5
x - 7 = 0
x = 7
All zeroes of the polynomial are 2 + √3, 2 - √3, -5 and 7.
```

Question-5 :-  If the polynomial x⁴ – 6x³ + 16x² – 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a, find k and a.

Solution :-
```    p(x) = x⁴ – 6x³ + 16x² – 25x + 10
g(x) = x² – 2x + k
r(x) = x + a
p(x) = g(x) x q(x) + r(x)
p(x) - r(x) = g(x) x q(x)
x⁴ – 6x³ + 16x² – 25x + 10 - x - a = (x² – 2x + k) x q(x)
x⁴ – 6x³ + 16x² – 26x + 10 - a = (x² – 2x + k) x q(x)

q(x) = x² – 4x + (8 - k)
r(x) = (-10 + 2k)x + (10 - a - 8k + k²)
-10 + 2k = 0
2k = 10
k = 5
10 - a - 8k + k² = 0
10 - a - 8 x 5 + 5² = 0
10 - a - 40 + 25 = 0
-a - 5 = 0
a = -5
a = -5, k = 5
```
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