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TOPICS
Exercise - 2.4

Question-1 :-  Determine which of the following polynomials has (x + 1) a factor :
(i) x³ + x² + x + 1,    (ii) x⁴ + x³ + x² + x + 1,    (iii) x⁴ + 3x³ + 3x² + x + 1,    (iv) x³ - x² - (2 + √2)x + √2.

Solution :-
```
(i) x³ + x² + x + 1
Let x + 1 is a factor of given polynomial.
Now, x + 1 = 0
x = -1
p(-1) = (-1)³ + (-1)² + (-1) + 1
= -1 + 1 - 1 + 1
= 2 - 2 = 0
Therefore, It is confirmed that x + 1 is a factor of x³ + x² + x + 1.
```
```
(ii) x⁴ + x³ + x² + x + 1
Let x + 1 is a factor of given polynomial.
Now, x + 1 = 0
x = -1
p(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1) + 1
= 1 - 1 + 1 - 1 + 1
= 3 - 2 = 1
Therefore, x + 1 is not a factor of x⁴ + x³ + x² + x + 1.
```
```
(iii) x⁴ + 3x³ + 3x² + x + 1
Let x + 1 is a factor of given polynomial.
Now, x + 1 = 0
x = -1
p(-1) = (-1)⁴ + 3 x (-1)³ + 3 x (-1)² + (-1) + 1
= 1 - 3 + 3 - 1 + 1
= 5 - 4 = 1
Therefore, x + 1 is not a factor of x⁴ + 3x³ + 3x² + x + 1.
```
```
(iv) x³ - x² - (2 + √2)x + √2
Let x + 1 is a factor of given polynomial.
Now, x + 1 = 0
x = -1
p(-1) = (-1)³ - (-1)² - (2 + √2) x (-1) + √2
= -1 - 1 + 2 + √2 + √2
= -2 + 2 + 2√2
= 2√2
Therefore, x + 1 is not a factor of x³ - x² - (2 + √2)x + √2.
```

Question-2 :-  Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1,
(ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2,
(iii) p(x) = x³ – 4x² + x + 6, g(x) = x – 3 .

Solution :-
```(i) p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1,
By factor theorem (x + 1) is a factor of 2x³ + x² – 2x – 1.
If x + 1 = 0
x = -1
p(-1) = 2(-1)³ + (-1)² – 2(-1) – 1
= -2 + 1 + 2 - 1
= 3 - 3 = 0
Therefore, g(x) = x + 1 is a factor of p(x) = 2x³ + x² – 2x – 1.
```
```
(ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2
By factor theorem (x + 2) is a factor of x³ + 3x² + 3x + 1.
If x + 2 = 0
x = -2
p(-1) = 2 x (-2)³ + 3 x (-2)² + 3 x (-2) + 1
= 2 x (-8) + 3 x 4 - 6 + 1
= -16 + 12 - 6 + 1
= -22 + 13 = -9
Therefore, g(x) = x + 2 is not a factor of p(x) = x³ + 3x² + 3x + 1.
```
```
(iii) p(x) = x³ – 4x² + x + 6, g(x) = x – 3
By factor theorem (x – 3) is a factor of x³ – 4x² + x + 6.
If x – 3 = 0
x = 3
p(3) = (3)³ – 4 x (3)² + 3 + 6
= 27 - 36 + 9
= 36 - 36 = 0
Therefore, g(x) = x – 3 is a factor of p(x) = x³ – 4x² + x + 6.
```

Question-3 :-  Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:
(i) p(x) = x² + x + k,    (ii) p(x) = 2x² + kx + √2,    (iii) p(x) = kx² – √2x + 1,    (iv) p(x) = kx² – 3x + k.

Solution :-
```(i) p(x) = x² + x + k , k = ?
Given that x - 1 is a factor of p(x).
So, x - 1 = 0
x = 1
And p(1) = 0.
p(1) = (1)² + 1 + k
0  = 1 + 1 + k
0  = 2 + k
k  = -2
```
```
(ii) p(x) = 2x² + kx + √2, k = ?
Given that x - 1 is a factor of p(x).
So, x - 1 = 0
x = 1
And p(1) = 0.
p(1) = 2 x (1)² + k x 1 + √2
0  = 2 + k + √2
0  = 2 + √2 + k
k  = -2 - √2
k  = -(2 + √2)
```
```
(iii) p(x) = kx² – √2x + 1, k = ?
Given that x - 1 is a factor of p(x).
So, x - 1 = 0
x = 1
And p(1) = 0.
p(1) = k x (1)² – √2 x 1 + 1
0  = k - √2 + 1
0  = -√2 + 1 + k
k  = √2 - 1
```
```
(iv) p(x) = kx² – 3x + k, k = ?
Given that x - 1 is a factor of p(x).
So, x - 1 = 0
x = 1
And p(1) = 0.
p(1) = k x (1)² - 3 + k
0  = k - 3 + k
0  = -3 + 2k
2k = 3
k = 3/2
```

Question-4 :-  Factorise :
(i) 12x² – 7x + 1,
(ii) 2x² + 7x + 3,
(iii) 6x² + 5x – 6,
(iv) 3x² – x – 4.

Solution :-
```
(i) 12x² – 7x + 1
By Spiliting method :   [x² + (a + b)x + ab]
a + b = -7 , ab = 12 x 1 = 12
= 12x² – 4x - 3x + 1
= 4x(3x - 1) - 1(3x - 1)
= (3x - 1)(4x - 1)
```
```
(ii) 2x² + 7x + 3
By Spiliting method :   [x² + (a + b)x + ab]
a + b = 7 , ab = 2 x 3 = 6
= 2x² + 6x + x + 3
= 2x(x + 3) + 1(x + 3)
= (x + 3)(2x + 1)
```
```
(iii) 6x² + 5x – 6
By Spiliting method :   [x² + (a + b)x + ab]
a + b = 5 , ab = 6 x (-6) = -36
= 6x² + 9x - 4x - 6
= 3x(2x + 3) - 2(2x + 3)
= (2x + 3)(3x - 2)
```
```
(iv) 3x² – x – 4
By Spiliting method :   [x² + (a + b)x + ab]
a + b = -1 , ab = 3 x (-4) = -12
= 3x² – 4x + 3x - 4
= x(3x - 4) + 1(3x - 4)
= (3x - 4)(x + 1)
```

Question-5 :- Factorise :
(i) x³ – 2x² – x + 2,
(ii) x³ – 3x² – 9x – 5,
(iii) x³ + 13x² + 32x + 20,
(iv) 2y³ + y² – 2y – 1

Solution :-
```(i) x³ – 2x² – x + 2
Firstly, the factor of 2 is ±1, ±2.
p(1) = (1)³ – 2(1)² – 1 + 2
= 1 - 2 - 1 + 2
= 2 - 2 = 0
Then, x = 1 and x - 1 is a factor of x³ – 2x² – x + 2.
Now, x³ – 2x² – x + 2 is divide by x - 1

p(x) = x³ – 2x² – x + 2
= (x - 1)(x² – x - 2)
= (x - 1)(x² – 2x + x - 2)
= (x - 1)[x(x - 2) + 1(x - 2)]
= (x - 1)(x - 2)(x + 1)
```
```
(ii) x³ – 3x² – 9x – 5
Firstly, the factor of 5 is ±1, ±5.
p(1) = (1)³ – 3(1)² – 9 x 1 - 5
= 1 - 3 - 9 - 5
= 6 - 17 = -11 ≠ 0
p(-1)= (-1)³ – 3 x (-1)² – 9 x (-1) - 5
= -1 - 3 + 9 - 5
= 9 - 9 = 0
Then, x = -1 and x + 1 is a factor of x³ – 3x² – 9x - 5.
Now, x³ – 3x² – 9x - 5 is divide by x + 1

p(x) = x³ – 3x² – 9x - 5
= (x + 1)(x² – 4x - 5)
= (x + 1)(x² – 5x + x - 5)
= (x + 1)[x(x - 5) + 1(x - 5)]
= (x + 1)(x - 5)(x + 1)
```
```
(iii) x³ + 13x² + 32x + 20
Firstly, the factor of 20 is ±1, ±2, ±4, ±5.
p(-1)= (-1)³ + 13 x (-1)² + 32 x (-1) + 20
= -1 + 13 - 32 + 20
= 33 - 33 = 0
Then, x = -1 and x + 1 is a factor of x³ + 13x² + 32x + 20.
Now, x³ + 13x² + 32x + 20 is divide by x + 1

p(x) = x³ + 13x² + 32x + 20
= (x + 1)(x² + 12x + 20)
= (x + 1)(x² + 10x + 2x + 20)
= (x + 1)[x(x + 10) + 2(x + 10)]
= (x + 1)(x + 10)(x + 2)
```
```
(iv) 2y³ + y² – 2y – 1
Firstly, the factor of 1 is ±1.
p(1) = 2(1)³ + (1)² – 2 x 1 - 1
= 2 + 1 - 2 - 1
= 3 - 3 = 0
Then, y = 1 and y - 1 is a factor of 2y³ + y² – 2y – 1.
Now, 2y³ + y² – 2y – 1 is divide by y - 1

p(y) = 2y³ + y² – 2y – 1
= (y - 1)(2y² + 3y + 1)
= (y - 1)(2y² + 2y + y + 1)
= (y - 1)[2y(y + 1) + 1(y + 1)]
= (y - 1)(2y + 1)(y + 1)
```
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