﻿ Class 9 NCERT Math Solution
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TOPICS
Unit-2(Examples)

Example-1 :-  Find the degree of each of the polynomials given below:
(i) x⁵ – x⁴ + 3,    (ii) 2 – y² – y³ + 2y⁸,    (iii) 2 .

Solution :-
```(i) x⁵ – x⁴ + 3,
The highest power of the variable is 5. So, the degree of the polynomial is 5.

(ii) 2 – y² – y³ + 2y⁸,
The highest power of the variable is 8. So, the degree of the polynomial is 8.

(iii) 2 .
There is no any variable with 2. Therefore, the degree of the polynomial is 0.
```

Example-2 :-  Find the value of each of the following polynomials at the indicated value of variables:
(i) p(x) = 5x² – 3x + 7 at x = 1.
(ii) q(y) = 3y³ – 4y + √11 at y = 2.
(iii) p(t) = 4t⁴ + 5t³ – t² + 6 at t = a.

Solution :-
```(i) p(x) = 5x² – 3x + 7 at x = 1.
p(1) = 5 x (1)² - 3 x 1 + 7
= 5 - 3 + 7
= 12 - 3
= 9

(ii) q(y) = 3y³ – 4y + √11 at y = 2.
q(2) = 3 x (2)³ – 4 x 2 + √11
= 3 x 8 - 8 + √11
= 24 - 8 + √11
= 16 - √11

(iii) p(t) = 4t⁴ + 5t³ – t² + 6 at t = a.
p(a) = 4 x (a)⁴ + 5 x (a)³ – (a)² + 6
= 4a⁴ + 5a³ – a² + 6
```

Example-3 :-  : Check whether –2 and 2 are zeroes of the polynomial x + 2.

Solution :-
```   Let p(x) = x + 2 at x = -2
p(-2) = -2 + 2 = 0

Now, p(x) = x + 2 at x = 2
p(2) = 2 + 2 = 4
So, -2 is zeroe of polynomial x + 2, but 2 is not.
```

Example-4 :-  Find a zero of the polynomial p(x) = 2x + 1.

Solution :-
```  p(x) = 2x + 1. Finding for zero p(x) = 0
0 = 2x + 1
-1 = 2x
-1/2 = x
x = -1/2
The zero of polynomial p(x) = 2x + 1 is -1/2.
```

Example-5 :-  Verify whether 2 and 0 are zeroes of the polynomial x² – 2x.

Solution :-
```   Let p(x) = x² – 2x at x = 2
p(2) = (2)² + 2
= 4 + 2
= 6
Now, p(x) = x² – 2x at x = 0
p(2) = 0² + 2 x 0
= 0 + 0
= 0
So, 0 is zero of polynomial x² – 2x, but 2 is not.
```

Example-6 :-  : Divide p(x) by g(x), where p(x) = x + 3x² – 1 and g(x) = 1 + x.

Solution :-
```   p(x) = x + 3x² – 1 and
g(x) = 1 + x Here, q(x) = 3x - 2 and r(x) = 1
Verification :
g(x) = 0
1 + x = 0
x = -1
Now, x = -1 put in p(x) = x + 3x² – 1
p(-1) = -1 + 3 x (-1)² - 1
= -1 + 3 x 1 - 1
= -1 + 3 - 1
= 3 - 2
= 1 = r(x)
```

Example-7 :-  Divide the polynomial 3x⁴ – 4x³ – 3x – 1 by x – 1.

Solution :-
```   p(x) = 3x⁴ – 4x³ – 3x – 1 and
g(x) = x – 1 Here, q(x) = 3x³ – x² – x – 4 and r(x) = -5
Verification :
g(x) = 0
x - 1 = 0
x = 1
Now, x = 1 put in p(x) = 3x⁴ – 4x³ – 3x – 1
p(1) = 3x⁴ – 4x³ – 3x – 1
= 3 x (1)⁴ - 4 x (1)³ - 3 x 1 - 1
= 3 - 4 - 3 - 1
= 3 - 8
= -5 = r(x)
```

Example-8 :-  Find the remainder obtained on dividing p(x) = x³ + 1 by x + 1.

Solution :-
```   p(x) = x³ + 1 and
g(x) = x + 1 Here, q(x) = x² – x + 1 and r(x) = 0
Verification :
g(x) = 0
x + 1 = 0
x = -1
Now, x = -1 put in p(x) = x³ + 1
p(1) = x³ + 1
= (-1)³ + 1
=  -1 + 1
= 0 = r(x)
```

Example-9 :-  : Find the remainder when x⁴ + x³ – 2x² + x + 1 is divided by x – 1.

Solution :-
```   p(x) = x⁴ + x³ – 2x² + x + 1 and
g(x) = x - 1
Now, g(x) = 0
x - 1 = 0
x = 1
Now, x = 1 put in p(x) = x⁴ + x³ – 2x² + x + 1
p(1) =  (1)⁴ + (1)³ – 2(1)² + 1 + 1
= 1 + 1 - 2 + 1 + 1
= 4 - 2
= 2 = r(x)
So, by the Remainder Theorem, 2 is the remainder when x⁴ + x³ – 2x² + x + 1 is divided by x – 1.
```

Example-10 :-  Check whether the polynomial q(t) = 4t³ + 4t² – t – 1 is a multiple of 2t + 1.

Solution :-
```   q(t) = 4t³ + 4t² – t – 1
g(t) = 2t + 1
Now, g(t) = 0
2t + 1 = 0
2t = -1
t = -1/2
Now, t = -1/2 put in q(t) = 4t³ + 4t² – t – 1
q(-1/2) = 4(-1/2)³ + 4(-1/2)² – (-1/2) – 1
= 4 x (-1/8) + 4 x 1/4 + 1/2 - 1
= -1/2 + 1 + 1/2 - 1
= 0 = r(t)
So the remainder obtained on dividing q(t) by 2t + 1 is 0.
So, 2t + 1 is a factor of the given polynomial q(t), i.e., q(t) is a multiple of 2t + 1.
```

Example-11 :-  Examine whether x + 2 is a factor of x³ + 3x² + 5x + 6 and of 2x + 4.

Solution :-
```   p(x) = x³ + 3x² + 5x + 6 and
g(x) = x + 2
Now, we take g(x) = 0
x + 2 = 0
x = -2
Now, x = -2 put in p(x) = x³ + 3x² + 5x + 6
p(1) = (-2)³ + 3(-2)² + 5 x (-2) + 6
= -8 + 12 - 10 + 6
=  18 - 18
= 0
So, by the Factor Theorem, x + 2 is a factor of x³ + 3x² + 5x + 6.
Again, s(–2) = 2(–2) + 4 = 0.
So, x + 2 is a factor of 2x + 4.
```

Example-12 :-  : Find the value of k, if x – 1 is a factor of 4x³ + 3x² – 4x + k.

Solution :-
```   p(x) = 4x³ + 3x² – 4x + k and
g(x) = x - 1
Now, g(x) = 0
x - 1 = 0
x = 1
Now, x = 1 put in p(x) = 4x³ + 3x² – 4x + k and p(1) = 0
p(1) = 4(1)³ + 3(1)² – 4(1) + k
0  = 4 + 3 - 4 + k
0  = 3 + k
k = -3
```

Example-13 :-  Factorise 6x² + 17x + 5 by splitting the middle term, and by using the Factor Theorem.

Solution :-
```   By splitting method :
If we can find two numbers a and b such that a + b = 17 and ab = 6 × 5 = 30,
then we can get the factors.
So, 6x² + 17x + 5
= 6x² + (2 + 15)x + 5
= 6 x² + 2x + 15x + 5
= 2 (3x + 1) + 5 (3x + 1)
= (3x + 1) (2x + 5)

By Factor method :
6x² + 17x + 5
= 6(x² + 17x/6 + 5/6)
= 6 p(x)
If a and b are the zeroes of p(x), then
So, 6x² + 17x + 5
= 6(x – a)(x – b).
So, ab = 5/6 and factors of 5 = 5 x 1, 6 = 2 x 3 x 1
Let us look at some possibilities for a and b.
They could be ±1/2, ±1/3, ±5/2, ±5/3, ±1.
Now, p(1/2) = 6[(1/2)² + 17/12 + 5/6] ≠ 0.
p(-1/3) = 0 and p(-5/2) = 0.
So, (x + 1/3) and (x + 5/2) are factors of p(x).
Therefore,
= 6(x + 1/3)(x + 5/2)
= 6 x (3x + 1)/3 x (2x + 5)/2
= 2(3x + 1)(2x + 5)/2
= (3x + 1)(2x + 5)
```

Example-14 :-  Factorise y² – 5y + 6 by using the Factor Theorem.

Solution :-
```   By Factor method :
p(x) = y² – 5y + 6
Factor of 6 = 2 x 3 x 1
Some possibilities are ±1, ±2, ±3.
Now, p(1) = (1)² – 5 x (1) + 6 = 1 - 5 + 6 ≠ 0.
p(2) = (2)² - 5 x 2 + 6 = 4 - 10 + 6 = 10 - 10 = 0
p(3) = (3)² - 5 x 3 + 6 = 9 - 15 + 6 = 15 - 15 = 0
There are two factors of p(x). i.e., (y - 2) and (y - 3).
p(x) = y² – 5y + 6
= (y - 2)(y - 3)
```

Example-15 :-  : Factorise x³ – 23x² + 142x – 120.

Solution :-
```   p(x) = x³ – 23x² + 142x – 120
All factors of -120 is  ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±60.
Now,  p(1) = 0. So x – 1 is a factor of p(x).
p(x) = x³ – 23x² + 142x – 120 is divide by x - 1 Now, we can take x³ – 23x² + 142x – 120 = (x - 1)(x² – 22x + 120)
p(x) = (x - 1)(x² – 22x + 120)
= (x - 1)(x² – 12x – 10x + 120)
= (x - 1)[x(x – 12) – 10(x – 12)]
= (x - 1)(x – 12)(x – 10)
So, x³ – 23x² + 142x – 120 = (x – 1)(x – 10)(x – 12)
```

Example-16 :-  Find the following products using appropriate identities:
(i) (x + 3) (x + 3)
(ii) (x – 3) (x + 5)

Solution :-
```(i) (x + 3) (x + 3)
= (x + 3)²               [(a + b)² = a² + b² + 2ab]
= x² + 3² + 2 x X x 3
= x² + 9 + 6x

(ii) (x – 3) (x + 5)
= x² + (-3 + 5)x + (-3)(5)   [(x + a)(x + b) = x² + (a + b)x + ab]
= x² + 2x - 15

```

Example-17 :-  Evaluate 105 × 106 without multiplying directly.

Solution :-
```    105 × 106
= (100 + 5)(100 + 6)
= (100)² + (5 + 6)100 + 5 x 6   [(x + a)(x + b) = x² + (a + b)x + ab]
= 10000 + 11 x 100 + 30
= 10000 + 1100 + 30
= 11130
```

Example-18 :-  : Factorise:
(i) 49a² + 70ab + 25b²
(ii) 25x²/4 - y²/9

Solution :-
```(i) 49a² + 70ab + 25b²
= (7a)² + 2 x 7a x 5b + (5b)²    [(a + b)² = a² + b² + 2ab]
= (7a + 5b)²
= (7a + 5b)(7a + 5b)

(ii) 25x²/4 - y²/9
= (5x/2)² - (y/3)²    [a² - b² = (a + b)(a - b)]
= (5x/2 + y/3)(5x/2 - y/3)
```

Example-19 :-  Write (3a + 4b + 5c)² in expanded form.

Solution :-
```    (3a + 4b + 5c)²
By using [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= (3a)² + (4b)² + (5c)² + 2 x 3a x 4b + 2 x 4b x 5c + 2 x 3a x 5c
= 9a² + 16b² + 25c² + 24ab + 40bc + 30ac
```

Example-20 :-  Expand (4a – 2b – 3c)².

Solution :-
```    (4a – 2b – 3c)²
By using [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= (4a)² + (-2b)² + (-3c)² + 2 x 4a x (-2b) + 2 x (-2b) x (-3c) + 2 x 4a x (-3c)
= 16a² + 4b² + 9c² - 16ab + 12bc - 24ac
```

Example-21 :-  : Factorise 4x² + y² + z² – 4xy – 2yz + 4xz.

Solution :-
```    4x² + y² + z² – 4xy – 2yz + 4xz
By using [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= (2x)² + (-y)² + z² + 2 x (2x) x (-y) + 2 x (-y) x z + 2 x (2x) x z
= (2x - y + z)²
= (2x - y + z)(2x - y + z)
```

Example-22 :-  Write the following cubes in the expanded form:
(i) (3a + 4b)³
(ii) (5p – 3q)³

Solution :-
```(i) (3a + 4b)³
By Using [(a + b)³ = a³ + b³ + 3a²b + 3ab²]
= (3a)³ + (4b)³ + 3 x (3a)² x 4b + 3 x 3a x (4b)²
= 27a³ + 64b³ + 3 x 9a² x 4b + 3 x 3a x 14b²
= 27a³ + 64b³ + 108a²b + 126ab²

(ii) (5p – 3q)³
By Using [(a - b)³ = a³ - b³ - 3a²b + 3ab²]
= (5p)³ - (3q)³ - 3 x (5p)² x 3q + 3 x 5p x (3q)²
= 125p³ - 27q³ - 3 x 25p² x 3q + 3 x 5p x 9q²
= 125p³ - 27q³ - 225p² + 135pq²
```

Example-23 :-  Evaluate each of the following using suitable identities:
(i) (104)³,    (ii) (999)³

Solution :-
```(i) (104)³
= (100 + 4)³         [(a + b)³ = a³ + b³ + 3a²b + 3ab²]
= (100)³ + (4)³ + 3 x (100)² x 4 + 3 x 100 x (4)²
= 1000000 + 64 + 3 x 10000 x 4 + 3 x 100 x 16
= 1000000 + 64 + 120000 + 4800
= 1124864

(ii) (999)³
= (1000 - 1)³         [(a - b)³ = a³ - b³ - 3a²b + 3ab²]
= (1000)³ - (1)³ - 3 x (1000)² x 1 + 3 x 1000 x (1)²
= 1000000000 - 1 - 3 x 1000000 x 1 + 3 x 1000 x 1
= 1000000000 - 1 - 3000000 + 3000
= 1000003000 - 3000001
= 997002999
```

Example-24 :-  : Factorise 8x³ + 27y³ + 36x²y + 54xy².

Solution :-
```     8x³ + 27y³ + 36x²y + 54xy²
By using [(a + b)³ = a³ + b³ + 3a²b + 3ab²]
= (2x)³ + (3y)³ + 3 x (2x)² x 3y + 3 x 2x x (3y)²
= (2x + 3y)³
= (2x + 3y)(2x + 3y)(2x + 3y)
```

Example-25 :-  Factorise : 8x³ + y³ + 27z³ – 18xyz

Solution :-
```    8x³ + y³ + 27z³ – 18xyz
By using [x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx) ]
= (2x)³ + y³ + (3z)³ – 3 x (2x) x (y) x (3z)
= (2 x + y + 3z)[(2x)² + y² + (3z)² – (2x) x (y) – (y) x (3z) – (2x) x (3z)]
= (2 x + y + 3z) (4x² + y² + 9z² – 2xy – 3yz – 6xz)
```
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