 # Polynomial Exercise 2.4 and 2.5|NCERT Solutions for Class 9 Maths

In this page we have NCERT Solutions for Class 9 Maths:Chapter 2 Polynomial for Exercise 2.4 and 2.5 . Hope you like them and do not forget to like , social share and comment at the end of the page.

## 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
We know from remainder theorem, if (x-a) is a factor of polynomial p(x) then p(a)=0
(i) If (x + 1) is a factor of p(x) = x3 + x2 + x + 1
Then   p (-1) must be zero.
Here, p(x) = x3 + x2 + x + 1
p (-1) = (-1)3 + (-1)2 + (-1) + 1
= -1 + 1 - 1 + 1 = 0
Hence, x + 1 is a factor of this polynomial
(ii) If (x + 1) is a factor of p(x) = x4 + x3 + x2 + x + 1
Then p (-1) must be zero.
Here, p(x) = x4 + x3 + x2 + x + 1
p (-1) = (-1)4 + (-1)3 + (-1)2 + (-1) + 1
= 1
As, p (-1) ≠ 0
Hence, x + 1 is not a factor of this polynomial
(iii)If (x + 1) is a factor of polynomial p(x) = x4 + 3x3 + 3x2 + x + 1
Then p (- 1) must be 0.
p (-1) = (-1)4 + 3(-1)3 + 3(-1)2 + (-1) + 1
= 1
As, p (-1) ≠ 0
Hence, x + 1 is not a factor of this polynomial.
(iv) If (x + 1) is a factor of polynomial
p(x) = x3 - x2 - (2 + √2) x + √2
Then p (- 1) must be 0.
p (-1) = (-1)3 - (-1)2 - (2 + √2) (-1) + √2
= -1 - 1 + 2 + √2 + √2
=2√2
As, p (-1) ≠ 0
Hence, x + 1 is not a factor of this polynomial.
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 - 4 x2 + x + 6, g(x) = x - 3
Solution
(i) If g(x) = x + 1 is a factor of given polynomial p(x)
Then, p (- 1) must be zero.
p(x) = 2x3 + x2 - 2x - 1
p (- 1) = 2(-1)3 + (-1)2 - 2(-1) - 1
= 2(- 1) + 1 + 2 - 1 = 0
Hence, g(x) = x + 1 is a factor of given polynomial.
(ii) If g(x) = x + 2 is a factor of given polynomial p(x)
Then, p (- 2) must be 0.
p(x) = x3 +3x2 + 3x + 1
p (-2) = (-2)3 + 3(- 2)2 + 3(- 2) + 1
= -8 + 12 - 6 + 1
= -1
As, p (-2) ≠ 0
Hence g(x) = x + 2 is not a factor of given polynomial.

(iii) If g(x) = x - 3 is a factor of given polynomial p(x)
Then, p (3) must be 0.
p(x) = x3 - 4x2 + x + 6
p (3) = (3)3 - 4(3)2 + 3 + 6
= 27 - 36 + 9 = 0
Hence, g(x) = x - 3 is a factor of given polynomial.

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) If x - 1 is a factor of polynomial p(x) = x2 + x + k, then
p (1) = 0
(1)2 + 1 + k = 0
2 + k = 0
k = - 2
So, value of k is -2.

(ii) If x - 1 is a factor of polynomial p(x) = 2x2 + kx +  √2, then
p(1) = 0
2(1)2 + k(1) + √2 = 0
k = -2 - √2 = -(2 + √2)
So, value of k is -(2 + √2).

(iii) If x - 1 is a factor of polynomial p(x) = kx2 - √2x + 1, then
p (1) = 0
k(1)2 - √2(1) + 1 = 0
k = √2 - 1
So, value of k is √2 - 1.

(iv) If x - 1 is a factor of polynomial p(x) = kx2 - 3x + k, then
p (1) = 0
k(1)2 + 3(1) + k = 0
k - 3 + k = 0
k = 3/2
So, value of k is 3/2.

Question 4
Factorise:
(i) 12x2 + 7x + 1
(ii) 2x2 + 7x + 3
(iii) 6x2 + 5x - 6
(iv) 3x2 - x - 4
Solution
Here we would be using splitting the middle term to factorize the polynomial
To factorise ax2 +bx +c, we should write b as the sum of two numbers whose product is ac
(i) 12x2 + 7x + 1
Here a=12, c=1 and b=7 So 7=3+4 ,3X4=12X1=12
= 12x2 +4x + 3x+ 1
= 4x (3x + 1) + 1 (3x + 1)
= (3x +1) (4x + 1)
(ii) 2x2 + 7x + 3
Here a=2, c=3 and b=7 So b=7=6+1 ,6X1=2X3=6

= 2x2 + 6x + x + 3
= 2x (x + 3) + 1 (x + 3)
= (x + 3) (2x + 1)
(iii) 6x2 + 5x – 6
Here a=6, c=-6 and b=5 So b=5=9+(-4) ,9X (-4) =6X (-6) =-36

= 6x2 + 9x - 4x - 6
= 3x (2x + 3) - 2 (2x + 3)
= (2x + 3) (3x - 2)

(iv) 3x2 - x – 4
Here a=3, c=-4 and b=-1 So b=-1=-4+3, (-)4X3=3X (-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) Let p(x) = x3 - 2x2 - x + 2
Factors of 2 are ±1 and ± 2
By trial method, we find that
p (-1) = 0
So, (x+1) is factor of p(x) Now, Dividend = Divisor × Quotient + Remainder
(x+1) (x2 - 3x + 2)
Factorizing the second part by split middle term method
= (x+1) (x2 - x - 2x + 2)
= (x+1) [x(x-1) -2(x-1)]
= (x+1) (x-1) (x+2)

(ii) Let p(x) = x3 - 3x2 - 9x - 5
Factors of 5 are ±1 and ±5
By trial method, we find that
p (5) = 0
So, (x-5) is factor of p(x) Now, Dividend = Divisor × Quotient + Remainder
(x-5) (x2 + 2x + 1)
Factorizing the second part by split middle term method
= (x-5) (x2 + x + x + 1)
= (x-5) {x(x+1) +1(x+1)}
= (x-5) (x+1) (x+1)
(iii) Let p(x) = x3 + 13x2 + 32x + 20
Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20
By trial method, we find that
p (-1) = 0
So, (x+1) is factor of p(x) Now, Dividend = Divisor × Quotient + Remainder
(x+1) (x2 + 12x + 20)
Factorizing the second part by split middle term method
= (x+1) (x2 + 2x + 10x + 20)
= (x-5) [x(x+2) +10(x+2)]
= (x-5) (x+2) (x+10)
(iv) Let p(y) = 2y3 + y2 - 2y - 1
Factors of ab = 2× (-1) = -2 are ±1 and ±2
By trial method, we find that
p (1) = 0
So, (y-1) is factor of p(y) Now, Dividend = Divisor × Quotient + Remainder
(y-1) (2y2 + 3y + 1)
Factorizing the second part by split middle term method
= (y-1) (2y2 + 2y + y + 1)
= (y-1) [2y(y+1) +1(y+1)]
= (y-1) (2y+1) (y+1)

## Exercise 2.5

Question 1
Use suitable identities to find the following products:
(i) (x + 4) (x + 10)
(ii) (x + 8) (x – 10)
(iii) (3x + 4) (3x – 5)
(iv) (y+ 3/2) (y- 3/2)
(v) (3 - 2x) (3 + 2x)
Solution
(i) Using identity, (x + a) (x + b) = x2 + (a + b) x + ab
Here a = 4 and b = 10
Therefore,
(x + 4) (x + 10) = x2 + (4 + 10)x + (4 × 10)
= x2 + 14x+ 40

(ii) (x + 8) (x – 10)
Using identity, (x + a) (x + b) = x2 + (a + b) x + ab
Here, a = 8 and b = –10
Therefore,
(x + 8) (x – 10) = x2 + {8 +(– 10)}x + {8× (– 10)}
= x2 + (8 – 10) x – 80
= x2 – 2x – 80
(iii) (3x + 4) (3x – 5)
Using identity, (x + a) (x + b) = x2 + (a + b) x + ab
Here, x = 3x, a = 4 and b = -5
Therefore,
(3x + 4) (3x – 5) = (3x) 2 + {4 + (-5)}3x + {4× (-5)}
= 9x2 + 3x (4 - 5) - 20
= 9x2 - 3x - 20
(iv) (y+ 3/2) (y- 3/2)
Using identity, (x + y) (x -y) = x2 - y2
Here, x =y2 and y = 3/2
Therefore,
(y+ 3/2) (y- 3/2) = (y2)- (3/2)2
= y4- 9/4
(v) (3 - 2x) (3 + 2x)
Using identity, (x + y) (x -y) = x2 - y2
Here, x = 3 and y = 2x
Therefore,
(3 - 2x) (3 + 2x) = 32 - (2x)2
=9- 4x2
Question 2
Evaluate the following products without multiplying directly:
(i) 103 × 107
(ii) 95 × 96
(iii) 104 × 96
Solution
(i) 103 × 107 can be written as
= (100 + 3) (100 + 7)
Using identity, (x + a) (x + b) = x2 + (a + b) x + ab
Here, x = 100, a = 3 and b = 7
Therefore,
103 × 107 = (100 + 3) (100 + 7) = (100)2 + (3 + 7)10 + (3 × 7)
= 10000 + 100 + 21
= 10121
(ii) 95 × 96 = (90 + 5) (90 + 4)
Using identity, (x + a) (x + b) = x2 + (a + b) x + ab
Here, x = 90, a = 5 and b = 4
Therefore,

95 × 96 = (90 + 5) (90 + 4) = 902 + 90(5 + 6) + (5 × 6)
= 8100 + (11 × 90) + 30
= 8100 + 990 + 30 = 9120

(iii) 104 × 96 = (100 + 4) (100 - 4)
Using identity, (x + y) (x -y) = x2 - y2
Here, x = 100 and y = 4
Therefore,

104 × 96 = (100 + 4) (100 - 4) = (100)2 -(4)= 10000 - 16 = 9984

Question 3
Factorise the following using appropriate identities:
(i) 9x2 + 6xy + y2
(ii) 4y2 - 4y + 1
(iii) x- y2/100
Solution
(i) 9x2 + 6xy + y2   = (3x) 2 + (2×3x×y) + y2
Using identity, (a + b)2 = a2 + 2ab + b2
Here, a = 3x and b = y
9x2 + 6xy + y2 = (3x) 2 + (2×3x×y) + y= (3x + y)= (3x + y) (3x + y)
(ii) 4y2 - 4y + 1 = (2y)2 - (2×2y×1) + 12
Using identity, (a - b)2 = a2 - 2ab + b2
Here, a = 2y and b = 1
4y2 - 4y + 1 = (2y)2 - (2×2y×1) + 1= (2y - 1)= (2y - 1) (2y - 1)
(iii) x- y2/100 = x- (y/10)2
Using identity, a2 - b2 = (a + b) (a - b)
Here, a = x and b = (y/10)
x- y2/100 = x- (y/10)= (x- y/10) (x+ y/10)
Question 4
Expand each of the following, using suitable identities:
(i) (x + 2y + 4z)2
(ii) (2x – y + z)2
(iii) (–2x + 3y + 2z)2
(iv) (3a – 7b – c)2
(v) (–2x + 5y – 3z)2
(vi) [1/4 a - 1/2 b + 1]2
We will be using the identity
(a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca

(i) (x + 2y + 4z)2
Here, a = x, b = 2y and c = 4z
(x + 2y + 4z)= x2 + (2y)2 + (4z)2 + (2×x×2y) + (2×2y×4z) + (2×4z×x)
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz
(ii) (2x – y + z)2
Here, a = 2x, b = -y and c = z
(2x – y + z)= (2x)2 + (-y)2 + z2 + (2×2x×-y) + (2×-y×z) + (2×z×2x)
= 4x2 + y2 + z2 - 4xy - 2yz + 4xz
(iii) (–2x + 3y + 2z)2
Here, a = -2x, b = 3y and c = 2z
(–2x + 3y + 2z)= (-2x)2 + (3y)2 + (2z)2 + (2×-2x×3y) + (2×3y×2z) + (2×2z×-2x)
= 4x2 + 9y2 + 4z2 - 12xy + 12yz - 8xz
(iv) (3a – 7b – c)2
Here, a = 3a, b = -7band c = -c
(3a – 7b – c)2= (3a)2 + (-7b)2 + (-c)2 + (2×3a×-7b) + (2×-7b×-c) + (2×-c×3a)
= 9a2 + 49b2 + c2 - 42ab + 14bc - 6ac
(v) (–2x + 5y – 3z)2
Here, a = -2x, b = 5y and c = -3z
(–2x + 5y – 3z)2= (-2x)2 + (5y)2 + (-3z)2 + (2×-2x×5y) + (2×5y×-3z) + (2×-3z×-2x)
= 4x2 + 25y2 + 9z2 - 20xy - 30yz + 12xz
(vi) [1/4 a - 1/2 b + 1]2
Here, a = 1/4 a, b = -1/2 band c = 1
[1/4 a - 1/2 b + 1]2= (1/4 a)2 + (-1/2 b)2 + 12 + (2×1/4 a×-1/2 b) + (2×-1/2 b×1) + (2×1×1/4 a)
= 1/16 a2 + 1/4 b2 + 1 - 1/4 ab - b + 1/2 a
Question 5
Factorise:
(i) 4x2 + 9y2 + 16z2 + 12xy - 24yz - 16xz
(ii) 2x2 + y2 + 8z2 - 2√2 xy + 4√2 yz - 8xz

Solution

(i) 4x2 + 9y2 + 16z2 + 12xy - 24yz - 16xz
Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca
4x2 + 9y2 + 16z2 + 12xy - 24yz - 16xz
= (2x)2 + (3y)2 + (-4z)2 + (2×2x×3y) + (2×3y×-4z) + (2×-4z×2x)
= (2x + 3y - 4z)2
= (2x + 3y - 4z) (2x + 3y - 4z)
(ii) 2x2 + y2 + 8z2 - 2√2 xy + 4√2 yz - 8xz
Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca
2x2 + y2 + 8z2 - 2√2 xy + 4√2 yz - 8xz
= (-√2x)2 + (y)2 + (2√2z)2 + (2×-√2x×y) + (2×y×2√2z) + (2×2√2z×-√2x)
= (-√2x + y + 2√2z)2
= (-√2x + y + 2√2z) (-√2x + y + 2√2z)
Question 6
Write the following cubes in expanded form:
(i) (2x + 1)3
(ii) (2a – 3b)3
(iii) [3/2 x + 1]3
(iv) [x - 2/3 y]3
Solution
We will be using the identity
(a + b)= a3 + b3 + 3ab (a + b)
(a - b)= a3 - b3 - 3ab (a - b)

(i) (2x + 1)3
=(2x)3 + 13 + (3×2x×1) (2x + 1)
= 8x3 + 1 + 6x (2x + 1)
= 8x3 + 12x2 + 6x + 1
(ii) (2a – 3b)3
= (2a)3 - (3b)3 - (3×2a×3b) (2a - 3b)
= 8a3 - 27b3 - 18ab (2a - 3b)
= 8a3 - 27b3 - 36a2b + 54ab2
(iii) [3/2 x + 1]3
= (3/2 x)3 + 13 + (3×3/2 x×1) (3/2 x + 1)
= 27/8 x+ 1 + 9/2 x (3/2 x + 1)
= 27/8 x+ 1 + 27/4 x2 + 9/2 x
= 27/8 x+ 27/4 x2 + 9/2 x + 1
(iv) [x - 2/3 y]3 = (x)3 - (2/3 y)3 - (3×x×2/3 y) (x - 2/3 y)
= x3 - 8/27y3 - 2xy (x - 2/3 y)
= x3 - 8/27y3 - 2x2y + 4/3xy2

Question 7
Find the value of the following using suitable identities:
(i) (99)3
(ii) (102)3
(iii) (998)3
We will be using the identity
(a + b)= a3 + b3 + 3ab (a + b)
(a - b)= a3 - b3 - 3ab (a - b)

(i) (99)3 = (100 - 1)3
= (100)3 - 13 - (3×100×1) (100 - 1)
= 1000000 - 1 - 300(100 - 1)
= 1000000 - 1 - 30000 + 300
= 970299
(ii) (102)3 = (100 + 2)3
= (100)3 + 23 + (3×100×2) (100 + 2)
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200
= 1061208
(iii) (998)3
= (1000)3 - 23 - (3×1000×2) (1000 - 2)
= 100000000 - 8 - 6000(1000 - 2)
= 100000000 - 8- 600000 + 12000
= 994011992
Question 8
Factorise each of the following:
(i) 8a3 + b3 + 12a2b + 6ab2
(ii) 8a3 - b3 - 12a2b + 6ab2
(iii) 27 - 125a3 - 135a + 225a2
(iv) 64a3 - 27b3 - 144a2b + 108ab2
(v) 27p3 - 1/216 - 9/2 p2 + 1/4 p

Solution
(i) 8a3 + b3 + 12a2b + 6ab2
Using identity, (a + b)= a3 + b3 + 3a2b + 3ab2
8a3 + b3 + 12a2b + 6ab2
= (2a)3 + b3 + 3(2a)2b + 3(2a) (b)2
= (2a + b)3
= (2a + b) (2a + b) (2a + b)
(ii) 8a3 - b3 - 12a2b + 6ab2
Using identity, (a - b)= a3 - b3 - 3a2b + 3ab2
8a3 - b3 - 12a2b + 6ab2= (2a)3 - b3 - 3(2a)2b + 3(2a) (b)2
= (2a - b)3
= (2a - b) (2a - b) (2a - b)
(iii) 27 - 125a3 - 135a + 225a2
Using identity, (a - b)= a3 - b3 - 3a2b + 3ab2
27 - 125a3 - 135a + 225a2= 33 - (5a)3 - 3(3)2(5a) + 3(3) (5a)2
= (3 - 5a)3
= (3 - 5a) (3 - 5a) (3 - 5a)
(iv) 64a3 - 27b3 - 144a2b + 108ab2
Using identity, (a - b)= a3 - b3 - 3a2b + 3ab2
64a3 - 27b3 - 144a2b + 108ab2= (4a)3 - (3b)3 - 3(4a)2(3b) + 3(4a) (3b)2
= (4a - 3b)3
= (4a - 3b) (4a - 3b) (4a - 3b)
(v) 27p3 - 1/216 - 9/2 p2 + 1/4 p

Using identity, (a - b)= a3 - b3 - 3a2b + 3ab2
27p3 - 1/216 - 9/2 p2 + 1/4 p
= (3p)3 - (1/6)3 - 3(3p)2(1/6) + 3(3p) (1/6)2
= (3p - 1/6)3
= (3p - 1/6) (3p - 1/6) (3p - 1/6)
Question 9
Verify the below identities
(i) x3 + y3 = (x + y) (x2 - xy + y2)
(ii) x3 - y3 = (x - y) (x2 + xy + y2)
Solution

(i) x3 + y3 = (x + y) (x2 - xy + y2)
We know that,
(x + y)= x3 + y3 + 3xy (x + y)
Taking 3xy (x + y) on another side, we get
x3 + y= (x + y)-3xy (x + y)
Taking (x + y) common
= (x + y) [(x + y)2 -3xy]
= (x + y)[(x2 + y+ 2xy) -3xy]
= (x + y)(x2 + y- xy)
(ii) x3 - y3 = (x - y) (x2 + xy + y2 )
We know that,
(x - y)= x3 - y3 - 3xy(x - y)
x3 - y= (x - y)+3xy(x - y)
Taking (x-y) common
= (x - y)[(x - y)2 +3xy]
= (x - y)[(x2 + y- 2xy) +3xy]
= (x + y)(x2 + y+ xy)
Question 10
Factorise each of the following:
(i) 27y3 + 125z3
(ii) 64m3 - 343n3
Solution
(i) 27y3 + 125z3
Using identity proved in the above question x3 + y3 = (x + y) (x2 - xy + y2)
27y3 + 125z3 = (3y)3 + (5z)3
= (3y + 5z) {(3y)2 - (3y)(5z) + (5z)2}
=(3y + 5z) (9y2 - 15yz + 25z)2
(ii) 64m3 - 343n3
Using identity proved in the above question, x3 - y3 = (x - y) (x2 + xy + y2
64m3 - 343n3= (4m)3 - (7n)3
= (4m + 7n) {(4m)2 + (4m) (7n) + (7n)2}
= (4m + 7n) (16m2 + 28mn + 49n)2
Question 11
Factorise: 27x3 + y3 + z3 - 9xyz

Solution
27x3 + y3 + z3 - 9xyz = (3x)3 + y3 + z3 - 3×3xyz
Using identity
x3 + y3 + z3 - 3xyz = (x + y + z) (x2 + y2 + z2 - xy - yz - xz)
Therefore,
27x3 + y3 + z3 - 9xyz
= (3x + y + z) {(3x)2 + y2 + z2 - 3xy - yz - 3xz}
= (3x + y + z) (9x2 + y2 + z2 - 3xy - yz - 3xz)
Question 12
Verify that: x3 + y3 + z3 - 3xyz = 1/2(x + y + z) [(x - y)+ (y - z)+(z - x)2]
Solution
We know that,
x3 + y3 + z3 - 3xyz = (x + y + z) (x2 + y2 + z2 - xy - yz - xz)
ultiplying and dividing by 2 on Right hand side
= 1/2×(x + y + z) 2(x2 + y2 + z2 - xy - yz - xz)
= 1/2(x + y + z) (2x2 + 2y2 + 2z2 - 2xy - 2yz - 2xz)
= 1/2(x + y + z) [(x2 + y2 -2xy) + (y+ z2 - 2yz) + (x2 + z- 2xz)]
= 1/2(x + y + z) [(x - y)+ (y - z)+(z - x)2]
Question 13
If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.
Solution
We know that,
x3 + y3 + z3 - 3xyz = (x + y + z) (x2 + y2 + z2 - xy - yz - xz)
Now put (x + y + z) = 0,
x3 + y3 + z3 - 3xyz = (0) (x2 + y2 + z2 - xy - yz - xz)
x3 + y3 + z3 - 3xyz = 0
Question 14
Without calculating the cubes, find the value of each of the following:
(i) (-12)3 + (7)3 + (5)3
(ii) (28)3 + (–15)3 + (-13)3
From previous question, we know that
If x + y + z = 0, then x3 + y3 + z3 = 3xyz

(i) (-12)3 + (7)3 + (5)3
Let x =-12, y = 7 and z = 5
We observed that, x + y + z = -12 + 7 + 5 = 0

Therefore,
(-12)3 + (7)3 + (5)3 = 3(-12)(7)(5) = -1260
(ii)(28)3 + (–15)3 + (-13)3
Let x =28, y = -15 and z = -13
We observed that, x + y + z = 28 - 15 - 13 = 0

Therefore.
(28)3 + (–15)3 + (-13)3 = 3(28)(-15)(-13) = 16380
Question 15
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) Area: 25a2 - 35a + 12
(ii) Area: 35 y2 + 13y - 12

Solution
(i) Area: 25a2 - 35a + 12

So, factorizing Area
25a2 - 35a + 12
= 25a2 - 15a -20a + 12
= 5a (5a - 3) - 4(5a - 3)
= (5a - 4) (5a - 3)
Possible expression for length = 5a - 4
Possible expression for breadth = 5a - 3
(ii)Area: 35 y2 + 13y - 12
35 y2 + 13y - 12
= 35y2 - 15y + 28y - 12
= 5y (7y - 3) + 4(7y - 3)
= (5y + 4) (7y - 3)
Possible expression for length = (5y + 4)
Possible expression for breadth = (7y - 3)
Question 16
What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume: 3x2 - 12x
(ii) Volume: 12ky2 + 8ky - 20k
(i) Volume: 3x2 - 12x
Volume of Cuboids= Length X Breadth X height
Now, 3x2 - 12x
= 3x (x - 4)
Possible expression for length = 3
Possible expression for breadth = x
Possible expression for height = (x - 4)
(ii) Volume: 12ky2 + 8ky - 20k
Volume of Cuboids= Length X Breadth X height
Now,
12ky2 + 8ky - 20k
= 4k (3y2 + 2y - 5)
= 4k (3y2 +5y - 3y - 5)
= 4k [y (3y +5) - 1(3y + 5)]
= 4k (3y +5) (y - 1)
Possible expression for length = 4k
Possible expression for breadth = (3y +5)
Possible expression for height = (y - 1)

Reference Books for class 9 Math

Given below are the links of some of the reference books for class 9 Math.

1. Mathematics for Class 9 by R D Sharma One of the best book for studying class 9 level mathematics. It has lot of problems to be solved.
2. Secondary School Mathematics for Class 9 by R S Aggarwal This is also as good as R.D. Sharma. Either this or the book by R.D. Sharma will do. I find book R.S. Aggarwal little bit more challenging than the one by R.D. Sharma.
3. Pearson IIT Foundation Series - Maths - Class 9 Buy this book if you want to challenge yourself further and want to prepare for JEE foundation.
4. Pearson IIT Foundation Physics, Chemistry & Maths combo for Class 9 Only buy if you are prepared to study extra topics and want to take your studies a step further. You might need help to understand topics in these books.

You can use above books for extra knowledge and practicing different questions.

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