NCERT Solutions for Class 9 Math Chapter 2 - Polynomials

Answer:

4x² – 3x + 7 is a polynomial in x because exponent of x in each of the terms is a whole number.

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x¹⁰ + y³ + t⁵⁰ is a polynomial in three variables x, y and t, because exponent of each variable is a whole number.

Answer:

Coefficient of x² in 2 + x² + x is 1.

Answer:

Coefficient of x² in 2 – x² + x³ is –1.

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Note: We can write some more polynomials with different coefficients.

Answer:

Highest power term in p(x) is 5x³ and the exponent is 3. So, the degree is 3.

Answer:

Highest power term in p(y) is – y² and the exponent is 2. So, the degree is 2.

Answer:

Highest power term in f(t) is 5t and the exponent is 1. So, the degree is 1.

Answer:

The only term here is 3 which can be written as 3x⁰ and so the exponent is 0. Therefore, the degree is 0.

Answer:

The polynomial x² + x is of degree 2. Therefore, it is a quadratic polynomial.

Answer:

The polynomial x – x³ is of degree 3. Therefore, it is a cubic polynomial.

Answer:

The polynomial y + y² + 4 is of degree 2. Therefore, it is a quadratic polynomial.

Answer:

Polynomial 1 + x is of degree 1. Therefore, it is a linear polynomial.

Answer:

3t is a polynomial of degree 1. Therefore, it is a linear polynomial.

Answer:

r² is a polynomial of degree 2. Therefore, it is a quadratic polynomial.

Answer:

7x³ is a polynomial of degree 3. Therefore, it is a cubic polynomial.

Answer:

The value of polynomial
p(x) = 5x – 4x² + 3 at x = 0 is given by
= 5(0) – 4(0)² + 3 = 0 – 0 + 3 = 3

Answer:

The value of polynomial
p(x) = 5x – 4x² + 3 at x = –1 is given by
= 5 (–1) – 4 (–1)² + 3
= – 5 – 4 + 3 = –6

Answer:

The value of polynomial
p(x) = 5x – 4x² + 3 at x = 2 is given by
= 5(2) – 4(2)² + 3 = 10 – 4 × 4 + 3 = 10 – 16 + 3 = –3

Answer:

p(y) = y² – y + 1
p(0) = (0)² – 0 + 1
$\Rightarrow$ p(0) = 1
p(1) = (1)² – 1 + 1
$\Rightarrow$ p(1) = 1 – 1 + 1
$\Rightarrow$ p(1) = 1
p(2) = (2)²– 2 + 1
$\Rightarrow$ p(2) = 4 – 2 + 1
$\Rightarrow$ p(2) = 3

Answer:

p(t) = 2 + t + 2t² – t³
p(0) = 2 + 0 + 2(0)² – (0)³
$\Rightarrow$ p(0) = 2 + 0 + 0 – 0
$\Rightarrow$ p(0) = 2
p(1) = 2 + 1 + 2(1)² – (1)³
$\Rightarrow$ p(1) = 2 + 1 + 2 – 1
$\Rightarrow$ p(1) = 4
p(2) = 2 + 2 + 2(2)² – (2)³
$\Rightarrow$ p(2) = 2 + 2 + 8 – 8
$\Rightarrow$ p(2) = 4

Answer:

p(x) = x³
p(0) = (0)3
$\Rightarrow$ p(0) = 0
p(1) = (1)³
$\Rightarrow$ p(1) = 1
p(2) = (2)³
$\Rightarrow$ p(2) = 8

Answer:

p(x) = (x – 1) (x + 1)
p(0) = (0 – 1) (0 + 1)
$\Rightarrow$ p(0) = (– 1) × (1)
$\Rightarrow$ p(0) = –1
p(1) = (1 – 1) (1 + 1)
$\Rightarrow$ p(1) = 0 × 2
$\Rightarrow$ p(1) = 0
p(2) = (2 – 1) (2 + 1)
$\Rightarrow$ p(2) = 1 × 3
$\Rightarrow$ p(2) = 3

Answer:

Answer:

Answer:

Here, p(x) = x² – 1
Substitute x = 1 in p(x)
So, p(1) = (1)² – 1
= 1 – 1
$\Rightarrow$ p(1) = 0
Therefore, x = 1 is the zero of the polynomial
x² – 1. Now substitute x = – 1, in the given
polynomial.
So, p( – 1) = ( – 1)² – 1
= 1 – 1
$\Rightarrow$ p( – 1) = 0
Therefore, x = – 1 is the zero of the
polynomial x² – 1.

Answer:

Here, p(x) = (x + 1) (x – 2)
Substitute x = – 1 in p(x)
So, p( – 1) = ( – 1 + 1) (– 1 – 2)
= 0 ( – 3)
= 0
Therefore, it is verified that x = – 1
is the zero of the polynomial
(x + 1) (x – 2)
Now substitute x = 2 in p(x)
So, p(2) = (2 + 1) (2 – 2)
= 3 × 0
$\Rightarrow$ p(2) = 0
Therefore, it is verified that x = 2 is the zero
of the polynomial (x + 1) (x – 2)

Answer:

Here, p(x) = x²
Substitute x = 0 in p(x)
So, p(0) = (0)²
$\Rightarrow$ p(0) = 0
Therefore, it is verified that x = 0 is the zero
of the polynomial x².

Answer:

Answer:

Answer:

Answer:

As finding the zero of the polynomial p(x) amounts to solve the equation p(x) = 0,
We have, x + 5 = 0
$\Rightarrow$ x = – 5
So, – 5 is the zero of the polynomial x + 5

Answer:

As finding the zero of the polynomial p(x) amounts to solve the equation p(x) = 0
We have, x – 5 = 0
$\Rightarrow$ x = 5
So, 5 is the zero of the polynomial x – 5

Answer:

As finding the zero of the polynomial p(x) amounts to solve the equation p(x) = 0.
We have,
2x + 5 = 0
$\Rightarrow$ 2x = – 5

Answer:

As finding the zero of the polynomial p(x) amounts to solve the equation p(x) = 0.
We have,
3x – 2 = 0
$\Rightarrow$ 3x = 2

Answer:

As finding the zero of the polynomial p(x) amounts to solve the equation p(x) = 0.
We have, 3x = 0

$\Rightarrow$ x = 0
So, 0 is the zero (root) of the polynomial 3x.

Answer:

As finding the zero of the polynomial p(x) amounts to solve the equation p(x) = 0.
We have, ax = 0

Answer:

As finding the zero of the polynomial p(x) amounts to solve the equation p(x) = 0
We have, cx + d = 0
where c $\ne$ 0, c, d are real numbers
$\Rightarrow$ cx = – d

$\Rightarrow$ x = 0
So, 0 is the zero (root) of the polynomial ax.

Answer:

Let p(x) be x³ + 3x² + 3x + 1 and divisor is x + 1.
Remainder by long division method is as follows:

Answer:

Answer:

Let p(x)= x³ + 3x² + 3x + 1
Here, divisor is x + $\mathrm{\pi }$.
We find the remainder by using remainder
theorem.
So, take x + $\mathrm{\pi }$ = 0
$\Rightarrow$ x = – $\mathrm{\pi }$
Now put x = – $\mathrm{\pi }$ in p(x), we get:
p($\mathrm{\pi }$) = (– $\mathrm{\pi }$)³ + 3(– $\mathrm{\pi }$)² + 3(–$\mathrm{\pi }$) + 1
= – $\mathrm{\pi }$³ + 3$\mathrm{\pi }$² – 3$\mathrm{\pi }$ + 1

Answer:

Let p(x) = x³ + 3x² + 3x + 1
Here, divisor is 5 + 2x
We find the remainder by using the remainder theorem

Answer:

Let p(x) = x³ – ax² + 6x – a
Divisor is x – a
So, take x – a = 0
$\Rightarrow$ x = a
put x = a in p(x), we get:
p(a) = a³ – a(a)² + 6a – a
= a³ – a³ + 6a – a
= 5a
$\Rightarrow$ p(a) = 5a
Hence, remaninder is 5a.

Answer:

Answer:

Let p (x) = x³ + x² + x + 1
Put x = – 1 in p (x), we get:
p (–1) = (–1)³ + (–1)² + (–1) + 1
= – 1 + 1 – 1 + 1
= 0
Hence, by factor theorem x + 1 is a factor of
x³ + x² + x + 1.

Answer:

Let p (x) = x⁴ + x³ + x² + x + 1
Put x = –1 in p (x), we get:
p (–1) = (–1)⁴ + (–1)³ + (–1)² + (–1) + 1
= 1 – 1 + 1 – 1 + 1
= 1 ≠ 0
Hence, by factor theorem x + 1 is not a factor
of x⁴ + x³ + x² + x + 1.

Answer:

Let p (x) = x⁴ + 3x³ + 3x² + x + 1
Put x = –1 in p (x), we get:
p (–1) = (–1)⁴ + 3(–1)³ + 3(–1)² + (–1) + 1
= 1 – 3 + 3 – 1 + 1
= 1 ≠ 0
Hence, by factor theorem x + 1 is not a factor
of x⁴ + 3x³ + 3x² + x + 1.

Answer:

Answer:

We have,
p (x) = 2x³ + x² – 2x – 1
and divisor; g (x) = x + 1
Take (x + 1) = 0 $\Rightarrow$ x = –1
Put x = –1 in p (x), we get:
p (–1) = 2 (–1)³ + (–1)² – 2(–1) – 1
= – 2 + 1 + 2 – 1 = 0
Hence, by factor theorem x + 1 i.e. g (x) is
a factor of 2x³ + x² – 2x – 1.

Answer:

We have,
p (x) = x³ + 3x² + 3x + 1
and divisor; g (x) = x + 2
Take x + 2 = 0
$\Rightarrow$ x = –2
Put x = – 2 in p (x), we get:
p (–2) = (–2)³ + 3(–2)² + 3(–2) + 1
= – 8 + 12 – 6 + 1
= – 1 ≠ 0
Hence, by factor theorem x + 2 i.e. g (x) not
a factor of x³ + 3x² + 3x + 1.

Answer:

We have,
p (x) = x³ – 4x² + x + 6
and divisor; g(x) = x – 3
Take x – 3 = 0
$\Rightarrow$ x = 3
Put x = 3 in p (x), we get:
p (3) = (3)³ – 4 (3)² + 3 + 6
= 27 – 36 + 3 + 6
= 36 – 36
= 0
Hence, by factor theorem x – 3, i.e. g (x) is a
factor of x³ – 4x² + x + 6.

Answer:

As x – 1 is a factor of
p (x) = x² + x + k
$\therefore$ By factor theorem, p (1) = 0
$\Rightarrow$ (1)² + 1 + k = 0
$\Rightarrow$ 1 + 1 + k = 0
$\Rightarrow$ 2 + k = 0
$\Rightarrow$ k = –2

Answer:

Answer:

Answer:

As x – 1 is a factor of
p (x) = kx² – 3x + k
$\therefore$ By factor theorem,
p (1) = 0
$\Rightarrow$ k (1)2 – 3(1) + k = 0
$\Rightarrow$ k – 3 + k = 0
$\Rightarrow$ 2k = 3

Answer:

Answer:

2x² + 7x + 3 = 2x² + 6x + x + 3
= 2x(x + 3) + 1(x + 3)
= (x + 3)(2x + 1)

Answer:

6x² + 5x – 6 = 6x² + 9x – 4x – 6
= 3x (2x + 3) – 2 (2x + 3)
= (2x + 3) (3x – 2)

Answer:

3x² – x – 4 = 3x² – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4)
= (3x – 4) (x + 1)

Answer:

Let us denote the given polynomial as p(x) So, p(x) = x³ – 2x² – x + 2
All factors of +2 are $\pm$ 1, $\pm$ 2
By inspection,
p(–1) = (–1)³ –2(–1)² – (–1) + 2
= – 1 – 2 + 1 + 2
$\Rightarrow$ p(–1) = 0
So, by factor theorem (x + 1) is a factor of p(x)
Now divide p(x) by (x + 1)

Thus x³ – 2x²– x + 2
= (x + 1) (x² – 3x + 2)
The factors of x² – 3x + 2 can be solved by
splitting the middle term or using factor
theorem.
By splitting the middle term,
we have
x² – 3x + 2 = x² – x – 2x + 2
= x(x – 1) –2(x – 1)
= (x – 1) (x – 2)
So, x³ – 2x² – x + 2 = (x + 1) (x – 1) (x –2)

Answer:

Let us denote the given polynomial as
p(x) = x³ – 3x² – 9x – 5
All factors of –5 are $\pm$ 1, $\pm$ 5
By inspection,
p(–1) = (–1)3 –3 (–1)2 –9 (–1) – 5
= –1 – 3 + 9 – 5
= 9 – 9
= 0
So, by factor theorem x + 1 is a factor of p(x)
Now divide p(x) by x + 1.

Thus x³ – 3x² – 9x – 5
= (x + 1) (x² – 4x – 5)
Factors of x² – 4x – 5 by splitting the middle
terms are as follows:
x² – 4x – 5 = x² – 5x + x – 5
= x(x – 5) + 1(x –5)
= (x – 5) (x + 1)
So, x³ – 3x² – 9x – 5 = (x + 1) (x – 5) (x +1)
or = (x + 1) (x + 1) (x – 5)

Answer:

Let us denote the given polynomial as
p(x) = x³ + 13x² + 32x + 20
All factors of 20 are
$\pm$ 1, $\pm$ 2, $\pm$ 4, $\pm$ 5, $\pm$ 10, $\pm$ 20
By inspection,
p(–1) = (–1)³ + 13(–1)² + 32(–1) + 20
= – 1 + 13 – 32 + 20
= 33 – 33
$\Rightarrow$ p(–1) = 0
So, by factor theorem x + 1 is a factor of p(x).
Now divide p(x) by (x + 1)

Thus x³ + 13x² + 32x + 20
= (x + 1) (x² + 12x + 20)
Factors of x² + 12x + 20 by splitting the
middle terms are as follows:
x² + 12x + 20 = x² + 2x + 10x + 20
= x(x + 2) + 10 (x + 2)
= (x + 2) (x + 10)
So, x³ + 13x² + 32 x + 20
= (x + 1) (x + 2) (x + 10)

Answer:

Let us denote the given polynomial as
p(y) = 2y³ + y² – 2y – 1
All factors of –1 are $\pm$ 1
By inspection,
p(1) = 2(1)³ + (1)² – 2(1) – 1
= 2 + 1 – 2 – 1
= 3 – 3
$\Rightarrow$ p(1) = 0
So, by factor theorem; y – 1 is a factor of
p(y).
Now divide p(y) by y – 1.

Thus 2y³ + y² – 2y – 1
= (y – 1) (2y² + 3y + 1)
Factors of 2y² + 3y + 1 by splitting the middle
terms are as follows:
2y² + 3y + 1 = 2y² + 2y + y + 1
= 2y (y + 1) + 1 (y + 1)
= (y + 1) (2y + 1)
So, 2y³ + y² – 2y – 1
= (y – 1) (y + 1) (2y + 1)

Answer:

(x + 4) (x + 10)
= x² + (4 + 10)x + 4 × 10
[Using the identity (x + a) (x + b) = x² +
(a + b)x + ab. Here a = 4, b = 10]
= x² + 14x + 40

Answer:

(x + 8) (x – 10) =x² + {8 + ( – 10)} x + 8 × ( – 10) =x² – 2x – 80 [Using the identity (x + a) (x + b)= x² + (a + b) x + ab. Here a = 8, b = – 10]

Answer:

(3x + 4) (3x – 5)
Put 3x = y, we get:
(y + 4) (y – 5)
= y² + {4 + ( – 5)} y + 4 (–5)
[Using the identity(x + a) (x + b) = x²
+ (a + b) x + ab. Here a = 4, b = – 5]
= y² – y – 20
= (3x)² – 3x – 20 [$\because$ 3x = y]
= 9x² – 3x – 20

Answer:

Answer:

(3 – 2x) (3 + 2x)
or = – (2x – 3) (2x + 3)
or = – (2x + 3) (2x – 3)
Put 2x = y, we get:
– (y + 3) (y – 3)
= – [y2 + (3 – 3)y + 3 ( – 3)]
[Using the identity (x + a) (x + b) = x² +
(a + b)x + ab. Here a = 3, b = – 3]
= – (y² + 0y – 9)
= – (y² – 9)
= – [(2x)² – 9]
(where 2x = y as substituted above)
= – (4x² – 9) = 9 – 4x²

Answer:

103 × 107
= (100 + 3) (100 + 7)
[Writing 103 as 100 + 3 and
107 as 100 + 7]
= (100)² + (3 + 7) 100 + 3 × 7
= 10000 + 10 × 100 + 21
= 10000 + 1000 + 21
= 11021 [Using the identity (x + a) (x + b)
= x² + (a + b)x + ab. Here
x = 100, a = 3, b = 7]

Answer:

95 × 96 = (100 – 5) (100 – 4)
[Writing 95 as 100 – 5 and 96 as 100 – 4]
= (100)² + [( – 5) + ( – 4)] 100 + ( – 5) ( – 4)
[Using the identity (x + a) (x + b) = x² +
(a + b) x + ab. Here x = 100, a = – 5, b = – 4]
= 10000 + (– 9) × 100 + 20
= 10000 – 900 + 20
= 9120

Answer:

104 × 96
= (100 + 4) (100 – 4)
[Writing 104 as 100 + 4 and 96 as 100 – 4]
(100)² – (4)²
[Using identity a² – b² = (a + b) (a – b).
Here a = 100 and b = 4]
= 10000 – 16
= 9984

Answer:

9x² + 6xy + y²
=(3x)² + 2(3x)y + y²
=(3x + y)²
[Using identity (a + b)² = a² + 2ab + b².
Here a = 3x and b = y]

Answer:

4y² – 4y + 1
= (2y)² – 2(2y) 1 + 1²
= (2y – 1)²
[Using the identity (a – b)² = a² – 2ab + b².
Here a = 2y and b = 1]

Answer:

Answer:

(x + 2y + 4z)²
Comparing the given expression with
(a + b + c)² we find that a = x, b = 2y
and c = 4z
Therefore using the identity
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
we write (x + 2y + 4z)² = x² + (2y)² + (4z)² +
2x(2y) + 2(2y) (4z) + 2(4z)x
= x² + 4y² + 16z² + 4xy + 16yz + 8xz

Answer:

(2x – y + z)²
= [2x + ( – y) + z]²
Comparing the given expression with
(a + b + c)² we find that a = 2x, b = – y
and c = z.
Therefore using the identity
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
we write
{(2x + (– y) + z)²} = (2x)² + (–y)² + z²
+ 2(2x) (– y) + 2( –y) z + 2z (2x)
= 4x² + y² + z² – 4xy – 2yz + 4zx

Answer:

(– 2x + 3y + 2z)²
Comparing the given expression with
(a + b + c)² we find that a = – 2x, b = 3y and
c = 2z
Therefore using the identity
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
we write
(– 2x + 3y + 2z)² = {(– 2x)² + (3y)² + (2z)²
+ 2( – 2x) (3y) + 2 (3y) (2z) + 2(2z) (– 2x)}
= 4x² + 9y² + 4z² – 12xy + 12yz – 8zx.

Answer:

(3a – 7b – c)²
= [(3a) + (–7b) + (–c)]²
Comparing the given expression with
(a + b + c)² we find that a = 3a, b = –7b, and
c = –c
Using the identity
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
we get:
[3a + (–7b) + (–c)]² = (3a)² + (–7b)² + (–c)²
+ 2 (3a) (–7b) + 2 (–7b) (–c) + 2(–c) (3a)
= 9a² + 49b² + c² – 42ab + 14bc – 6ca

Answer:

(–2x + 5y – 3z)² = [–2x + 5y + (–3z)]²
Comparing the given expression with
(a + b + c)² we find that a = –2x, b = 5y and c = –3z
Therefore using the identity
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
we write
[–2x + 5y + (–3z)]²
= (–2x)² + (5y)² + (–3z)² + 2(–2x) (5y)
+ 2(5y) (–3z) + 2 (–3z) (–2x)
= 4x² + 25y² + 9z² – 20xy – 30yz + 12zx

Answer:

Answer:

Answer:

Answer:

(2x + 1)³
Compare the expression (2x + 1)³ with (a + b)³,
we find that a = 2x, b = 1
Therefore using the identity
(a + b)³ = a³ + b³ + 3ab (a + b)
we have:
(2x + 1)³ = (2x)³ + (1)³ + 3(2x) 1 (2x + 1)
= 8x³ + 1 + 12x² + 6x
= 8x³ + 12x² + 6x + 1
[Re-arranging in descending power of x.]

Answer:

(2a – 3b)³
Comparing the expression (2a – 3b)³
with (x – y)³ we find that x = 2a, y = 3b
Therefore using the identity
(x – y)³=x³ – y³ – 3xy (x – y), we get:
(2a – 3b)³=(2a)³ – (3b)³ –3 (2a) (3b) (2a – 3b)
=8a³ – 27b³ – 36a²b + 54ab²

Answer:

Answer:

Answer:

(99)³ = (100 – 1)³
= (100)³ – 1³ – 3(100) 1(100 – 1)
[Using the identity
(a – b)³ = a³ – b³ – 3ab (a – b)] = 1000000 – 1 – 30000 + 300
= 970299

Answer:

(102)³= (100 + 2)³
= (100)³ + 2³ + 3(100) (2) (100 + 2)
[Using the identity
(a + b) ³ = a³ + b³ + 3ab (a + b)]
= 1000000 + 8 + 60000 + 1200
= 1061208

Answer:

(998)³= (1000 – 2)³
= (1000)³ – 2³ – 3(1000) (2)(1000 – 2)
[Using the identity
(a – b)³ = a³ – b³ – 3ab (a – b)]
= 1000000000 – 8 – 6000000 + 12000
= 1000012000 – 6000008
= 994011992

Answer:

8a³ + b³ + 12a²b + 6ab²
= (2a)³ + b³ + 3 (2a) b (2a + b)
= (2a + b)³
[Using the identity
(x + y)³ = x³ + y³ + 3xy (x + y).
Here x = 2a and y = b]

Answer:

8a³ – b³ – 12a²b + 6ab²
= (2a)³ – b³ – 3 (2a) b (2a – b)
= (2a – b)³
[Using the identity
(x – y)³ = x³ – y³ – 3xy (x – y).
Here x = 2a and y = b]

Answer:

27 – 125a³ – 135a + 225a²
= (3)³ – (5a)³ – 3 (3) (5a) (3 – 5a)
= (3 – 5a)³
[Using the identity
(x – y)³ = x³ – y³ – 3xy (x – y).
Here x = 3, y = 5a]

Answer:

64a³ – 27b³ – 144a²b + 108ab²
= (4a)³ – (3b)³ – 3(4a) (3b)
(4a – 3b)
= (4a – 3b)³
[Using the identity
(x – y)³ = x³ – y³ – 3xy (x – y).
Here x = 4a, y = 3b]

Answer:

Answer:

Answer:

Answer:

27y³ + 125z³
= (3y)³ + (5z)³
= (3y + 5z) [(3y)² – (3y) (5z) + (5z)²]
[Using identity
a³ + b³ = (a + b) (a² – ab + b²)
Here a = 3y, b = 5z]
= (3y + 5z) (9y² – 15yz + 25z²)

Answer:

64m³ – 343n³
= (4m)³ – (7n)³
= (4m – 7n) [(4m)² + (4m) (7n) + (7n)²]
[Using identity
a³ – b³ = (a – b) (a² + ab + b²).
Here a = 4m, b = 7n]
= (4m – 7n) (16m² + 28mn + 49n²)

Answer:

Answer:

Answer:

Answer:

Let a = –12, b = 7 and c = 5
When a + b + c = 0
then a³ + b³ + c³ = 3abc
Here a + b + c = – 12 + 7 + 5 = 0
$\therefore$(–12)³ + (7)³ + (5)³
= 3(–12) (7) (5)
= –1260

Answer:

Let a = 28, b = –15 and c = –13
Now a + b + c = 28 + (–15) + (–13)
= 28 – 15 – 13
= 28 – 28
= 0
$\therefore$a³ + b³ + c³ = 3abc
$\Rightarrow$ (28)³ + (–15)³ + (–13)³
= 3(28) (–15) (–13)
= 16380

Answer:

Area of rectangle
= 25a² – 35a + 12 (given)
$\Rightarrow$ length × breadth = 25a² – 15a – 20a + 12
= 5a (5a – 3) – 4 (5a – 3)
= (5a – 4) (5a – 3)
If length = (5a – 4)
then breadth = 5a – 3
and If length = 5a – 3
then breadth = (5a – 4)

Answer:

Area of rectangle = 35y² + 13y – 12
$\Rightarrow$ length × breadth = 35y² + 28y – 15y – 12
= 7y(5y + 4) – 3(5y + 4)
$\Rightarrow$ Length × Breadth= (5y + 4) (7y – 3)
If length = 5y + 4
then breadth = 7y – 3
and If length = 7y – 3
then breadth = 5y + 4

Answer:

Volume of cuboid = 3x² – 12x (given)
$\Rightarrow$ Length × Breadth × Height = 3x (x – 4)
$\therefore$ One possible dimensions of cuboid are
3, x and x – 4

Answer:

Volume of cuboid = 12ky² + 8ky – 20k
$\Rightarrow$ Length × Breadth × Height
= 4k (3y² + 2y – 5)
= 4k [3y² + 5y – 3y – 5]
= 4k [y(3y + 5) – 1 (3y + 5)]
= 4k [(3y + 5) (y – 1)]
$\Rightarrow$ Length × Breadth × Height
= 4k (3y + 5) (y – 1)
$\therefore$ One possible dimensions of cuboid are
4k, (3y + 5) and (y –1).