Algebra

The coefficient of a variable is a multiplicative factor or factors.

In the above question coefficient of x⁴ is its factor i.e. .

The coefficient of a variable is a multiplicative factor or factors.

In the above question coefficient of xy² is its factor i.e. + 14.

The monomials in the polynomial are called the terms. The highest power of the terms is the degree of the polynomial.

It is a polynomial in variables x,y and z

The power of is 3, power of is 2 and the power of is 2

So the power of x³ y² z² is 3 + 2 + 2 = 7

The monomials in the polynomial are called the terms. The highest power of the terms is the degree of the polynomial.

x² – 5x⁴ + x⁷ – 73x + 5 is a polynomial in variable x. Here we have 5 monomials x², – 5x⁴, + , x⁷, –73x, and + 5 which are called the terms of the polynomial.

The highest power is 7 so the degree of the polynomial is 7

x² – 5x²y³ + 30x³ y⁴ – 576xy is a polynomial in variable x and y.

Term 1: x² variable x, power of x is 2. Hence the power of the term is 2.

Term 2: – 5x²y³ the variables are x and y; the power of x is 2 and the power of y is 3.

Hence the power of the term –5x²y³ is 2 + 3 = 5 [Sum of the exponents of variables x and y ].

Term 3: 30x³y⁴ the variables are x and y; the power of x is 3 and the power of y is 4. Hence the power of the term 30x³y⁴ is 3 + 4 = 7 [Sum of the exponents of variables x and y ].

Term 4: – 576xy the variables are x and y; the power of x is 1 and the power of y is 1. Hence the power of the term -576xy is 1 + 1 = 2 [Sum of the exponents of variables x and y ].

So the highest power is 7,hence the degree of the polynomial is 7.

Expression that contains only one term is called a monomial.

Expression that contains two terms is called a binomial.

Expression that contains three terms is called a trinomial.

Expression that contains one or more terms with non-zero coefficient is called a polynomial.

.

i. 3abc - 5ca

The terms are 3ab and -5ca

Coefficient of 3ab = 3

Coefficient of -5ca = -5

ii. 1 + x + y²

There are 3 terms 1,x and y²

Coefficient of 1 = 1

Coefficient of x = 1

Coefficient of y² = 1

iii. 3x² y² - 3xyz + z³

There are 3 terms 3x² y², - 3xyz, and + z³

Coefficient of 3x² y² = 3

Coefficient of- 3xyz = -3

Coefficient of + z³ = 1

iv. -7 + 2pq – qr + rp

v. 0.3xy

Thus, we can conclude

3x²

Expression that contains only one term is called a monomial.

The above question has only one term i.e. 3, so it is monomial.

3x + 2

Expression that contains two terms is called a binomial.

The above question has two terms i.e. 3x and + 2 , so it is binomial.

x² – 4x + 2

Expression that contains three terms is called a trinomial.

The above question has 3 terms i.e. , -4x and + 2, so it is trinomial.

x⁵ – 7

Expression that contains two terms is called a binomial.

The above question has two terms i.e. and -7, so it is a binomial.

x² + 3xy + y²

Expression that contains three terms is called a trinomial.

The above question has 3 terms i.e., + 3xy and , so it is trinomial.

s² + 3st – 2t²

Expression that contains three terms is called a trinomial.

The above question has 3 terms i.e., + 3st and -, so it is trinomial.

xy + yz + zx

Expression that contains three terms is called a trinomial.

The above question has 3 terms i.e. xy, + yz and + zx, so it is trinomial.

a²b + b²c

Expression that contains two terms is called a binomial.

The above question has 2 terms i.e. and + , so it is a binomial.

2l + 2m

Expression that contains two terms is called a binomial.

The above question has 2 terms i.e. 2l and + 2m, so it is a binomial.

i. 2x² + 3x + 5 and 3x² - 4x -7

Column method of addition

5 - x -2

ii. -2 x -3 and + 3x + 1

Column method of addition

+ x -2

iii. + t -4 and

Row method of addition

( + t -4) + ()

Now combine the like terms

= () + (t-3t) + (-4 + 1)

= - 2t - 3

iv. xy - yz, yz - xz and zx - xy

(xy-yz) + (yz-xz) + ( zx-xy)

Now combine the like terms

= (xy-xy) + (-yz + yz) + ( -xz + zx)

= 0 + 0 + 0

= 0

V. + , + , + and 2ab + 2bc + 2ca

Observe we have written the term- of the second polynomial below the corresponding term of the first polynomial. Since the term in the second polynomial and the term 2ab + 2bc + 2ca in the fourth polynomial do not exist, so their respective places have been left blank to facilitate the process of addition.

i. Subtract 2 a – b from 3a – b

Answer = a

ii. Subtract -3x + 8y from -7x - 10y

Answer = -4x-18y

iii. Subtract 2ab + 5bc - 3ca from 7ab - 2bc + 10ca

Solution:

iv. Subtract x⁵ - 2x² - 3x from x³ - 3x² + 1

Answer = - x⁵ + x³ - x² + 3x + 1

v. Subtract 3x²y - 2xy + 2xy² + 5x -7y-10 from 15-2x + 5y-11xy + 2xy² + 8x²y

Row method of subtraction

= (15-2x + 5y-11xy + 2xy² + 8x²y) – (3x²y - 2xy + 2xy² + 5x -7y-10)

= 15-2x + 5y-11xy + 2xy² + 8x²y - 3x²y + 2xy - 2xy² - 5x + 7y + 10

Now combining the like terms

= (2xy²-2xy²) + (8x²y- 3x²y) + (-11xy + 2xy) + (-2x-5x) + (5y + 7y) + (15 + 10)

= 5 x²y + (-9xy) + (-7x) + 12y + 25

= 5 x²y – 9xy – 7x + 12y + 25

Answer = 5 x²y – 9xy – 7x + 12y + 25

(i) The monomials in the polynomial are called the terms. The highest power of the terms is the degree of the polynomial.

x² – 2x³ + 5x⁷ – x³ – 70x – 8 is a polynomial in x. Here we have 6 monomials x², – 2x³, + 5x⁷, –x³, –70x and –8 which are called the terms of the polynomial.

The highest power is 7 so the degree of the polynomial is 7.

(ii) 13x³ – x¹³ – 113 is a polynomial in x. Here we have 3 monomials and the highest power is 13 so the degree of the polynomial is 13.

(iii) -77 + 7x² – x⁷ is a polynomial in x. Here we have 3 monomials and the highest power is 7 so the degree of the polynomial is 7.

(iv) -181 + 0.8y – 8y² + 115y³ + y⁸ is a polynomial in x. Here we have 5 monomials and the highest power is 8 so the degree of the polynomial is 8.

(v) x⁷ – 2x³y⁵ + 3xy⁴ – 10xy + 10 is a polynomial in x and y, Here we have 5 monomials.

Term 1: x⁷ variable x, power of x is 7. Hence the power of the term is 7.

Term 2: – 2x³y⁵ the variables are x and y; the power of x is 3 and the power of y is 5.

Hence the power of the term – 2x³y⁵ is 3 + 5 = 8 [Sum of the exponents of variables x and y ].

Term 3: 3xy⁴ the variables are x and y; the power of x is 1 and the power of y is 4.

Hence the power of the term 3xy⁴ is 1 + 4 = 5 [Sum of the exponents of variables x and y].

Term 4: – 10xy the variables are x and y; the power of x is 1 and the power of y is 1.

Hence the power of the term -10xy is 1 + 1 = 2 [Sum of the exponents of variables x and y].

Term 5: 10 the constant term and it can be written as 10x⁰y⁰. The power of the variables x⁰y⁰ is zero. Hence the power of the term 10 is 0.

So the highest power is 8, hence the degree of the polynomial is 8.

i. 3, 7x

= 3×7x

= 21x

ii. – 7x, 3y

= -7 ×3×x ×y

= -21xy

iii. – 3a, 5ab

= -3 ×5 ×a ×a ×b

= -15 b

= -15

= -15 b

iv. 5a², – 4a

= (5

= -20)

= -20

v.

= )

= ( )

= ( )

=

vi. Xy², x²y

= y^{2 + 1}

=

vii. x³y⁵, xy²

=

=

viii. abc, abc

= a× b× c × a × b × c

= (a × a) × (b × b) x ( c × c)

= × ×

=

ix. xyz. X²yz

= ( x × x² ) x (y × y) × (z× z)

= x³

x. a²b²c³, abc²

= (a² × a) x ( b² × b)x ( c³ × c)

= × ×

For finding B1A2

2X x -3y

= -6xy

B1A4

2X x -5xy

= -10 y

= -10 y

B2A1

-3Y x 2x

= -6xy

B2A6

-3Y x -6x²y²

= 18 x (y^{(1 + 2)}) x x²

= 18 y³ x²

B3A3

4x² x 4x²

= 16x^{(2 + 2)}

= 16x⁴

B3A5

4x² x 7x²y

= (4x7) x ( x^{2 + 2}) x y

= 28 x⁴y

B4A1

-5xy X 2x

= (-5x2) x ( ) x y

= -10 y

= -10 y

So the final table will be

i. 2a, 3a², 5a⁴

= (2 3 5) (aa²a⁴)

= (30) (a^{1 + 2 + 4})

= 30 a⁷

ii. 2x, 4y, 9z

= (24(x

= 72xyz

iii. ab, bc, ca

= abbc

= axaxbxbxcxc

= a² b² c²

iv. m, 4m, 3m², - 6m³

= mx4mx3 m²x- 6m

= (43 m² m³)

= -72( m^{1 + 1 + 2 + 3})

= -72 m⁷

v. xyz, y²z, yx²

= (x x²)

= (x^{1 + 2}) ( ) ( )

= (x³) (y⁴) ()

= x³y⁴

vi. lm², mn², ln²

= ( l

=

=

vii. -2p, -3q, -5p²

= (-2 p²)

= (-30) p^{1 + 2}

= -30 p³q

(i) (2× 4) (a³ × a⁵ × a¹⁵)

⇒ 8 a^{3 + 5 + 15}

⇒ 8 a²³ (∵ a^{n + m} = aⁿ × a^m)

ii. 5 × (4 + x) -2x × (4 + x)

⇒ (5× 4 + 5x) – (2x × 4 + 2x × x)

⇒ (20 + 5x) – (8x + 2x^{1 + 1} ) (∵ a^{n + m} = aⁿ × a^m)

⇒ (20 + 5x) – (8x + 2x² )

⇒ 20 + 5x – 8x - 2x²

⇒ -2x² -3x + 20

iii. x × (3x -y) + 3y × (3x -y)

⇒ (x × 3x + x × (-y)) + (3y × 3x + 3y × (-y))

⇒ (3x^{1 + 1} - xy) + ((3× 3)xy – 3y^{1 + 1} ) (∵ a^{n + m} = aⁿ × a^m)

⇒ (3x² - xy ) + (9xy – 3y² )

⇒ 3x² - xy + 9xy - 3y²

⇒ 3x² + 8xy - 3y²

iv. 3x × (4x-3) + 2 × (4x-3)

⇒ (3x × 4x + 3x × (-3)) + (2× 4x + 2 × (-3))

⇒ ((3× 4)x^{1 + 1} -9x) + (8x -6) (∵ a^{n + m} = aⁿ × a^m)

⇒ 12x² -9x + 8x -6

⇒ 12x² -x -6

v. ()x( abX

= x()

=

a × (2a² – 5ab + 3b²) + b × (2a² – 5ab + 3b²)

⇒( a × 2a² - a × 5ab + a × 3b²) + (b × 2a² – b × 5ab + b × 3b²)

⇒( 2a^{2 + 1} – 5a^{1 + 1}b + 3ab²) + (2a²b – 5ab² + 3b^{2 + 1})

(∵ a^{n + m} = aⁿ × a^m)

⇒( 2a³ – 5a²b + 3ab²) + (2a²b – 5ab² + 3b³)

⇒ 2a³ – 5a²b + 2a²b + 3ab²– 5ab² + 3b³

⇒ 2a³ + (– 5 + 2)a²b + (3-5)ab² + 3b³

⇒ 2a³ - 3a²b - 2ab² + 3b³

2x × (x² – xy + y²) + 3y × (x² – xy + y²)

⇒ (2x × x² – 2x × xy + 2x × y²) + (3y × x² -3y × xy + 3y× y²)

⇒ (2x^{2 + 1} – 2x^{1 + 1}y + 2xy²) + (3yx² -3xy^{1 + 1} + 3y^{2 + 1})

(∵ a^{n + m} = aⁿ × a^m)

⇒ (2x^{2 + 1} – 2x^{1 + 1}y + 2xy²) + (3yx² -3xy^{1 + 1} + 3y^{2 + 1})

⇒ (2x³ – 2x²y + 2xy²) + (3yx² -3xy² + 3y³)

⇒ 2x³ – 2x²y + 3x²y + 2xy² -3xy² + 3y³

⇒ 2x³ (- 2 + 3)x²y + (2-3)xy² + 3y³

⇒ 2x³ + x²y - xy² + 3y³

x(x + y-z) + y(x + y-z) + z(x + y-z)

⇒ (x× x + xy -xz ) + (yx + y × y -yz) + (zx + zy -z × z)

⇒ (x^{1 + 1} + xy -xz ) + (yx + y^{1 + 1} -yz) + (zx + zy -z^{1 + 1})

(∵ a^{n + m} = aⁿ × a^m)

⇒ x² + xy -xz + xy + y² -yz + xz + yz -z²

⇒ x² + y² -z² + xy + xy -xz + xz -yz + yz

⇒ x² + y² -z² + (1 + 1) xy + (1-1)xz + (1-1)yz

⇒ x² + y² -z² + 2xy

a(a² + 2ab + b²) + b(a² + 2ab + b²)

⇒ (a × a² + a × 2ab + a × b²) + (b × a² + b × 2ab + b× b²)

⇒ (a^{2 + 1} + 2a^{1 + 1}b + ab²) + (ba² + 2ab^{1 + 1} + b^{2 + 1})

(∵ a^{n + m} = aⁿ × a^m)

⇒ (a³ + 2a²b + ab²) + (ba² + 2ab² + b³)

⇒ a³ + 2a²b + a²b + ab² + 2ab² + b³

⇒ a³ + (2 + 1)a²b + (1 + 2)ab² + b³

⇒ a³ + 3a²b + 3ab² + b³

m(m² + mn + n²) -n(m² + mn + n²)

⇒ (m× m² + m× mn + mn²) –( nm² + n× mn + n× n²)

⇒ (m^{2 + 1} + m^{1 + 1}n + mn²) –( nm² + mn^{1 + 1} + n^{2 + 1})

(∵ a^{n + m} = aⁿ × a^m)

⇒ (m³ + m²n + mn²) –( nm² + mn² + n³)

⇒ m³ + m²n + mn² – nm² - mn² - n³

⇒ m³ + m²n– m²n + mn² - mn² - n³

⇒ m³ + (1-1)m²n + (1-1)mn² - n³

= m³ - n³

2xXx – 2xy – 2xz and z-y-x

(3a X a) + ( 3ax -2b) + (3ax3c) from (4ax5a) + (4ax2b) + (4ax(-3c))

= 3 – 6ab + 9ac from 20 + 8ab -12ac

Squaring a term means multiplying it with itself.

(a + b)² = (a + b) × (a + b)

Squaring a term means multiplying it with itself.

(a – b)² = (a – b) × (a – b)

(a – b) × (a + b) = a² + ab – ab – b²

⇒ (a – b) × (a + b) = a² – b²

So, (a² – b²) = (a – b) × (a + b)

Given 9.6² = (10 – 0.4)²

We know that

(a– b)² = a² + b² – 2a×b

⇒ 9.6² = 10² + 0.4² – 2(10)×(0.4)

⇒ 9.6² = 100 + 0.16 – 8

⇒ 9.6² = 92.16

We know that

(a– b)² = a² + b² – 2a×b

and (a+ b)² = a² + b² + 2a×b

⇒ (a+ b)² – (a– b)² = (a² + b² + 2a×b) – (a² + b² – 2a×b)

⇒ (a+ b)² – (a– b)² = 4ab

Given m² + (c + d) m + cd = m² + cm + dm + cd

⇒ m² + (c + d) m + cd =m(m + c) + d(m + c)

⇒ m² + (c + d) m + cd = (m + d)(m + c)

Given (x + 3) (x + 3) = (x + 3)²

We know that

(a+ b)² = a² + b² + 2a×b

⇒ (x + 3) (x + 3) = (x² + 3² + 2x×3)

⇒ (x + 3) (x + 3) = x² + 9 + 6x

Given (2m + 3) (2m + 3) = (2m + 3)²

We know that

(a+ b)² = a² + b² + 2a×b

⇒ (2m + 3) (2m + 3) = ((2m)² + 3² + 2× 2m×3)

⇒ (2m + 3) (2m + 3)= 4m² + 9 + 12m

Given (2x – 5) (2x – 5) = (2x – 5)²

We know that

(a– b)² = a² + b² – 2a×b

⇒ (2x – 5) (2x – 5) = ((2x)² + 5² – 2× 2x×5)

⇒ (2x – 5) (2x – 5) = 4x² + 25 – 20x

Given

We know that

(a– b)² = a² + b² – 2a×b

∵ (a² – b²) = (a – b) × (a + b)

∴ (3x + 2) (3x – 2) = (3x)² – (2²)

⇒ (3x + 2) (3x – 2) = (9x² – 4)

Given (5a – 3b) (5a – 3b) = (5a – 3b)²

We know that

(a– b)² = a² + b² – 2a×b

⇒ (5a – 3b) (5a – 3b) = ((5a)² + (3b)² – 2×5a×3b)

⇒ (5a – 3b) (5a – 3b) = 25a² + 9b² – 30ab

∵ (a² – b²) = (a – b) × (a + b)

∴ (2l – 3m) (2l + 3m) = (2l)² – (3m)²

⇒ (2l – 3m) (2l + 3m) = 4l² – 9m²

∵ (a² – b²) = (a – b) × (a + b)

∵ (a² – b²) = (a – b) × (a + b)

∵ (a² – b²) = (a – b) × (a + b)

∴ (100 + 3) (100 – 3) = (100)² – (3)²

⇒ (100 + 3) (100 – 3) = 10000 – 9 = 9991

Given (x + 4)(x + 7)

Using (x + a)(x + b) = x² + (a + b)x + ab

(x + 4)(x + 7) = x² + (4 + 7)x + 4× 7

⇒ (x + 4)(x + 7) = x² + 11x + 28

Given (5x + 3)(5x + 4)

Using (x + a)(x + b) = x² + (a + b)x + ab

(5x + 3)(5x + 4)= (5x)² + (3 + 4)5x + 3×4

⇒ (5x + 3)(5x + 4)= 25x² + 35x + 12

Given (7x + 3y) (7x – 3y)

Using (x + a)(x + b) = x² + (a + b)x + ab

(7x + 3y) (7x – 3y)= (7x)² + (3y – 3y)7x + 3y×3y

⇒ (7x + 3y) (7x – 3y) = 49x² – 9y²

Given (8x – 5) (8x – 2)

Using (x + a)(x + b) = x² + (a + b)x + ab

(8x – 5) (8x – 2) = (8x)² + ( – 5 – 2)8x + ( – 5)×( – 2)

⇒ (8x – 5) (8x – 2) = 64x² – 56x + 10

Given (2m + 3n) (2m + 4n)

Using (x + a)(x + b) = x² + (a + b)x + ab

(2m + 3n) (2m + 4n)= (2m)² + (3n + 4n)(2m) + 3n×4n

⇒ (2m + 3n) (2m + 4n)= 4m² + 14mn + 12n²

Given (xy – 3) (xy – 2)

Using (x + a)(x + b) = x² + (a + b)x + ab

(xy – 3) (xy – 2)= (xy)² + ( – 3 – 2)xy + ( – 3)×( – 2)

⇒ (xy – 3) (xy – 2)= x²y² – 5xy + 6

Given

Using (x + a)(x + b) = x² + (a + b)x + ab

Given (2 + x) (2 – y)

Using (x + a)(x + b) = x² + (a + b)x + ab

(2 + x) (2 – y) = (2)² + (x – y)2 – x×y

⇒ (2 + x) (2 – y) = 4 + 2(x – y) – xy

Given (p – q)²

We know that

(a– b)² = a² + b² – 2a×b

⇒ (p – q)² = ((p)² + (q)² – 2×p×q)

⇒ (p – q)² = p² + q² – 2pq

Given (a – 5)²

We know that

(a– b)² = a² + b² – 2a×b

⇒ (a – 5)² = ((a)² + (5)² – 2×a×5)

⇒ (a – 5)² = a² + 25 – 10a

Given (3x + 5)²

We know that

(a+ b)² = a² + b² + 2a×b

⇒ (3x + 5)² = ((3x)² + (5)² + 2×3x×5)

⇒ (3x + 5)² = 9x² + 25 + 30x

Given (5x – 4)²

We know that

(a– b)² = a² + b² – 2a×b

⇒ (5x – 4)² = ((5x)² + (4)² – 2×5x×4)

⇒ (5x – 4)² = 25x² + 16 – 40x

Given (7x + 3y)²

We know that

(a+ b)² = a² + b² + 2a×b

⇒ (7x + 3y)² = ((7x)² + (3y)² + 2×7x×3y)

⇒ (7x + 3y)² = 49x² + 9y² + 42xy

Given (10m – 9n)²

We know that

(a– b)² = a² + b² – 2a×b

⇒ (10m – 9n)² = ((10m)² + (9n)² – 2×10m×9n)

⇒ (10m – 9n)² = 100m² + 81n² – 180mn

Given (0.4a – 0.5b)²

We know that

(a– b)² = a² + b² – 2a×b

⇒ (0.4a – 0.5b)² = ((0.4a)² + (0.5b)² – 2×0.4a×0.5b)

⇒ (0.4a – 0.5b)² = 0.16a² + 0.25b² – 0.4ab

Given

We know that

(a– b)² = a² + b² – 2a×b

Given

We know that

(a– b)² = a² + b² – 2a×b

∵ (a² – b²) = (a – b) × (a + b)

∴ (0.54 × 0.54 – 0.46 × 0.46) = (0.54)² – (0.46)²

⇒ 0.54 × 0.54 – 0.46 × 0.46 = (0.54 – 0.46) (0.56 + 0.46)

⇒ 0.54 × 0.54 – 0.46 × 0.46 = 0.08×1.02

⇒ 0.54 × 0.54 – 0.46 × 0.46 = 0.0816

∵ 103² = (100 + 3)²

We know that

(a+ b)² = a² + b² + 2a×b

⇒ (100 + 3)² = ((100)² + (3)² + 2×100×3)

⇒ (100 + 3)² = 10000 + 9 + 600

⇒ 103² = 10609

∵ 48² = (50 – 2)²

We know that

(a– b)² = a² + b² – 2a×b

⇒ (50 – 2)² = ((50)² + (2)² – 2×50×2)

⇒ (50 – 2)² = 2500 + 4 – 200

⇒ 48² = 2304

∵ 54² = (50 + 4)²

We know that

(a+ b)² = a² + b² + 2a×b

⇒ (50 + 4)² = ((50)² + (4)² + 2×50×4)

⇒ (50 + 4)² = 2500 + 16 + 400

⇒ 54² = 2916

∵ 92² = (100 – 8)²

We know that

(a– b)² = a² + b² – 2a×b

⇒ (100 – 8)²= ((100)² + (8)² – 2×100×8)

⇒ (100 – 8)² = 10000 + 64 – 1600

⇒ 92² = 8464

∵ 998² = (1000 – 2)²

We know that

(a– b)² = a² + b² – 2a×b

⇒ (1000 – 2)² = ((1000)² + (2)² – 2×1000×2)

⇒ (1000 – 2)² = 1000000 + 4 – 4000

⇒ 998² = 996004

∵ 53 × 47 = (50 + 3) (50 – 3)

We know that

∵ (a² – b²) = (a – b) × (a + b)

∴ (50 + 3) (50 – 3) = (50)² – (3)²

⇒ (50 + 3) (50 – 3) = 2500 – 9 = 2491

∵ 96 × 104= (100 – 4) (100 + 4)

We know that

∵ (a² – b²) = (a – b) × (a + b)

∴ (100 – 4) (100 + 4) = (100)² – (4)²

⇒ (100 – 4) (100 + 4)= 10000 – 16 = 9984

∵ 28 × 32= (30 – 2) (30 + 2)

We know that

∵ (a² – b²) = (a – b) × (a + b)

∴ (30 – 2) (30 + 2) = (30)² – (2)²

⇒ (30 – 2) (30 + 2) = 900 – 4 = 896

∵ 81 × 79= (80 + 1) (80 – 1)

We know that

∵ (a² – b²) = (a – b) × (a + b)

∴ (80 + 1) (80 – 1) = (80)² – (1)²

⇒ (80 + 1) (80 – 1) = 6400 – 1 = 6399

∵ 2.8² = (3 – 0.2)²

We know that

(a– b)² = a² + b² – 2a×b

⇒ (3 – 0.2)²= ((3)² + (0.2)² – 2×3×0.2)

⇒ (3 – 0.2)² = 9 + 0.04 – 1.2

⇒ 2.8² = 7.84

We know that

∵ (a² – b²) = (a – b) × (a + b)

∴ 12.1² – 7.9² = (12.1 + 7.9) (12.1 – 7.9)

⇒ 12.1² – 7.9² = 20 × 4.2 = 84

Given 9.7 × 9.8 = (9 + 0.7)(9 + 0.8)

Using (x + a)(x + b) = x² + (a + b)x + ab

(9 + 0.7)(9 + 0.8) = (9)² + (0.7 + 0.8)9 + (0.7)×(0.8)

⇒ 9.7 × 9.8 = 81 + 13.5 + 0.56

⇒ 9.7 × 9.8 =95.06

Solving L.H.S. first,

(3x + 7)² – 84x

We know that

(a+ b)² = a² + b² + 2a×b

⇒ (3x + 7)² – 84x = ((3x)² + (7)² + 2×3x×7) – 84x

⇒ (3x + 7)² – 84x = 9x² + 49 + 42x – 84x

⇒ (3x + 7)² – 84x = 9x² + 49 – 42x

⇒ (3x + 7)² – 84x = ((3x)² + (7)² – 2×3x×7)

Using (a– b)² = a² + b² – 2a×b

⇒ (3x + 7)² – 84x = (3x – 7)²

∵ L.H.S. = R.H.S.

Hence, proved.

To Prove: (a – b) (a + b) + (b – c) (b + c) + (c – a)(c + a) = 0
Proof:
Solving L.H.S. first,

(a – b)(a + b) + (b – c)(b + c) + (c – a)(c + a)

We know that

∵ (a² – b²) = (a – b) × (a + b)

⇒ (a – b)(a + b) + (b – c)(b + c) + (c – a)(c + a) = (a² – b²) + (b² – c²) + (c² – a²)

⇒ (a – b)(a + b) + (b – c)(b + c) + (c – a)(c + a) = 0

∵ L.H.S. = R.H.S.

Hence, proved.

Given: a + b = 5 and a + b = 5

Using (a– b)² = a² + b² – 2a×b and (a+ b)² = a² + b² + 2a×b

Similarly,

(i) Given (a + b) = 12 and ab = 32

a² + b² = (a+ b)² – 2ab

⇒ a² + b² = (12)² – 2(32)

⇒ a² + b² = 144 – 64

⇒ a² + b² = 80

(a– b)² = (a+ b)² – 4ab

⇒ (a– b)² = (12)² – 4(32)

⇒ (a– b)² = 144 – 128

⇒ (a– b)² = 16

(ii) Given (a – b) = 6 and ab = 40

a² + b² = (a– b)² + 2ab

⇒ a² + b² = (6)² + 2(40)

⇒ a² + b² = 36 + 80

⇒ a² + b² = 116

(a+ b)² = (a– b)² + 4ab

⇒ (a+ b)² = (6)² + 4(40)

⇒ (a+ b)² = 36 + 160

⇒ (a+ b)² = 196

Given: (x + a) (x + b) = x² – 5x – 300

Using (x + a)(x + b) = x² + (a + b)x + ab

⇒ x² + (a + b)x + ab = x² – 5x – 300

⇒ (a + b) = – 5 and ab = – 300

Also, a² + b² = (a+ b)² – 2ab

⇒ a² + b² = ( – 5)² – 2( – 300)

⇒ a² + b² = 25 + 600

⇒ a² + b² = 625

Given (x + a)(x + b)(x + c) = (x + a)[(x + b)(x + c)]

⇒ (x + a)(x + b)(x + c) = (x + a)[x (x + c) + b(x + c)]

⇒ (x + a)(x + b)(x + c) = (x + a)[x² + cx + bx + bc]

⇒ (x + a)(x + b)(x + c) = (x + a)[x² + x(c + b) + bc]

⇒ (x + a)(x + b)(x + c) = x[x² + x(c + b) + bc] + a[x² + x(c + b) + bc]

⇒ (x + a)(x + b)(x + c) = x³ + x²(c + b) + xbc + ax² + xa(c + b) + abc

⇒ (x + a)(x + b)(x + c) = x³ + x²(a + b + c) + x(ab + bc + ca) + abc

Given 3a + 21ab can be written as 3× a× (1 + 7b)

Given x² – x – 12

Using (x + a)(x + b) = x² + (a + b)x + ab … (I)

⇒ x² – x – 12 = x² + ( – 4 + 3)x + ( – 4)×3

On comparing with (I),

a = – 4 and b = 3

So, x² – x – 12 = (x – 4)(x + 3)

The factors of x² – x – 12 are (x – 4) and (x + 3)

Given 6x² – x – 15

⇒ 6x² – x – 15 = 6x² – (10 – 9)x – 15

⇒ 6x² – x – 15 = 6x² – 10x + 9x – 15

⇒ 6x² – x – 15 = 2x(3x – 5) + 3(3x – 5)

⇒ 6x² – x – 15 = (3x – 5)(2x + 3)

The factors of 6x² – x – 15 are(2x + 3) and (3x – 5)

Given 169l² – 441m²

∵ (a² – b²) = (a – b) × (a + b)

⇒ 169l² – 441m² = (13l)² – (21m)²

⇒ 169l² – 441m² = (13l – 21m)(13l + 21m)

The factors of 169l² – 441m² are (13l – 21m) and (13l + 21m).

Given (x – 1) (2x – 3)

⇒ (x – 1) (2x – 3) = x(2x – 3) + ( – 1)(2x – 3)

⇒ (x – 1) (2x – 3) = 2x² – 3x – 2x + 3

⇒ (x – 1) (2x – 3) = 2x² – 5x + 3

Given 3x – 45 = (3x – 3×15)

Taking 3 common,

⇒ 3x – 45 = 3(x – 15)

Given 7x – 14y = (7x – 7×2)

Taking 7 common,

⇒ 7x – 14y = 7(x – 2)

Given 5a² + 35a = (5a² + 5a×7)

Taking 5a common,

⇒ 5a² + 35a = 5a(a + 7)

Given – 12y + 20y³= (4y×( – 3) + 4y×5y²)

Taking 4y common,

⇒ – 12y + 20y³= 4y( – 3 + 5y²)

Given 15a²b + 35ab = (5ab×3a + 5ab×7)

Taking 5ab common,

⇒ 15a²b + 35ab = 5ab(3a + 7)

Given (pq – prq) = (pq×1 – pq×r)

Taking pq common,

(pq – prq) = pq(1 – r)

Given 18m³ – 45mn²= (9m×2m² – 9m×5n²)

Taking 9m common,

18m³ – 45mn²= 9m×(2m² – 5n²)

Given 17 l² + 85m²= (17×l² + 17×5m²)

Taking 17 common,

17 l² + 85m²= 17× (l² + 5m²)

Given 6x³y – 12x²y + 15x⁴= (3x²×2xy – 3x²×4y + 3x²×5x²)

Taking 3x² common,

6x³y – 12x²y + 15x⁴ = 3x² (2xy – 4y + 5x²)

Given 2a⁵b³ – 14a²b² + 4a³b = (2a²b×a³b² – 2a²b×7b + 2a²b×2a)

Taking 2a²b common,

2a⁵b³ – 14a²b² + 4a³b= 2a²b (a³b² – 7b + 2a)

Given 2ab + 2b + 3a = (a×2b + 2b + 3a)

Taking 2b common from 1^st and 2^nd term,

2ab + 2b + 3a = 2b (a + 1) + 3a

Given 6xy – 4y + 6 – 9x = (2y×3x – 2y×2 + 3×2 – 3×3x)

Taking 2y common from 1^st and 2^nd term and – 3 from 3^rd and 4^th,

6xy – 4y + 6 – 9x = 2y(3x – 2) + ( – 3)(3x – 2)

⇒ 6xy – 4y + 6 – 9x = (3x – 2) (2y – 3)

Given 2x + 3xy + 2y + 3y² = (x×2 + x×3y + y×2 + y×3y)

Taking x common from 1^st and 2^nd term and y from 3^rd and 4^th,

2x + 3xy + 2y + 3y² = x(2 + 3y) + (y)(2 + 3y)

⇒ 2x + 3xy + 2y + 3y²= (x + y) (2 + 3y)

Given 15b² – 3bx2 – 5b + x² = (3b×5b – 3b×x² + ( – 1)× 5b + x² )

Taking 3b common from 1^st and 2^nd term and ( – 1) from 3^rd and 4^th,

15b² – 3bx2 – 5b + x²= 3b(5b – x²) + ( – 1)(5b – x²)

⇒ 15b² – 3bx2 – 5b + x²= (5b – x²) (3b – 1)

Given a²x² + axy + abx + by = (ax × ax + ax × y + b × ax + b × y )

Taking ax common from 1^st and 2^nd term and b from 3^rd and 4^th,

a²x² + axy + abx + by = ax (ax + y) + b (ax + y)

⇒ a²x² + axy + abx + by = (ax + y) (ax + b)

Given a²x + abx + ac + aby + b²y + bc = (ax × a + ax × b + by × a + by × b + c × a + c × b)

Taking ax common from 1^st and 2^nd term and by from 3^rd and 4^th and c from 5^th and 6^th,

a²x + abx + ac + aby + b²y + bc = ax (a + b) + by (a + b) + c (a + b)

⇒ a²x + abx + ac + aby + b²y + bc = (ax + by + c) (a + b)

Given ax³ – bx² + ax – b = (x² × ax – x² × b + ax – b)

Taking x² common from 1^st and 2^nd term and 1 from 3^rd and 4^th,

ax³ – bx² + ax – b = (x² × ax – x² × b + ax – b)

⇒ ax³ – bx² + ax – b = x² (ax – b) + (ax – b)

⇒ ax³ – bx² + ax – b = (x² + 1)(ax – b)

Given mx – my – nx + ny =(m × x – m × y + ( – n) × x + n × y)

Taking m common from 1^st and 2^nd term and ( – n) from 3^rd and 4^th,

⇒ mx – my – nx + ny = m (x – y) + ( – n)(x – y)

⇒ mx – my – nx + ny = (m – n)(x – y)

Given 2m³ + 3m – 2m² – 3=(m × 2m² + m × 3 + ( – 1) × 2m² + ( – 1) × 3)

Taking m common from 1^st and 2^nd term and ( – 1) from 3^rd and 4^th,

⇒ 2m³ + 3m – 2m² – 3= m (2m² + 3) + ( – 1)(2m² + 3)

⇒ 2m³ + 3m – 2m² – 3= (2m² + 3)(m – 1)

Given a² + 11b + 11ab + a =(a × a + a × 1 + (11b) × 1 + (11b) × a)

Taking a common from 1^st and 2^nd term and (11b) from 3^rd and 4^th,

⇒ a² + 11b + 11ab + a = a (a + 1) + (11b)(1 + a)

⇒ a² + 11b + 11ab + a = (a + 1)(a + 11b)

Given a² + 14a + 49 = a² + 2× 7× a + (7)²

Comparing with (a+ b)² = a² + b² + 2a×b

a = a and b = 7

So, a² + 14a + 49 = (a + 7)(a + 7)

Given x² – 12x + 36= x² – 2× 6× x + (6)²

Comparing with (a– b)² = a² + b² – 2a×b

a = x and b = 6

So, x² – 12x + 36= (x – 6)(x – 6)

Given 4p² – 25q²= (2p)² – (5q)²

Comparing with a² – b² = (a + b) (a – b)

a = 2p and b = 5q

So, 4p² – 25q² = (2p + 5q)(2p – 5q)

Given 25x² – 20xy + 4y² = (5x)² – 2× 5x ×2y x + (2y)²

Comparing with (a– b)² = a² + b² – 2a×b

a = 5x and b = 2y

So, 25x² – 20xy + 4y² = (5x – 2y)(5x – 2y)

Given 169m² – 625n² = (13m)² – (25n)²

Comparing with a² – b² = (a + b) (a – b)

a = 13m and b = 25n

So, 169m² – 625n² = (13m + 25n)(13m – 25m)

Given

Comparing with (a+ b)² = a² + b² + 2a×b

a =x and

So,

Given 121a² + 154ab + 49b² = (11a)² + 2× (11a)× (7b) + (7b)²

Comparing with (a+ b)² = a² + b² + 2a×b

a = 11a and b = 7b

So, 121a² + 154ab + 49b² = (11a + 7b)(11a + 7b)

Given 3x³ – 75x = 3x × x² – 3x × (5)²

Taking 3x common,

3x³ – 75x = 3x (x² – (5)²)

Comparing with a² – b² = (a + b) (a – b)

a = x and b = 5

So, 3x³ – 75x = 3x(x + 5)(x – 5)

Given 36 – 49x²= (6)² – (7x)²

Comparing with a² – b² = (a + b) (a – b)

a = 6 and b = 7x

So, 36 – 49x²= (6 + 7x)(6 – 7x)

Given 1 – 6x + 9x²= (1)² – 2× 1 ×3x + (3x)²

Comparing with (a– b)² = a² + b² – 2a×b

a = 1 and b = 3x

So, 1 – 6x + 9x² = (1 – 3x)(1 – 3x)

Given x² + 7x + 12

Using (x + a)(x + b) = x² + (a + b)x + ab … (I)

⇒ x² + 7x + 12= x² + (4 + 3)x + (4)×3

On comparing with (I),

a = 4 and b = 3

So, x² + 7x + 12= (x + 4)(x + 3)

Given p² – 6p + 8

Using (x + a)(x + b) = x² + (a + b)x + ab … (I)

⇒ p² – 6p + 8= p² + ( – 4 – 2)p + ( – 4)×( – 2)

On comparing with (I),

a = – 4 and b = – 2

So, p² – 6p + 8= (p – 4)(p – 2)

Given m² – 4m – 21

Using (x + a)(x + b) = x² + (a + b)x + ab … (I)

⇒ m² – 4m – 21= m² + ( – 7 + 3)m + ( – 7)×(3)

On comparing with (I),

a = – 7 and b = 3

So, m² – 4m – 21= (m – 7)(m + 3)

Given x² – 14x + 45

Using (x + a)(x + b) = x² + (a + b)x + ab … (I)

⇒ x² – 14x + 45= x² + ( – 9 – 5)x + ( – 9)×( – 5)

On comparing with (I),

a = – 9 and b = – 5

So, x² – 14x + 45= (x – 9)(x – 5)

Given x² – 24x + 108

Using (x + a)(x + b) = x² + (a + b)x + ab … (I)

⇒ x² – 24x + 108= x² + ( – 18 – 6)x + ( – 18)×( – 6)

On comparing with (I),

a = – 18 and b = – 6

So, x² – 24x + 108= (x – 18)(x – 6)

Given a² + 13a + 12

Using (x + a)(x + b) = x² + (a + b)x + ab … (I)

⇒ a² + 13a + 12= a² + (12 + 1)a + (12)×(1)

On comparing with (I),

a = 12 and b = 1

So, a² + 13a + 12= (a + 12)(a + 1)

Given x² – 5x + 6

Using (x + a)(x + b) = x² + (a + b)x + ab … (I)

⇒ x² – 5x + 6= x² + ( – 2 – 3)x + ( – 2)×( – 3)

On comparing with (I),

a = – 2 and b = – 3

So, x² – 5x + 6= (x – 2)(x – 3)

Given x² – 14xy + 24y²

Using (x + a)(x + b) = x² + (a + b)x + ab … (I)

⇒ x² – 14xy + 24y² = x² + ( – 12y – 2y)x + ( – 12y)×( – 2y)

On comparing with (I),

a = – 12 and b = – 2

So, x² – 14xy + 24y² = (x – 12y)(x – 2y)

Given m² – 21m – 72

Using (x + a)(x + b) = x² + (a + b)x + ab … (I)

⇒ m² – 21m – 72 = m² + ( – 24 + 3)m + ( – 24)×(3)

On comparing with (I),

a = – 24 and b = 3

So, m² – 21m – 72 = (m – 24)(m + 3)

Given x² – 28x + 132

Using (x + a)(x + b) = x² + (a + b)x + ab … (I)

⇒ x² – 28x + 132= x² + ( – 22 – 6)x + ( – 22)×( – 6)

On comparing with (I),

a = – 22 and b = – 6

So, x² – 28x + 132 = (x – 22)(x – 6)

= -6y

= 7m – 6

= 9l²m³n⁵

= 5y² – 4y + 3

= 3x³ – 5x² – 7

= x – 7 + y

= 8x³ – 4y² + 3xz³

Factorize the numerator,

x² +7x + 10 = x² + 5x + 2x + 10

= x(x + 5) + 2(x + 5)

= (x + 2)(x + 5)

Now,

= x + 5

Factorize the numerator,

a² + 24a + 144 = a² + 12a + 12a + 144

= a(a + 12) + 12(a + 12)

= (a + 12)(a + 12)

Now,

= a + 12

Factorize the numerator,

m² + 5m – 14 = m² + 7m – 2m – 14

= m(m + 7) – 2(m + 7)

= (m – 2)( m + 7)

Now,

= m – 2

Factorize the numerator,

25m² – 4n² = (5m)² – (2n)²

= (5m + 2n)(5m – 2n) [∵ a² – b² = (a + b)(a – b)]

Now,

= 5m – 2n

Factorize the numerator,

4a² – 4ab – 15b² = 4a² + 6ab – 10ab – 15b²

= 2a(2a + 3b) – 5b(2a + 3b)

= (2a + 3b)(2a – 5b)

Now,

= (2a + 3b)

Factorize the numerator,

a⁴ – b⁴ = (a²)² – (b²)²

= (a² + b²)(a² – b²) [∵ a² – b² = (a + b)(a – b)]

= (a² + b²)(a + b)(a – b) [∵ a² – b² = (a + b)(a – b)]

Now,

= (a² + b²)(a + b)

Subtracting 5 from both sides,

⇒ 3x + 5 – 5 = 23 – 5

⇒ 3x = 18

Dividing both sides by 3,

⇒ x = 6

Subtracting 10 from both sides,

⇒ 17 - 10 = 10 – y – 10

⇒ 7 = -y

Dividing both sides by -1,

⇒ y = -7

Adding 7 to both sides,

⇒ 2y – 7 + 7 = 1 + 7

⇒ 2y = 8

Dividing both sides by 2,

⇒ y = 4

Dividing both sides by 6,

⇒ x = 12

Multiplying both sides by 11,

⇒ y = -77

9x – 21 = 10x – 15

⇒ 10x – 15 – 9x + 21 = 0

⇒ x + 6 = 0

⇒ x = -6

⇒ 8x – 12 + 15x – 20 = 14

⇒ 8x + 15x = 14 + 12 + 20

⇒ 23x = 46

Dividing both sides by 23,

⇒ x = 2

⇒ 7(x – 7) = 5(x – 5)

⇒ 7x – 49 = 5x – 25

⇒ 7x – 5x = 49 – 25

⇒ 2x = 24

Dividing both sides by 2,

⇒ x = 12

⇒ 5(2x + 3) = 3(3x + 7)

⇒ 10x + 15 = 9x + 21

⇒ 10x – 9x = 21 – 15

⇒ x = 6

⇒ 7m × 2 = 12

⇒ 14m = 12

Dividing both sides by 2,

Let the number be x

Then according to question,

⇒ x = 18

Hence, the number is 18.

Let the numbers be x, x+1 and x+2

Then according to the question,

x + (x + 1) + (x + 2) = 90

⇒ x + x + 1 + x + 2 = 90

⇒ 3x + 3 = 90

⇒ 3x = 90 – 3

⇒ 3x = 87

⇒ x = 29

⇒ x + 1 = 29 + 1 = 30

⇒ x + 2 = 29 + 2 = 31

Hence, the numbers are 29, 30 and 31.

Let breadth of rectangle = x

Then length of rectangle = x + 8

Perimeter = 60 cm

We know that,

Perimeter of rectangle = 2 (length of rectangle + breadth of rectangle)

⇒ 60 = 2(x + (x + 8))

⇒ 60 = 2(x + x + 8)

⇒ 60 = 2(2x + 8)

⇒ 60 = 4x + 16

⇒ 4x = 60 – 16

⇒ 4x = 44

⇒ x = 11

⇒ x + 8 = 11 + 8 = 19

Hence, breadth of rectangle = 11 cm

length of rectangle = 19 cm

Let the smaller number be x

Then bigger number = 4x

Then according to the question,

x + 4x = 60

⇒ 5x = 60

⇒ x = 12

⇒ 4x = 4 × 12 = 48

Hence, the numbers are 12 and 48.

Let the bigger number be x

Then the smaller number = x – 3 [∵ the difference is 3]

Then according to the question,

x + (x – 3) = 21

⇒ 2x – 3 = 21

⇒ 2x = 21 + 3

⇒ x = 12

⇒ x – 3 = 12 – 3 = 9

Hence, the numbers are 12 and 9.

Let the numbers be 5x and 3x

Then according to the question,

5x – 3x = 18

⇒ 2x = 18

⇒ x = 9

⇒ 5x = 5 × 9 = 45

⇒ 3x = 3 × 9 = 27

Hence, the numbers are 45 and 27.

Let the number be x

Then according to the question,

⇒ x = 4000

Hence, the number is 4000.

Let the numerator be x

Then the denominator = x + 2

Then according to the question,

⇒ 3(x + 1) = 2(x + 3)

⇒ 3x + 3 = 2x + 6

⇒ 3x – 2x = 6 – 3

⇒ x = 3 (Numerator)

⇒ x + 2 = 3 + 2 = 5 (Denominator)

Hence, the fraction is

At present,

Let age of Nandini = x

Then age of Mary = 3x

After 10 years,

Age of Nandini = x + 10

Age of Mary = 3x + 10

Then according to the question,

[x + 10] + [3x + 10] = 80

⇒ x + 10 + 3x + 10 = 80

⇒ 4x + 20 = 80

⇒ 4x = 80 – 20

⇒ 4x = 60

⇒ x = 15

⇒ 3x = 3 × 15 = 45

Hence, Present age of Nandini = 15 years

Present age of Mary = 45 years

Let the savings of Murali be x

According to the question,

⇒ x = 50000 × 6

⇒ x = 300000

Hence, Share of his wife = Rs 1,50,000

Share of his son = Rs 1,00,000