Algebraic Expressions

In the given question

The symbols which has a fixed value are constant such as 15, 3, –3/7 and 7

The symbols which do not have any fixed value, but may be assigned the value (values) according to the requirement are called the variables such as

Where z, x and y are variables which may acquire different values according to the situation.

And a combination of a constant and a variable is a variable.

A polynomial which contains only one term is monomials

Such as

A Polynomial which contains two terms is called a Polynomial.

Such as 9–4y, 4y² – xz, 7x +z²

A polynomial which contains three terms is called as trinomial

Such as x –2y +3z, 4+5y –6z

The terms having same variable with same exponents are called like terms

Here like terms are

(4x²,3x²) with same variable x and exponential power as 2

( has no like term with x as variable and exponential power as 1)

(–8x³, 6x³, –74x³ with variable as x and exponential power as 3 )

7xyz has no like term as no other term has three variables x, y, z

(xy, 8xy has same variables x, y and exponential power of both x and y as 1)

While adding or subtracting, the like term`s numerical coefficient are added or subtracted

(i) 7x – 9y + 3 – 3x – 5y + 8

⇒ (7– 3)x +(–9 –5)y +3+8

⇒ 4x – 14y +11

(ii) 3x² + 5xy – 4y² + x² – 8xy – 5y²

⇒ (3+1) x² +( –4 –5) y² +( 5–8) x y

⇒ 4x² – 9y² – 3xy

(i) While adding or subtracting, the like term`s numerical coefficient are added or subtracted

⇒ (5 + 1+ 3)a + (3 – 2 +5) b

⇒ 9 a + 6b

(ii) While adding or subtracting, the like term`s numerical coefficient are added or subtracted

⇒ (1–1) x³ + (–1+3) x²y + (5–9 +9) xy² + (1+1) y

⇒ 0x³ + 2 x²y +5xy² +2y

⇒ 2 x²y +5xy² +2y

Given problem asks to subtract

(i) –2xy + 3xy² from 8xy

⇒ 8xy – ( –2xy + 3xy² )

⇒ 8xy +2xy –3xy²

⇒ 10xy –3xy²

(ii) a – b – 2c from 4a + 6b – 2c

⇒ 4a +6b –2c – (a– b – 2c)

⇒ 4a +6b –2c –a + 2c +b

While adding or subtracting, the like term`s numerical coefficient are added or subtracted

⇒ (4–1) a + (6 +1) b +( –2 +2) c

⇒ 3a +7b

To find out product of monomials

We multiple the numerical coefficient together and variables together

The given table hence can be completed as :

(i) By using the distributive law

(5x +8) 3x

= 5x × 3x + 8× 3x

= 15x² +24x

(ii) By using the distributive law

= (–3pq)(–15 p³q²) – (–3pq)(q³)

= 45p⁴q³ + 3pq⁴

(iii) By using the distributive law

=

(iv) By using the distributive law

= –x² (x) – (–x²) (15)

= –x³ + 15x²

By using the distributive law

(i) (2xy – xy)(3xy – 5)

= 2xy (3xy – 5) –xy (3xy –5)

= 6x²y² –10xy – 3x²y² +5xy

= 3x²y² – 5xy (after adding and subtracting the like terns)

(ii) (3xy² + 1)(4xy – 6xy²)

By using the distributive law

3xy² ((4xy – 6xy²) +1((4xy – 6xy²)

= 3xy² × 4xy – 3xy² × 6xy² + 4xy – 6xy²

= 12x²y³ – 18 x²y⁴ + 4xy –6xy²

(iii) (3x² + 2x)(2x² + 3)

By using the distributive law

3x² (2x² + 3) +2x (2x² + 3)

= 3x² × 2x² + 3x² × 3 + 2x × 2x² + 2x × 3

= 6x⁴ + 9x² + 4x³ +6x

(iv) (2m³ + 3m)(5m – 1).

By using the distributive law

2m³ (5m –1) +3m (5m –1)

= 2m³ × 5m – 1× 2m³ + 3m × 5m – 1× 3m

= 10m⁴ – 2m³ + 15m² – 3m

(i) By using the distributive law

= a(a + 5) +3((a+5)

= a² +5a +3a +15

= a² +8a +15

(ii) By using the distributive law

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

= 9t² + 12t +3t + 4

= 9t² + 15t +4

(iii) By using the distributive law

= a( a+2) –8(a+2)

= a² + 2a – 8a – 16

= a² – 6a –16

(iv) By using the distributive law

= a (a–2) – 6 ( a – 2)

= a² –2a – 6a +12

= a² – 8a +12

(i) 53 × 55

We can re–write 53 and 55 as

(50+3)× (50+5)

Using the identity

(x+a) (x+b) = x² + (a+b) x + ab

(50+3)× (50+5) where x = 50, a = 3 and b = 5

= 50² + ( 3+5) 50 + 3× 5

=2500 + 400 + 15

= 2915

(ii) 102 × 106

= (100 + 2) (100 + 6)

Using the identity

(x+a) (x+b) = x² + (a+b) x + ab

Here x= 100, a = 2 and b = 6

⇒ (100 + 2) (100 + 6)

= 100² + ( 2+6) 100 + 2× 6

= 10000 + 800 +12

= 10812

(iii) 34 × 36

= (30+4) + (30+6)

Using the identity

(x+a) (x+b) = x² + (a+b) x + ab

Here x = 30, a= 4 and b = 6

So, 30² +( 4+6) 30 +(4× 6)

= 900 + 300 +24

= 1224

(iv) 103 × 96

= (90 + 13) (90 +6)

Using the identity

(x+a) (x+b) = x² + (a+b) x + ab

Here x = 90, a = 13 and b = 6

So, (90 + 13) (90 +6)

= 90² + ( 13+6) 90 +(13× 6)

= 8100 + 1710 +78

= 9888

first we will expand (x + a)(x + b)

Using the identity

(x+a) (x+b) = x² + (a+b)x + ab

= x² + (a+b)x + ab

Now multiplying the expansion with (x+c)

(x² + (a+b)x + ab) (x+c)

By using the distributive law

x(x² + (a+b) x + ab) + c(x² + (a+b) x + ab)

= x³ + (a + b)x² + abx + cx² + (a + b)cx + abc

Arranging the like terms

= x³ + (a + b + c)x² + abx + acx + bcx + abc

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

Given the identity

(a +b)² = a² + 2ab + b²

(i) (a + 6)²

Using the given identity

Here a = a, b = 6

= a² +2 × a × 6 +6²

= a² +12a +36

(ii) (3x + 2y)²

Using the given identity

Here a = 3x, b = 2y

= (3x)² + 2 × 3x × 2y +(2y)²

= 9x² + 12xy +4y²

(iii) (2p + 3q)²

Using the given identity

Here a = 2p, b = 3q

= (2p)² +2× 2p × 3q + (3q)²

= 4p² + 12pq +9q²

(iv) (x² + 5)²

Using the given identity

Here a = x², b = 5

= (x²)² +2 × x² × 5 +5²

= x⁴ + 10x² +25

Given identity (a + b)² = a² + 2ab + b²

(i) (34)²

= (30+4)²

Here a = 30 and b = 4

= 30² +2× 30× 4 + 4²

= 900 + 240 +16

= 1156

(ii)(10.2)²

= (10 +0.2)²

Using the given identity

Here a = 10, b = 0.2

= 10² + 2× 10 × 0.2 +(0.2)²

= 100 + 4 + 0.04

= 104.04

(iii) (53)²

= (50+3)²

Using the given identity

Here a = 50, b = 3

= (50)² +2× 50 × 3 + 3²

= 2500 + 300 +9

= 2809

(iv) (41)²

= (40+1) ²

Using the given identity

Here a = 40, b = 1

= 40² +2 × 40 × 1 + 1²

= 1600 + 80 +1

= 1681

Given identity

(a – b)² = a² – 2ab + b²

(i) (x – 6)²

Using the given identity

Here a = x, b = 6

= x² – 2× x × 6 +6²

= x² –12x +36

(ii)(3x – 5y)²

Using the given identity

Here a = 3x, b = 5y

= (3x)² – 2× 3x × 5y + (5y)²

= 9x² – 30xy +25y²

(iii)(5a – 4b)²

Using the given identity

Here a = 5a, b = 4b

= (5a)² –2× 5a × 4b + (4b)²

= 25a² – 40 ab + 16b²

(iv) (p² – q²)²

Using the given identity

Here a = p², b = q²

= (p²)² – 2 × p² × q² + (q²)²

= p⁴ – 2 p² q² + q⁴

The given identity is (a – b)² = a² – 2ab + b²

(i) (49)²

= (50 –1)²

Using the given identity

Here a = 50, b = 1

= 50² – 2× 50 × 1 +1²

= 2500 –100 +1

= 2401

(ii) (9.8)²

= (10 – 0.2)²

Using the given identity

Here a = 10, b = 0.2

= 10² – 2× 10 × 0.2 + (0.2)²

= 100 – 4 + 0.04

= 96.04

(iii) (59)2

= (60 – 1)²

Using the given identity

Here a = 60, b = 1

= 60² – 2× 60× 1 + 1²

= 3600 – 120 +1

= 3481

(iv) (198)2

= (200 –2)²

Using the given identity

Here a = 200, b = 2

= 200² – 2× 200 × 2 + 2²

= 40000 – 800 +4

= 39204

The given identity is (a+b) (a–b) = a² –b²

(i) (x – 6) (x + 6)

Using the given identity

Here a = x and b = 6

= x² – 6²

= x² – 36

(ii) (3x + 5)(3x + 5)

Using the given identity

Here a = 3x and b = 5

= (3x)² – 5²

= 9x² – 25

(iii) (2a + 4b)(2a – 4b)

Using the given identity

Here a = 2a and b = 4b

= (2a)² – (4b)²

= 4a² – 16b²

(iv)

Using the given identity

Here

=

(i) 55 × 45

We can split 55 as (50+5)

And 45 as (50–5)

Now 55 × 45

= (50+5) (50–5)

Using the identity (a +b) (a–b) = a² – b²

Here a = 50 and b = 5

= 50² – 5²

= 2500 –25

= 2475

(ii) 33 × 27

= (30+3) (30–3)

Using the identity (a +b) (a–b) = a² – b²

Here a = 30 and b = 3

= (30)² – 3²

= 900 – 9

= 891

(iii) 8.5 × 9.5

= (9 – 0.5) (9 + 0.5)

Using the identity (a +b) (a–b) = a² – b²

Here a = 9 and b = 0.5

= 9² – (0.5)²

= 81 – 0.25

= 80.75

(iv) 102 × 98

= (100 + 2) (100 – 2)

Using the identity (a +b) (a–b) = a² – b²

Here a = 100 and b = 2

= (100)² – 2²

= 10000 – 4

= 9996

First solving (x – 3)(x + 3)

Using the identity (a +b) (a–b) = a² – b²

Here a = x and b = 3

= x² –3²

= x² – 9

Now (x² – 9) (x² + 9)

Again, using the identity (a +b) (a–b) = a² – b²

Here a = x² and b = 9

= (x²)² – 9²

= x⁴ –81 ( ∵ (a^x)^m = a^xm)

First solving (2a + 3)(2a – 3)

Using the identity (a +b) (a–b) = a² – b²

Here a = 2a and b = 3

= (2a)² –3²

= 4a² – 9

Now (4a² – 9)(4a² + 9)

Using the identity (a +b) (a – b) = a² – b²

Here a = 4a and b = 9

= (4a)² – 9 ²

= 16a⁴ – 81

First solving (p+2) (p–2)

Using the identity (a +b) (a–b) = a² – b²

Here a = p and b = 2

= p² – 2²

= p² – 4

Now solving (p² + 4) (p² – 4)

Again Using the identity (a +b) (a–b) = a² – b²

Here a = p² and b = 4

= (p²)² – 4²

= p⁴ – 16 (∵ (a^x)^m = a ^xm)

First solving

Using the identity (a +b) (a–b) = a² – b²

Here

Now solving

Again, Using the identity (a +b) (a–b) = a² – b²

Here

=

First solving (2x – y) (2x+y)

Using the identity (a +b) (a–b) = a² – b²

Here a = 2x and b = y

= (2x)² – y²

= 4x² – y²

Now solving (4x² – y² ) (4x² + y²)

Again using the identity (a +b) (a–b) = a² – b²

Here a = 4x² and b = y²

= (4x²)² – (y²)²

= 16 x⁴ – y⁴

First solving for (2x – 3y)(2x + 3y)

Using the identity (a +b) (a–b) = a² – b²

Here a = 2x and b = 3y

= (2x)² – (3y)²

= 4x² – 9y²

Now solving for ( 4x² – 9y² ) (4x² + 9y² )

Again Using the identity (a +b) (a–b) = a² – b²

Here a = 4x² and b = 9y²

= (4x² )² – (9y²)²

= 16 x⁴ – 81 y⁴

Terms having the same literal factors with same exponents are called like terms

So B is the correct option

The coefficient is the term excluding ab in 2ab.

So B is the correct option

The exponential form of a × a × a is a³.

So C is the correct option.

Sum of two negative integers is always negative

So A is the correct option.

(a + b)² = a² + 2ab + b²

In the above expression we have to add b² to make it (a + b)²

So A is the correct option.

(x + 2)(x-3)

⇒ (x + 2)(x-3) = x(x-3) + 2(x-3)

⇒ (x + 2)(x-3) = x²-3x + 2x-6

⇒ (x + 2)(x-3) = x² – x – 6

So C is the correct option.

(7.2)² = (7 + 0.2)²

⇒ (7.2)² = 7² + 0.2² + 2× 7× 0.2

⇒ (7.2)² = 49 + 0.04 + 2.8

⇒ (7.2)² = 51.84

So C is the correct option.

(a + b)² = a² + 2ab + b²

⇒ (2x – 3y)² = 4x² + 9y² – 12xy

So B is the correct option.

58× 62 = (60-2)× (60 + 2)

We use the form (a + b)(a-b) = a² - b²

⇒ 58× 62 = 60²-2²

⇒ 58× 62 = 3600-4 = 3596

So C is the correct option.

We need to subtract the two expressions

10x + 10y – 7p + 9q-8x + 7y + 8p - 10q = 2x-17y + p-q

(i) We use (a + b)² = a² + 2ab + b²

⇒ (4x + 3)² = (4x)² + 2× 4x × 3 + 3²

⇒ (4x + 3)² = 16x² + 24x + 9

(ii) We use (a + b)² = a² + 2ab + b²

⇒(x + 2y)² = x² + 4xy + 4y²

(iii) We use (a + b)² = a² + 2ab + b²

(iv) We use (a - b)² = a²- 2ab + b²

(i) We use the form (a + b)(a-b) = a² - b²

⇒ (2t + 5)(2t – 5) = 4t²-25

(ii) We use the form (a + b)(a-b) = a² - b²

⇒(xy + 8)(xy – 8) = x²y²-64

(iii) We use the form (a + b)(a-b) = a² - b²

⇒(2x + 3y)(2x – 3y) = 4x²-9y²

Applying the formula (a + b)(a - b) = a² - b² on first two terms

(n – 1)(n + 1)(n² + 1) = (n² - 1) (n² + 1)

Applying the formula (a + b)(a - b) = a² - b² again

⇒ (n – 1)(n + 1)(n² + 1) = (n⁴ - 1)

Applying the formula (a + b)(a - b) = a² - b² on first two terms

Applying the formula (a + b)(a - b) = a² - b² again

Applying the formula (a + b)(a - b) = a² - b² on first two terms

(x – 1)(x + 1)(x² + 1)(x⁴ + 1) = (x² - 1) (x² + 1)(x⁴ + 1)

Applying the formula (a + b)(a - b) = a² - b² again

⇒ (x – 1)(x + 1)(x² + 1)(x⁴ + 1) = (x⁴ - 1)(x⁴ + 1)

Applying the formula (a + b)(a - b) = a² - b² again

⇒ (x – 1)(x + 1)(x² + 1)(x⁴ + 1) = (x⁸ - 1)

Applying the formula (a + b)(a - b) = a² - b² on first two terms

(2x – y)(2x + y)(4x² + y²) = (4x² - y²) (4x² + y²)

Applying the formula (a + b)(a - b) = a² - b² again

⇒ (2x – y)(2x + y)(4x² + y²) = (16x⁴ – y⁴)

(103)² = (100 + 3)²

We use (a + b)² = a² + 2ab + b²

⇒ (103)² = 100² + 2× 100× 3 + 3²

⇒ (103)² = 10000 + 600 + 9 = 10609

(96)² = (100-4)²

We use (a - b)² = a²- 2ab + b²

⇒ (96)² = 100²-2× 100× 4 + 4²

⇒ (96)² = 10000-800 + 16 = 9216

107 × 93 = (100 + 7)(100 – 7)

We use (a + b)(a - b) = a² - b²

⇒ 107 × 93 = 100²-7²

⇒ 107 × 93 = 10000-49

⇒ 107 × 93 = 9951

1008 × 992 = (1000 + 8)(1000 – 8)

We use (a + b)(a - b) = a² - b²

⇒ 1008 × 992 = 1000²-8²

⇒ 1008 × 992 = 1000000-64

⇒ 1008 × 992 = 999936

185² – 115²= (150 + 35)²-(150-35)²

(a + b)²-(a-b)² = 4ab

⇒ 185² – 115²= 4× 150× 35

⇒ 185² – 115² = 21000

x + y = 7

⇒ y = 7-x …Equation (i)

xy = 12

Putting the value of y from above Equation (i) we get

x(7-x) = 12

⇒ 7x-x² = 12

⇒x²-7x + 12 = 0

Solving the above equation by the method of factorization we get

⇒ x²-4x-3x + 12 = 0

⇒x(x-4)-3(x-4) = 0

⇒(x-4)(x-3) = 0

x = 4,3

When x = 4, y = 3

⇒ x² + y = 19

When x = 3, y = 4

⇒ x² + y = 13

x + y = 12

⇒ y = 12-x …Equation (i)

xy = 32

Putting the value of y from above Equation (i) we get

x(12-x) = 32

⇒ 12x-x² = 32

⇒x²-12x + 32 = 0

Solving the above equation by the method of factorization we get

⇒ x²-8x-4x + 12 = 0

⇒x(x-8)-4(x-8) = 0

⇒(x-4)(x-8) = 0

x = 4,8

When x = 4, y = 8

⇒ x² + y = 24

When x = 8, y = 4

⇒ x² + y = 68

4x² + y² = 40 …Equation (i)

xy = 6

⇒ 4xy = 24 …Equation (ii)

Adding Equation (i) and (ii)

4x² + y² + 2xy = 64

⇒ (2x + y)² = 8²

⇒ 2x + y = ±8

x - y = 3

⇒ y = x-3 …Equation (i)

xy = 10

Putting the value of y from above Equation (i) we get

x(x-3) = 10

⇒ x²-3x = 10

⇒x²-3x-10 = 0

Solving the above equation by the method of factorization we get

⇒ x²-5x + 2x + 10 = 0

⇒x(x-5) + 2(x-5) = 0

⇒(x + 2)(x-5) = 0

x = -2,5

Neglecting the negative value

When x = 5, y = 2

⇒ x² + y = 27

…Equation (i)

Squaring both sides of the equation we get

Cubing both sides of the equation (i) we get

Squaring both sides of the equation we get

Again Squaring both sides of the above equation we get

(i) We use the formula (a + b)² + (a - b)² = 2(a² + b²)

(x + y)² + (x – y)² = 2(x² + y²)

(ii) Applying the formula (a + b)²(a - b)² = (a² - b²)²

(x + y)² × (x – y)² = (x² - y²)²

⇒ (x + y)² × (x – y)² = x⁴-2x²y² + y⁴

(i) We use

(x + 2z)(2x + z) =

⇒ (x + 2z)(2x + z) =

(ii) 4(x + 2y)(2x + y) = (2x + 4y)(4x + 2y)

We use

⇒ 4(x + 2y)(2x + y) =

⇒ 4(x + 2y)(2x + y) = ( 3x + 3y)²-(2y – 2x)²

(iii) We use

(x + 98)(x + 102) =

⇒ (x + 98)(x + 102) = (x + 100)²-(x-100)²

(iv) 505 × 495 = (500 + 5)(500-5)

We use (a + b)(a - b) = a² - b²

⇒ 505 × 495 = 500²-5²

(i) a + b – c = 3x – 5y + 6x + 3y-2y + 4x

⇒ a + b – c = 13x - 4y

(ii) 2a–3b + 4c = 2(3x – 5y) + 3(6x + 3y)-4(2y – 4x)

⇒ 2a–3b + 4c = 6x-10y + 18x + 9y-8y + 16x

⇒ 2a–3b + 4c = 40x-9y

Perimeter = 15x² – 23x + 9

First side = 5x² + 8x – 1

Second side = 6x² – 9x + 4

Sum of first two side = 11x²-x + 3

Third side = (Perimeter- Sum of first two side)

⇒ Third side = 4x²-22x + 6

Two adjacent sides are 2x² – 5xy + 3z² and 4xy – x² – z.

Area of rectangle = (2x² – 5xy + 3z²)× (4xy – x² – z)

⇒ Area of rectangle = 2x²(4xy – x² – z)-5xy(4xy – x² – z) + 3z²(4xy – x² – z)

⇒ Area of rectangle = 8x³y-2x⁴-2x²z-20x²y² + 5x³y + 5xyz + 12xyz²-3x²z²-3z³

Perimeter of rectangle = 2× (2x² – 5xy + 3z² + 4xy – x² – z)

⇒ Perimeter of rectangle = 2× (x²-xy + 3z²-z)

⇒ Perimeter of rectangle = (2x²-2xy + 6z²-2z)

Base of triangle = (3x – 4y)

Altitude of triangle = (6x + 5y)

Area of triangle

⇒ Area of triangle

⇒ Area of triangle

⇒ Area of triangle

Length of rectangle = 2x + 3y

Breadth of rectangle = 3x + 2y

Area of rectangle = (Length× breadth) = (2x + 3y)× (3x + 2y)

⇒ Area of rectangle = 6x² + 13xy + 6y²

Side of square = x + y

Area of Square = (Side× Side) = (x + y)²

⇒ Area of Square = x² + y² + 2xy

Area of remaining region = ( Area of rectangle- Area of Square) = 5x² + 5y² + 11xy

a² + b² + c² – ab – bc – ca = 0

Multiplying both sides by 2 we get

2 (a² + b² + c² – ab – bc – ca ) = 0

⇒ (a² + b² - 2ab) + ( b² + c² - 2bc) + (c² + a² -2ac) = 0

The individual terms inside the brackets can be expressed as a whole square

⇒ (a – b)² + (b – c)² + (c – a)² = 0

Since a, b, c are rational and none of the term is equal to zero so each of the terms inside the bracket must individually be equal to zero

⇒ a – b = 0

⇒ a = b

⇒ b – c = 0

⇒ b = c

⇒ c – a = 0

⇒ c = a

So together we can say that

a = b = c