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49² = (50 – 1)² 

(a – b)² = a² – 2ab + b² 

Here a = 50, b = 1 

(50 – 1)² = 50² – (2)(50)(1) + 1² 

= 2500 – 100 + 1 

(49)² = 2401

(10.2)² = (10 + 0.2)² 

(a + b)² = a² + 2ab + b² 

Here a = 10, b=0.2 

[(10 + (0.2))]² = 10² + 2.(10).(0.2) + (0.2)² 

= 100 + 4 + 0.04 

(10.2)² = 104.04

34² = (30 + 4)² (a + b)² 

= a² + 2ab+ b² 

Here a = 30, b = 4 

(30 + 4)² = (30)² + 2 × 30 × 4 + (4)² 

= 900 + 240 + 16 

(34)² = 1156

(a – b)² = a² – 2ab + b² 

Here a = p² , b = q² 

(p² – q²)² = (p²)² – (2)(p²)(q²) + (q²)² 

= p⁴ – 2p²q² + q⁴

5a(6a – 3b)

Let p, q and r be three monomials.

Then, by distributive law of multiplication over subtraction, we have:

P × (q – r) = (p × q) – (p × r)

Now,

= (5a × 6a) – (5a × 3b)

= (30a² – 15ab) … [∵ a^m × aⁿ = a^m+n]

8a²(2a + 5b)

Let p, q and r be three monomials.

Then, by distributive law of multiplication over addition, we have:

P × (q + r) = (p × q) + (p × r)

Now,

= (8a² × 2a) + (8a² × 5b)

= (16a³ + 40a²b) … [∵ a^m × aⁿ = a^m+n]

9x²(5x + 7)

Let p, q and r be three monomials.

Then, by distributive law of multiplication over addition, we have:

P × (q + r) = (p × q) + (p × r)

Now,

= (9x² × 5x) + (9x² × 7)

= (45x³ + 63x²) … [∵ a^m × aⁿ = a^m+n]

ab(a² – b²)

Let p, q and r be three monomials.

Then, by distributive law of multiplication over subtraction, we have:

P × (q – r) = (p × q) – (p × r)

Now,

= (ab × a²) – (ab × b²)

= (a³b + ab³) … [∵ a^m × aⁿ = a^m+n]

((1/5)x + 2y) ((2/3)x – y)

Suppose (a + b) and (c – d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a + b) × (c – d) = a × (c – d) + b × (c – d) = (a × c – a × d) + (b × c – b × d)

= ac – ad + bc – bd

Let,

a= (1/5)x, b= 2y, c= (2/3)x, d= y

Now,

= (1/5)x × ((2/3)x – y) + 2y × ((2/3)x – y)

= [((1/5)x × (2/3)x) + ((1/5)x × -y)] + [(2y × (2/3)x) + (2y × -y)]

= [(2/15)x² – (1/5)xy + (4/3)yx – 2y²)]

= [(2/15)x² + (17/15) xy – 2y²]

(3/5)m²n(m + 5n)

Let p, q and r be three monomials.

Then, by distributive law of multiplication over addition, we have:

P × (q + r) = (p × q) + (p × r)

Now,

= ((3/5)m²n × m) + ((3/5)m²n × 5n)

= ((3/5)m³n + 3m²n²) … [∵ a^m × aⁿ = a^m+n]

((2/5)x – (1/2)y) (10x – 8y)

Suppose (a – b) and (c – d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) × (c – d) = a × (c – d) – b × (c – d) = (a × c – a × d) – (b × c – b × d)

= ac – ad – bc + bd

Let,

a= (2/5)x, b=(1/2)y, c= 10x, d= 8y

Now,

= (2/5)x × (10x – 8y) – (1/2)y × (10x – 8y)

= [((2/5)x × 10x) + ((2/5)x × -8y)] – [((1/2)y × 10x) + ((1/2)y × -8y)]

= [4x² – (16/5)xy – 5yx + 4y²]

= [4x² – (41/5)xy + 4y²]

2x²(3x – 4x²)

Let p, q and r be three monomials.

Then, by distributive law of multiplication over subtraction, we have:

P × (q – r) = (p × q) – (p × r)

Now,

= (2x² × 3x) – (2x² × 4x²)

= (6x³ – 8x⁴) … [∵ a^m × aⁿ = a^m+n]

False.

Consider the given expression,

(3a – b + 3) – (a + b)

3a – b + 3 – a – b

2a – 2b + 3

Therefore, the given expression contains 3 terms.

So, it is a trinomial.

True.

In general, an expression with one or more than one term (with nonnegative integral exponents of the variables) is called a ‘Polynomial’.

6x⁴ + 9x³ – 15x² – 10x³ – 15x² + 25x + 8x² + 12x – 20

= 6x⁴ – x³ – 22x² + 37x - 20

On adding a monomial 2x or -4y² or -z to – 2x + 4y² + z, the resulting expression becomes a binomial.

2x + (-2x + 4y² + z) = 2x – 2x + 4y² + z

= 4y² + z

-4y² + (-2x + 4y² + z) = -4y² – 2x + 4y² + z

= – 2x + z

-z + (-2x + 4y² + z) = -z – 2x + 4y² + z

= -2x + 4y²

If Rohit has 5xy toffees and Shantanu has 20yx toffees, then Shantanu has 15xy more toffees.

Froom the question,

Rohit has 5xy toffees

Shantanu has 20yx toffees

Then, difference between the toffees of both Rohith and shantanu = 20yx – 5xy = 15 xy

Then Shantanu has 15 xy more toffees.

(x² – 7x + 10) (6 – 5x)

= - 5x³ + 35x² – 50x + 6x² – 42x + 60

= - 5x² + 41x² – 92x + 60

B) 15x² + 4xy – 35y² 

Correct option is (B) 15x² + 4xy – 35y²

(3x + 5y) (5x – 7y) \(=15x^2-21xy+25xy–35y^2\)

= \(15x^2+4xy–35y^2\)

As given in the question, the Sum of 2x – x² + 5 and -4x – 3 + 7x² is given as:

= 2x – x² + 5 – 4x – 3 + 7x²

= 2x – 4x – x² + 7x² + 5 – 3

= - 2x + 6x² + 2 (i)

Now subtracting equation (i) from 5 we get,

Subtracting (ii) from (i), we get

= 5 - (-2x + 6x² + 2)

= 5 + 2x – 6x² – 2

= 3 + 2x – 6x²

Therefore, the resultant expression is 3 + 2x – 6x²

3x + 23x² + 6y² + 2x + y² + (-23x²) = 5x + 7y².

Let us consider the missing letter be p.

Then,

3x + 23x² + 6y² + 2x + y² + p = 5x + 7y²

By transposing 3x, 23x², 6y², 2x and y² to RHS

5x – 3x – 23x² + 7y² – 6y² – 2x – y² = p

2x – 2x – 23x² + y² – y² = p

P = 0 – 23x² – 0

P = – 23x²

6x² – 9x – 4x + 6 + 5x² + 5x – 3x – 3

= 11x² – 11x + 3

(5x² – 2x – 3) (3x – 2)

= 15x³ – 6x² – 9x – 10x² + 4x + 6

= 15x³ – 16x² – 5x + 6

(x – 6) (4x + 9)

Suppose (a – b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) × (c + d) = a × (c + d) – b × (c + d) = (a × c + a × d) – (b × c + b × d)

= ac + ad – bc – bd

Let,

a= x, b= 6, c= 24x, d= 9

Now,

= x × (4x + 9) -6 × (4x + 5)

= [(x × 4x) + (x × 9)] – [(6 × 4x) + (6 × 9)]

= [4x² + 9x – 24x – 54]

= [4x² – 15x – 54]

2x⁴ – 4x³ + 4x² – 14x – 3x³ + 6x² – 6x + 21

= 2x⁴ – 7x³ + 10x² – 20x + 21

(4x – 3) (2x + 5)

Suppose (a – b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) × (c + d) = a × (c + d) – b × (c + d) = (a × c + a × d) – (b × c + b × d)

= ac + ad – bc – bd

Let,

a= 4x, b= 3, c= 2x, d= 5

Now,

= 4x × (2x + 5) -3 × (2x + 5)

= [(4x × 2x) + (4x × 5)] – [(3 × 2x) + (3 × 5)]

= [8x² + 20x – 6x – 15]

= [8x² + 14x – 15]

(2a²b³) × (-3a³b)

The coefficient of the product of two monomials is equal to the product of their coefficients.

The variable part in the product of two monomials is equal to the product of the variables in the given monomials.

Then,

= (2 × -3) × (a² × a³) × (b³ × b)

= (-6) × (a²⁺³) × (b³⁺¹) … [∵ a^m × aⁿ = a^m+n]

= (-6) × (a⁵) × (b⁴)

= -6a⁵b⁴

x²y (3 – 2x²y) – 1 (3 – 2x²y)

= 3x²y- 2x⁴y² – 3 + 2x²y

= 2x⁴y² + 5x²y – 3

Putting x = -1 and y = -2, we have

= [(-1)² (-2) – 1] [3 – 2 (-1)² (-2) = [-2 (-1)⁴ (-2)² + 5 (-1)² (2) – 3]

= (-2 – 1) (3 + 4) = - 8 – 10 – 3

-21 = - 21

Therefore,

L.H.S = R.H.S

Hence, verified

x²y² (x – y) (x + 2y)

= x²y² (x² + 2xy – xy – 2y²)

= x²y² (x² + xy – 2y²)

= x⁴y² + x³y³ – 2x²y⁴

(5x + 7) (3x + 4)

Suppose (a + b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a + b) × (c + d) = a × (c + d) + b × (c + d) = (a × c + a × d) + (b × c + b × d)

= ac + ad + bc + bd

Let,

a= 5x, b= 7, c= 3x, d= 4

Now,

= 5x × (3x + 4) +7 × (3x + 4)

= [(5x × 3x) + (5x × 4)] + [(7 × 3x) + (7 × 4)]

= [15x² + 20x + 21x + 28]

= [15x² + 41x + 28]

9t²(t + 7t³)

Let p, q and r be three monomials.

Then, by distributive law of multiplication over addition, we have:

P × (q + r) = (p × q) + (p × r)

Now,

= (9t² × t) + (9t² × 7t³)

= (9t³ + 63t⁵) … [∵ a^m × aⁿ = a^m+n]

(3x – 5y) × (x + y)

= x (3x – 5y) + y (3x – 5y)

= 3x² – 5xy + 3xy – 5y²

= 3x² – 2xy – 5y²

Putting x = - 1 and y = - 2, we have

[3 (-1) – 5 (-2)] [(1) + (-2)] = 3 (-1)² – 2 (-1) (-2) – 5 (-2)²

(-3 + 10) (-1 – 2) = 3 – 4 – 20

- 21 = - 21

Therefore,

L.H.S = R.H.S

Hence, verified

a²b² (3a² + ab + 6ab + 2b²)

= a²b² (3a² + 7ab + 2b²)

= 3a⁴b² + 7a³b³ + 2a²b⁴

(-3d + 7f) × (5d + f)

= -3d (5d + f) + 7f (5d + f)

= -15d² – 3df + 35df + 7f²

= -15d² + 32df + 7f²

10a²(0.1a – 0.5b)

Let p, q and r be three monomials.

Then, by distributive law of multiplication over subtraction, we have:

P × (q – r) = (p × q) – (p × r)

Now,

= (10a² × 0.1a) – (10a² × 0.5b)

= (1a³ – 5a²b) … [∵ a^m × aⁿ = a^m+n]

(3x² + y²) × (2x² + 3y²)

= 3x² (2x² + 3y²) + y² (2x² + 3y²)

= 6x⁴ + 9x²y² + 2x²y² + 3y⁴

= 6x⁴ + 11x²y² + 3y⁴

(x² – y²) × (3a + 2b)

= x² (3a + 2b) – y² (3a + 2b)

= 3ax² + 2bx² – 3ay² – 2by²

(a – 1) × (0.1a² + 3)

= a (0.1a² + 3) – 1 (0.1a² + 3)

= 0.1a³ + 3a – 0.1a² - 3

(2x + 8) × (x – 3)

= 2x (x – 3) + 8 (x – 3)

= 2x² – 6x + 8x – 24

= 2x² – 2x - 24

(x⁶ – y⁶) × (x² + y²)

= x⁶ (x² + y²) – y⁶ (x² + y²)

= x⁸ + x⁶y² – x²y⁶ – y⁸

(7x + y) × (x + 5y)

= 7x (x + 5y) + y (x + 5y)

= 7x² + 35xy + xy + 5y²

= 7x² + 36xy + 5y²

(5x + 3) × (7x + 2)

= 5x (7x + 2) + 3 (7x + 2)

= 35x² + 10x + 21x + 6

= 35x² + 31x + 6

(2xy + 3y²) × (3y² – 2)

= 2xy (3y² – 2) + 3y² (3y² – 2)

= 6xy³ – 4xy + 3y⁴ – 6y²

(2x² – 1) × (4x³ + 5x²)

= 2x² (4x³ + 5x) – 1 (4x³ + 5x²)

= 8x⁵ + 10x³ – 4x³ – 5x²

= 8x⁵ + 6x³ – 5x²

Correct option is  B) x² – 1

We know that,

From formula, (a + b)(a – b) = a² – b²

(a + 1)(a – 1)(a² + 1) = (a² – 1)(a²+ 1)

Again applying the formula,

(a² – 1)(a²+ 1) = (a²)² – (12)² = a4 – 1

Correct option is  B) 0

Correct option is (B) 0

\((3x^2–7x+1)(4x–4x)\) \(=(3x^2–7x+1)\times0=0\)

Correct option is  A) 0

Given (5x + 7) × (3x + 4)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(5x + 7) × (3x + 4)

⇒5x (3x + 4) + 7 (3x + 4)

⇒15x² + 20x + 21x + 28

⇒ 15x² + 41x + 28

Given (82)²

But we can write 82 as 80+2

And also we know that (a + b)² = a²+2ab+b²

By applying the above identity we get

(82)² = (80+2)²= 80²+2(80) (2) +2²

(80+2)²=6400+320+4=6724