Operations On Algebraic Expressions

To add the expressions, we have to arrange the given expression in the form of rows and then add the expression column wise, so we have;

To add the expressions, we have to arrange the given expression in the form of rows and then add the expression column wise, so we have;

To add the expressions, we have to arrange the given expression in the form of rows and then add the expression column wise, so we have;

To add the expressions, we have to arrange the given expression in the form of rows and then add the expression column wise, so we have;

To add the expressions, we have to arrange the given expression in the form of rows and then add the expression column wise, so we have;

Let’s arrange the data in a table in the form of descending power of x,

We will get rows and columns; add the data column wise;

So, the answer after adding all the expressions will be;

5x³ - 11x² – 14x + 7

To add the expressions, we have to arrange the given expression in the form of rows and then add the expression column wise, so we have;

So, the answer is;

– 6p + 17q

By arrange the given expression in descending powers of x it will be easier To add the expressions,s,

So, we have;

We have to subtract 3a²b from – 5a²b.

According to the rule when both the expressions have negative sign so we add both the expression and put negative sign only.

So, by arranging the data in rows and columns form we have;

According to the rule of subtraction two negative becomes positive and the result will have negative sign.

So, To subtract the expression,we have to arrange the given expression in the form of rows and then subtract the expression column wise,

Therefore, we have;

According to the rule of subtraction two negative becomes positive and the result will have negative sign.

To subtract the expression, we have to arrange the given expression in the form of rows and then subtract the expression column wise, so we have;

According to the rule of subtraction two negative becomes positive and the result will have negative sign.

To subtract the expression,we have to arrange the given expression in the form of rows and then subtract the expression column wise, so we have;

According to the rule of subtraction two negative becomes positive and the result will have negative sign.

To subtract the expression,we have to arrange the given expression in the form of rows and then subtract the expression column wise, so we have;

According to the rule of subtraction two negative becomes positive and the result will have negative sign.

To subtract the expression,we have to arrange the given expression in the form of rows and then subtract the expression column wise, so we have;

According to the rule of subtraction two negative becomes positive and the result will have negative sign.

To subtract the expression,we have to arrange the given expression in the form of rows and then subtract the expression column wise, so we have;

According to the rule of subtraction two negative becomes positive and the result will have negative sign.

To subtract the expression,we have to arrange the given expression in the form of rows and then subtract the expression column wise, so we have;

According to the rule of subtraction two negative becomes positive and the result will have negative sign.

To subtract the expression,we have to arrange the given expression in the form of rows and then subtract the expression column wise, so we have;

Let’s suppose the required number be x,

So we have;

(3a² – 6ab – 3b² – 1) – x = 4a² – 7a – 4b² + 1

(3a² – 6ab – 3b² – 1) – (4a² – 7a – 4b² + 1) = x

So,

To get the required number we have to subtract 4a² – 7a – 4b² + 1 from 3a² – 6ab – 3b² - 1

So, the required number is - a² + ab + b² – 2

We know that;

Two adjacent sides of a rectangle are l and b;

l = 5x² – 3y²

b = x² + 2xy

Perimeter of rectangle = (2l+2b)

Which is;

2 (5x² – 3y²) + 2(x² + 2xy)

= (10x² – 6y²) + (2x² + 4xy)

Perimeter of the triangle = 6p² – 4p + 9

Two sides are;

Side one = p² – 2p + 1 and

Side two = 3p² – 5p + 3

Let’s take third side be = x

As we know perimeter of a triangle = sum of all the sides

So, we have

6p² – 4p + 9 = {(p² – 2p + 1) + (3p² – 5p + 3) + (x)}

6p² – 4p + 9 = p² – 2p + 1 + 3p² – 5p + 3 + x

6p² – 4p + 9 - p² + 2p – 1 - 3p² + 5p - 3 = x

Let’s make the pairs;

(6p² – p² – 3p²) + (- 4p + 2p + 5p) + (9 – 1 – 3) = x

2p² + 3p + 5 = x

The required side is 2p² + 3p + 5.

To find the product of the given expression we have to Horizontal method;

Horizontal method is the method where each term of one expression is multiplied with each term of other expression.

So, by using horizontal method,

We have;

= (5x + 7) × (3x + 4)

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

= 15x² + 20x + 21x + 28

= 15x² 41x + 28

By using horizontal method,

We have;

= (4x + 9) × (x - 6)

= 4x(x – 6) + 9(x - 6)

= 4x² – 24x + 9x – 54

= 4x² – 15x – 54

By using horizontal method,

We have;

= (2x + 5) × (4x - 3)

= 2x (4x – 3) + 5 (4x – 3)

= 8x² - 6x + 20x – 15

= 8x² + 14x - 15

By using horizontal method,

We have;

= (3y - 8) × (5y - 1)

= 3y(5y – 1) – 8(5y – 1)

= 15y² – 3y – 40y + 8

= 15y² – 43y +8

By using horizontal method,

We have;

= (7x + 2y) × (x + 4y)

= 7x(x +4y) + 2y(x + 4y)

= 7x² + 28xy + 2xy + 8y²

= 7x² + 30xy + 8y²

By using horizontal method,

We have;

= (9x + 5y) × (4x + 3y)

= 9x(4x + 3y) + 5y(4x + 3y)

= 36x² + 27xy + 20xy + 15y²

= 36x² + 47xy + 15y²

By using horizontal method,

We have;

= (3m – 4n) × (2m – 3n)

= 3m(2m – 3n) – 4n(2m – 3n)

= 6m² – 9mn – 8mn + 12n²

= 6m² – 17mn + 12n²

By using horizontal method,

We have;

= (x² – a²) × (x – a)

= x²(x – a) – a²(x – a)

= x³ – ax² – a²x + a³

By using horizontal method,

We have;

= (x² – y²) × (x + 2y)

= x² (x + 2y) – y²(x + 2y)

= x³ + 2x²y – xy² – 2y³

By using horizontal method,

We have;

= (3p² + q²) × (2p² – 3q²)

= 3p²(2p² – 3q²) + q²(2p² – 3q²)

= 6p⁴ – 9p²q² + 2p²q² – 3q⁴

= 6p⁴ – 7p²q² – 3q⁴

By using horizontal method,

We have;

= (2x² – 5y²) × (x² + 3y²)

= 2x²(x² + 3y²) – 5y²(x² + 3y²)

= 2x⁴ + 6x²y² – 5x²y² – 15y⁴

= 2x⁴ + x²y² – 15y⁴

By using horizontal method,

We have;

= (x³ – y³) × (x² + y²)

= x³(x² + y²) – y³(x² + y²)

= x⁵ + x³y² – x²y³ - y⁵

By using horizontal method,

We have;

= (x⁴ + y⁴) × (x² – y²)

= x⁴(x² – y²) + y⁴(x² – y²)

= x⁶ – x⁴y² + x²y⁴ – y⁶

By using horizontal method,

We have;

By using horizontal method,

We have;

= (x² – 3x + 7) × (2x + 3)

= 2x(x² – 3x + 7) + 3(x² – 3x + 7)

= 2x³ - 6x² + 14x + 3x² – 9x + 21

By arranging the expression in the form of descending powers of x,

We get;

= 2x³ – 6x² + 3x² + 14x – 9x + 21

= 2x³ – 3x² + 5x + 21

By using horizontal method,

We have;

= (3x² + 5x – 9) × (3x – 5)

= 3x(3x² + 5x – 9) – 5(3x² + 5x – 9)

= 9x³ + 15x² – 27x – 15x² - 25x + 45

By arranging the expression in the form of descending powers of x,

We get;

= 9x³ + 15x² – 15x² – 27x – 25x + 45

= 9x³ – 52x + 45

By using horizontal method,

We have;

= (x² – xy + y²) × (x + y)

= x(x² – xy + y²) + y(x² – xy + y²)

= x³ - x²y + xy² + x²y – xy² + y³

By arranging the expression in the form of descending powers of x,

We get;

= (x³ + y³)

By using horizontal method,

We have;

= (x² + xy + y²) × (x - y)

= x(x² + xy + y²) - y(x² + xy + y²)

= x³ + x²y + xy² - x²y – xy² - y³

By arranging the expression in the form of descending powers of x,

We get;

= (x³ - y³)

By using horizontal method,

We have;

= (x³ – 2x² + 5) × (4x – 1)

= 4x(x³ – 2x² + 5) – 1(x³ – 2x² + 5)

= 4x⁴ – 8x³ + 20x – 1x³ + 2x² – 5

By arranging the expression in the form of descending powers of x,

We get;

= 4x⁴ – 8x³ – x³ + 2x² + 20x – 5

= 4x⁴ – 9x³ + 2x² + 20x – 5

By using horizontal method,

We have;

= (9x² – x + 15) × (x² – 3)

= x²(9x² – x +15) – 3(9x² – x + 15)

= 9x⁴ – x³ + 15x² – 27x² + 3x – 45

= 9x⁴ - x³ – 12x² + 3x – 45

By using horizontal method,

We have;

= (x² – 5x + 8) × (x² + 2)

= x²(x² – 5x + 8) + 2(x² – 5x + 8)

= x⁴ – 5x³ + 8x² + 2x² – 10x + 16

= x⁴ – 5x³ + 10x² – 10x +16

By using horizontal method,

We have;

= (x³ – 5x² + 3x + 1) × (x² – 3)

= x²(x³ – 5x² + 3x + 1) – 3(x³ – 5x² + 3x +1)

= x⁵ – 5x⁴ + 3x³ + x² – 3x³ + 15x² – 9x – 3

By arranging the expression in the form of descending powers of x,

We get;

= x⁵ – 5x⁴ + 3x³ – 3x³ + x² + 15x² – 9x – 3

= x⁵ – 5x⁴ +16x² – 9x – 3

By using horizontal method,

We have;

= (3x + 2y – 4) × (x – y + 2)

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

= 3x² + 2xy – 4x – 3xy – 2y² + 4y + 6x + 4y – 8

By arranging the expression in the form of descending powers of x,

We get;

= 3x² – 4x + 6x + 2xy – 3xy – 2y² + 4y + 4y – 8

= 3x² + 2x – xy – 2y² + 8y – 8

By using horizontal method,

We have;

= (x² – 5x + 8) × (x² + 2x – 3)

= x²(x² – 5x + 8) + 2x(x² – 5x + 8) – 3(x² – 5x + 8)

= x⁴ – 5x³ + 8x² + 2x³ – 10x² + 16x – 3x² + 15x – 24

By arranging the expression in the form of descending powers of x,

We get;

= x⁴ – 3x³ – 5x² + 31x – 24

By using horizontal method,

We have;

(2x²+ 3x – 7) ×(3x² – 5x + 4)

= 2x²(3x² – 5x + 4) + 3x(3x² – 5x + 4) – 7(3x² – 5x + 4)

= 6x⁴ – 10x³ + 8x² + 9x³ – 15x² + 12x– 21x² + 35x – 28

Now, putting equal power terms together, we get,

= 6x⁴ – 10x³ + 9x³ + 8x² – 15x²– 21x² + 35x + 12x– 28

= 6x⁴ – x³ – 28x² + 47x – 28

By using horizontal method,

We have;

(9x² – x + 15) × (x² – x – 1)

= 9x²(x² – x – 1) – x (x² – x – 1) +15(x² – x – 1)

= 9x⁴ – 9x³ – 9x² – x³ + x² + x + 15x² – 15x – 15

Putting equal power terms together, we get,

= 9x⁴ – 9x³ – x³ – 9x² + x² + 15x² – 15x + x – 15

= 9x⁴ – 10x³ + 7x² – 14x – 15

(i) By dividing 24x²y³ by 3xy

We get;

= 8xy²

(ii) By dividing 36xyz² by by – 9xz

We get;

= - 4yz

(iii) By dividing – 72x²y²z by – 12xyz

We get;

= 6xy

(iv) By dividing – 56mnp² by 7mnp

We get;

= - 8p

(i) By dividing 5m³ – 30m² + 45m by 5m

We get;

= m² – 6m + 9

(ii) By dividing 8x²y² – 6xy² + 10x²y³ by 2xy

We get;

= 4xy – 3y + 5xy²

(iii) If we divide 9x²y – 6xy + 12xy² by – 3xy

We get;

= - 3x + 2 – 4y

(iv) If we divide 12x⁴ + 8x³ – 6x² by – 2x²

We get;

= - 6x² – 4x + 3

If we divide x² – 4x + 4 by x -2;

So, we get;

Quotient = x – 2 and remainder = 0

If we divide (x² – 4) by (x + 2);

So, we get;

Quotient = x – 2 and remainder = 0

If we divide (x² + 12x + 35) by (x + 7)

So, we get;

Quotient = (x + 5) and remainder = 0

If we divide (15x² + x – 6) by (3x + 2)

We get;

Quotient = (5x – 3) and remainder = 0

If we divide (14x² – 53x + 45) by (7x – 9)

So we get;

Quotient = 2x - 5 and remainder = 0

By dividing the given expressions we get;

Quotient = (3x – 8) and remainder = 7

By dividing the given expressions we get;

Quotient = (x² – x – 1) and remainder = 1

By dividing the given expressions we get;

Quotient = (x² – x + 1) and remainder = 0

By dividing the given expressions we get;

Quotient = x² – 3x + 4 and remainder = 0

By dividing the given expressions we get;

Quotient = (x – 1) and remainder = 0

By dividing the given expressions we get;

Quotient = (5x + 3) and remainder = (x + 1)

By dividing the given expressions we get;

Quotient = (x – 1) and remainder = 0

If we divide (8x⁴ + 10x³ – 5x² – 4x + 1) by (2x² – 3x + 5)

We get,

So,

We get the quotient 4x² + 11x + 4

And the remainder – 47x – 19

(i) As we have (x + 6)(x+6)

(x + 6)(x + 6) = (x + 6)²

By using the formula;

[(a + b)² = a² + b² + 2ab]

We get,

(x + 6)² = x² + (6)² + 2× (x) × (6)

= x² + 36 + 12x

By arranging the expression in the form of descending powers of x we get;

= x² + 12x + 36

(ii) Given;

(4x + 5y)(4x + 5y)

By using the formula;

[(a + b)² = a² + b² + 2ab]

We get,

(4x + 5y)(4x + 5y) = (4x + 5y)²

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

= 16x² + 25y² + 40xy

(iii) Given,

(7a + 9b)(7a + 9b)

By using the formula;

[(a + b)² = a² + b² + 2ab]

We get,

(7a + 9b)(7a + 9b) = (7a + 9b)²

(7a + 9b)² = (7a)² + (9b)² + 2 × (7a) × (9b)

= 49a² + 81b² + 126ab

(iv)

By using the formula (a + b)²

We get;

(v) (x² + 7)(x² + 7)

By using the formula (a + b)²

We get;

(x² + 7)(x² + 7) = (x² + 7)²

= (x²)² +(7)² + 2 × (x²) × (7)

= x⁴ + 49 + 14x²

(vi)

By using the formula (a + b)²

We get;

(i) Given,

(x – 4)(x – 4)

By using the formula (a – b)² = a² – 2ab + b²

We get;

= (x – 4)²

= (x)² – 2 × (x) × 4 + (4)²

= x² – 8x + 16

(ii) Given,

(2x – 3y)(2x – 3y)

By using the formula (a – b)² = a² – 2ab + b²

We get;

= (2x – 3y)²

= (2x)² – 2 × (2x) × (3y) + (3y)²

= 4x² – 12xy + 9y²

(iii)

By using the formula (a – b)² = a² – 2ab + b²

We get;

(iv)

By using the formula (a – b)² = a² – 2ab + b²

We get;

(v)

By using the formula (a – b)² = a² – 2ab + b²

We get;

(vi)

By using the formula (a – b)² = a² – 2ab + b²

We get;

(i) Given,

(8a + 3b)²

By using the formula (a + b)² = a² + b² + 2ab

We get;

= (8a)² + (3b)² + 2 × 8a × 3b

= 64a² + 9b² + 48ab

(ii) (7x + 2y)²

By using the formula (a + b)² = a² + b² + 2ab

We get;

= (7x)² + (2y)² + 2 × (7x) × (2y)

= 49x² + 4y² + 28xy

(iii) (5x + 11)²

By using the formula (a + b)² = a² + b² + 2ab

We get;

= (5x)² + (11)² + 2×(5x) × 11

= 25x² + 121 + 110x

(iv)

By using the formula (a + b)² = a² + b² + 2ab

We get;

(v)

By using the formula (a + b)² = a² + b² + 2ab

We get;

(vi) (9x – 10)²

By using the formula (a - b)² = a² - 2ab + b²

We get;

(9x – 10)²

= (9x)² – 2 × (9x) × 10 + (10)²

= 81x² – 180x + 100

(vii) (x²y – yz²)²

By using the formula (a - b)² = a² - 2ab + b²

We get;

= (x²y – yz²)²

= (x²y)² – 2 × (x²y) × yz² + (yz²)²

= x⁴y² – 2x²y²z² + y²z⁴

(viii)

By using the formula (a - b)² = a² - 2ab + b²

We get;

(ix)

By using the formula (a - b)² = a² - 2ab + b²

We get;

=

(i) Given,

(x + 3)(x – 3)

By using the formula (a + b) (a – b) = a² – b²

We get;

= x(x + 3) – 3(x + 3)

= x² + 3x – 3x – 9

= x² – 9

(ii) Given,

(2x + 5)(2x – 5)

By using the formula (a + b) (a – b) = a² – b²

We get;

= 2x(2x + 5) – 5(2x + 5)

= 4x² + 10x – 10x – 25

= 4x² – 25

(iii) Given,

(8 + x)(8 – x)

By using the formula (a + b) (a – b) = a² – b²

We get;

= 8(8 + x) – x(8 + x)

= 64 + 8x – 8x – x²

= 64 – x²

(iv) Given,

(7x + 11y)(7x – 11y)

By using the formula (a + b) (a – b) = a² – b²

We get;

= 7x(7x + 11y) – 11y(7x + 11y)

= 49x² + 77xy – 77xy – 121y²

= 49x² – 121y²

(v) Given,

By using the formula (a + b) (a – b) = a² – b²

We get;

(vi)

By using the formula (a + b) (a – b) = a² – b²

We get;

(vii)

By using the formula (a + b) (a – b) = a² – b²

We get;

(viii)

By using the formula (a + b) (a – b) = a² – b²

We get;

(ix)

By using the formula (a + b) (a – b) = a² – b²

We get;

(i) Given,

(54)²

If we break the given number we get;

(50 + 4)²

Now we can use the (a + b)² = a² + b² + 2ab

So,

= (50 + 4)² = (50)² + (4)² + 2 × 50 × 4

= 2500 + 16 + 400

= 2916

(ii) (82)²

We can also write it as;

(80 + 2)²

By using the formula (a + b)² = a² + b² + 2ab

We get,

= (80 + 2)² = (80)² + (2)² + 2 × 80 × 2

= 6400 + 4 + 320

= 6724

(iii) (103)²

We can also write it as;

(100 + 3)²

By using the formula (a + b)² = a² + b² + 2ab

We get,

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

= 10000 + 9 + 600

= 10609

(iv) (704)²

We can also write it as;

(700 + 4)²

By using the formula (a + b)² = a² + b² + 2ab

We get,

= (700 + 4)² = (700)² + (4)² + 2 × 700 × 4

= 490000 + 16 + 5600

= 495616

(i) Given,

(69)²

We can also write it as;

(70 – 1)²

Now,

By using the formula (a - b)² = a² - 2ab + b²

We get,

= (70 – 1)² = (70)² – 2 × 70 × 1 + (1)²

= 4900 – 140 + 1

= 4761

(ii) Given = (78)²

We can also write it as;

(80 – 2)²

Now,

By using the formula (a - b)² = a² - 2ab + b²

We get,

(80 – 2)² = (80)² – 2 × 80 × 2 + (2)²

= 6400 – 320 + 4

= 6084

(iii) (197)²

We can also write it as;

(200 – 3)²

Now,

By using the formula (a - b)² = a² - 2ab + b²

We get,

(200 – 3)² = (200)² – 2 × 200 × 3 + (3)²

= 40000 – 1200 + 9

= 38809

(iv) (999)²

We can also write it as;

(1000 – 1)²

Now,

By using the formula (a - b)² = a² - 2ab + b²

We get,

(1000 - 1)² = (1000)² – 2 × 1000 × 1 + (1)²

= 1000000 – 2000 + 1

= 998001

(i) Given,

(82)² – (18)²

By using (a – b)(a + b) = a² – b²

= (82 – 18)(82 + 18)

= (64)(100)

= 6400

(ii) (128)² – (72)²

By using (a – b)(a + b) = a² – b²

= (128 – 72)(128 + 72)

= (56)(200)

=11200

(iii) 197 × 203

By converting the given number into the form of formula we get,

= (200 – 3)(200 + 3)

= (200)² –(3)²

= 40000 – 9

= 39991

(iv) Given,

By using the formula (a – b)(a + b) = a² – b²

We get;

= 300

(v) (14.7 × 15.3)

By using (a – b)(a + b) = a² – b²

We get;

= (15 – 0.3)(15 + 0.3)

= (15)² – (0.3)²

= 225 – 0.09

= 224.91

(vi) (8.63)² – (1.37)²

By using (a – b)(a + b) = a² – b²

We get;

= (8.63 – 1.37)(8.63 + 1.37)

= (7.26)(10)

= 72.6

Given,

(9x² + 24x + 16)

x = 12

So, we can also write it as;

= (3x)² + 2(3x)(4) + (4)²

→ By the formula (a + b)² we get,

= (3x + 4)²

= [3 (12) + 4]²

= [36 + 4]²

= [40]² = 1600

Hence the value of the expression is 1600 when x =12.

Given,

(64x² + 81y² + 144xy)

X = 11

Y =

By using the formula (a + b)² we get;

= (8x)² + (9y)² + 2(8x)(9y)

= (8x + 9y)²

= [8(11) + 9 ]²

= (88 + 12)²

= (100)² = 10000

Hence the value of the expression is 10000.

Given,

(36x² + 25y² – 60xy)

X =

Y =

With the help of the formula (a - b)² we get;

= (6x)² + (5y)² – 2(6x)(5y)

= (6x – 5y)²

=

= (4 – 1)²

= (3)² = 9

(i)

We know that,

From formula (a + b)² = a² + b² + 2ab

=

=

So, by putting the values , we get,

= 4² =

=

(ii)

We know that,

From formula (a + b)² = a² + b² + 2ab

=

=

So, by putting the values , we get,

= 14² =

=

(i)

We know that,

From formula (a – b)² = a² + b² – 2ab

=

=

So, by putting the values, we get,

= 5² =

=

(ii)

We know that,

From formula (a + b)² = a² + b² + 2ab

=

=

So, by putting the values, we get,

= 27² =

=

(i) (x +1)(x – 1)(x² + 1)

We know that, from formula,

(a + b)(a – b) = a² – b²

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

= (x²)² – 1 = x⁴ – 1

(ii) (x- 3)(x + 3)(x² + 9)

We know that, from formula,

(a + b)(a – b) = a² – b²

(x – 3)(x + 3)(x² + 9)

= (x² – 9)(x² + 9)

= (x²)² – 9² = x⁴ – 81

(iii) (3x – 2y)(3x + 2y)(9x² + 4y²)

We know that, from formula,

(a + b)(a – b) = a² – b²

(3x – 2y)(3x + 2y)(9x² + 4y²)

= (9x² – 4y²)(9x² + 4y²)

= 81x⁴ – 16y⁴

(iv) (2p + 3)(2p – 3)(4p² + 9)

We know that, from formula,

(a + b)(a – b) = a² – b²

(2p + 3)(2p – 3)(4p² + 9)

= (4p² – 9)(4p²+ 9)

= (4p²)² – 9² = 16p⁴ – 81

Given,

x + y = 12

Let’s square the both sides,

We get;

= (x + y)² = (12)²

= x² + y² + 2xy = 144

= x² + y² = 144 – 2xy

Also given,

xy = 14

= x² + y² = 144 – 2(14)

= x² + y² = 144 – 28

= x² + y² = 116

So, the value of (x² + y²) is 116.

x – y = 7 (given)

By squaring both the sides we get;

= (x – y)² = (7)²

= x² + y² – 2xy = 49

= x² + y² = 49 + 2xy

Also given,

xy = 9

= x² + y² = 49 + 2(9)

= x² + y² = 49 + 18

= x² + y² = 67

So, the value of x² + y² is 67.

After solving the bracket,

we get,

= 3q + 7p² – 2r³ + 4 – 4p² + 2q – 7r³ + 3 = 7p² – 4p² + 3q + 2q – 2r³ – 7r³ + 3 + 4

= 3p² + 5q – 9r³ + 7

After solving the equation,

we get,

(x + 5)(x – 3) = x(x – 3) + 5(x – 3)

= x² – 3x + 5x – 15

= x² + 2x – 15

After solving the equations,

we get,

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

= 6x² – 2x + 9x – 3

= 6x² + 7x – 3

We know that,

(x + 4)(x + 4) = (x + 4)²

From formula, (a + b)² = a² + b² + 2ab

(x + 4)² = x² + 4² + 2 × x × 4

= x² + 8x + 16

We know that,

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

(2x + 5)(2x – 5) = (2x)² – (5)²

= 4x² – 25

If we divide 8a²b³ by (-2ab)we get;

=

= - 4ab²

By dividing (2x² + 3x + 1) by (x + 1)

We get;

By dividing (x² - 4x + 4) by (x - 2)

We get;

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

We know that,

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

We know that,

From formula, (a + b)² = a² + b² + 2ab

……………….(i)

And

Putting value of in equation (i), we get,

(5)² =

= 25 – 2 = 23.

We know that,

From formula, (a – b)² = a² + b² – 2ab

……………….(i)

And

Putting value of in equation (i), we get,

(6)² =

= 36+ 2= 38.

(82)² – (18)²

By using (a – b)(a + b) = a² – b²

= (82 – 18)(82 + 18)

= (64)(100)

= 6400

We can write following problem such as,

(197 × 203) = (200 – 3)(200 + 3)

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

We get,

(200 – 3)(200 + 3) = 200² – 3² = 40000 – 9 = 39991.

From the formula,

(a + b)² = a² + b² + 2ab

We get,

= a+ b = 12 and ab = 14

By putting values, we get,

12² = a² + b² + 2× 14

= a² + b² = 144 – 28 = 116.

From the formula,

(a – b)² = a² + b² –2ab

We get,

= a – b = 7 and ab = 9

By putting values, we get,

7² = a² + b² - 2× 9

= a² + b² = 49 + 18 = 67

(4x² + 20x + 25)

By using (a + b)² = a² + b² + 2ab,

We get;

= (2x)² + (5)² + 2(2x)(5)

= (2x + 5)²

= (2(10) + 5)²

= (20 + 5)²

= (25)²

= 625