FIND EXPANSION USING THREE BASIC ALGEBRIAC IDENTITIES

Expand the following using algebraic identities given above.

Problem 1 :

(2x + y)²

Solution :

Here a = 2x and b = y

It looks like (a + b)²

= (2x)² + 2(2x) y + y²

= 2²x² + 2(2x) y + y²

= 4x² + 4xy + y²

Problem 2 :

(3a + 2b)²

Solution :

a = 3a and b = 2b

It looks like (a + b)²

= (3a)² + 2(3a) (2b) + (2b)²

= 3²a² + 12ab + 2²b²

= 9a² + 12ab + 4b²

Problem 3 :

(x² + 2)²

Solution :

a = x² and b = 2

It looks like (a + b)²

= (x²)² + 2(x²) (2) + 2²

= x⁴ + 4x² + 4

Problem 4 :

(√2 + x)²

Solution :

a = √2 and b = x

It looks like (a + b)²

= (√2)² + 2(√2) (x) + x²

= 2 + 2√2x + x²

Problem 5 :

(x - 1/x)²

Solution :

a = x and b = 1/x

It looks like (a - b)²

= x² - 2x (1/x) + (1/x)²

= x² - 2 + (1/x²)

Problem 6 :

(2 - x²)²

Solution :

a = 2 and b = x²

It looks like (a - b)²

= 2² - 2(2) (x²) + (x²)²

= 4 - 4x² + x⁴

Problem 7 :

(8y - 5z)²

Solution :

a = 8y and b = 5z

It looks like (a - b)²

= (8y)² - 2(8y) (5z) + (5z)²

= 8²y² - 16y (5z) + 5²z²

= 64y² - 80yz + 25z²

Problem 8 :

x² - 9

Solution :

Here 9 = 3²

x² - 9 = x² - 3²

It looks like a² - b²

Here a = x and b = 3

x² - 9 = (x + 3) (x - 3)

Problem 9 :

x²y² - 1

Solution :

= x²y² - 1

= (xy)² - 1²

= (xy + 1) (xy - 1)

Problem 10 :

3x² - 27y²

Solution :

= 3x² - 27y²

Factoring 3, we get

= 3(x² - 9y²)

= 3 (x² - 3²y²)

= 3(x² - (3y)²)

= 3(x + 3y) (x - 3y)

Problem 11 :

400x² - 169y²

Solution :

= 400x² - 169y²

We cannot factorize anything.

400 = 20² and 169 = 13²

= ((20x)² - (13y)²)

= (20x + 13y) (20x - 13y)

Problem 12 :

(a+b)² - c²

Solution :

Here a = a + b and b = c

Using the algebraic identity a² - b², we get

= (a + b + c) (a + b - c)

Problem 13 :

(3x + 1)² - 9

Solution :

= (3x + 1)² - 9

= (3x + 1)² - 3²

Here a = 3x + 1 and b = 3

= (3x + 1 + 3) (3x + 1 - 3)

= (3x + 4) (3x - 2)