Special Products and Expansions

Special Products and Expansions

Special products are the products of algebraic expressions which can be obtained by using many methods.

Product of two Binomials having One Common Term

1. (x + a) and (x + b)

(x + a) (x + b) = x (x + b) + a (x + b) = x² + bx + ax + ab = x² + x (a + b) + ab

Thus, (x + a) (x + b) = x² + (a + b) x + ab

2. (x – a) and (x + b)

(x – a) (x + b) = x (x + b) – a (x + b) = x² + bx – ax – ab = x² + x (b – a) – ab

Thus, (x – a) (x + b) = x² + (b – a) x – ab

3. (x + a) and (x – b)

(x + a) (x – b) = x (x – b) + a (x – b) = x² – bx + ax – ab = x² – x (b – a) – ab

Thus, (x + a) (x – b) = x² – (b – a) x – ab

4. (x – a) and (x – b)

(x – a) (x – b) = x (x – b) – a (x – b) = x² – bx – ax + ab = x² – x (b + a) + ab

Thus, (x – a) (x – b) = x² – (a + b) x + ab

Example 1: Using the special products, find the product of the following:

a. (x + 4) (x + 6)

b. (x + 3) (x – 2)

c. (x – 6.5) (x + 2.5)

Solution: a. (x + 4) (x + 6)

Here, a = 4 and b = 6

We know that, (x + a) (x + b) = x² + (a + b) x + ab

Replacing the values of a and b we get, (x + 4) (x + 6) = x² + (6 + 4) x + (6 × 4)

Thus, (x + 4) (x + 6) = x² + 10x + 24

b. (x + 3) (x – 2)

Here, a = 3 and b = 2

We know that (x + a) (x – b) = x² – (b – a) x – ab

Replacing the values of a and b we get, (x + 3) (x – 2) = x² – (2 – 3) x – (3 × 2) = x² + x – 6 Thus, (x + 3) (x – 2) = x2 + x – 6

c. (x – 6.5) (x + 2.5)

Here, a = 6.5 and b = 2.5

We know that (x – a) (x + b) = x² + (b – a) x – ab

Replacing the values, we get (x – 6.5) (x + 2.5) = x² + (2.5 – 6.5) x – (2.5 × 6.5) = x² – 4x – 16.25

Thus, (x – 6.5) (x + 2.5) = x² – 4x – 16.25

Example 2: Evaluate the following expressions:

a. 105 × 102                               b. 58 × 45

Solution: a. 105 × 102 = (100 + 5) (100 + 2) = (100)² + 100 (5 + 2) + 5 × 2 

= 10,000 + 700 + 10 = 10710

b. 58 × 45 = (50 + 8) (50 – 5) = (50)² + 50 (8 – 5) + 8 × (–5) = 2500 + 150 – 40 = 2610

Product of Sum and Difference of Two Terms

Consider (x + y) and (x – y).

(x + y) (x – y) = x (x – y) + y (x – y) = x² – xy + xy − y²

Thus, (x + y) (x – y) = x² – y²

Example 3: Using special products, find the following products.

a. (x + 6) (x – 6)                           b. (5x + 4y) (5x – 4y)

Solution: a. (x + 6) (x – 6)

Using the formula (x + y) (x – y) = x² – y²

(x + 6) (x – 6) = x² – 6² = x² – 36

Thus, (x + 6) (x – 6) = x² – 36

b. (5x + 4y) (5x – 4y) = (5x)² – (4y)²

Thus, (5x + 4y) (5x – 4y) = 25x² – 16y²

Squares of Binomials

When an algebraic expression is multiplied with itself by any number of times, the result obtained is known as an expansion.

1.      Consider (x + y)² = (x + y) (x + y).

(x + y) (x + y) = x (x + y) + y (x + y) = x² + xy + xy + y² = x² + 2xy + y²

Thus, (x + y)² = x² + 2xy + y²

2.      Now, consider (x – y)² = (x – y) (x – y).

(x – y) (x – y) = x (x – y) – y (x – y) = x² – xy – xy + y² = x² – 2xy + y²

Thus, (x – y) (x – y) = x² – 2xy + y²

Example 4: Using binomial expansion, find the following products.

a. (x + 5)²                          b. (2x – 3y)²

Solution: a. (x + 5)² = x² + 2 × x × 5 + 5² = x² + 10x + 25                        

b. (2x – 3y)² = (2x)² – 2 × 2x × 3y + (3y)² = 4x² – 12xy + 9y²     

Squares of Trinomials

Consider (a + b + c)² .

We can evaluate (a + b + c)² as follows:

(a + b + c)² = (a + k)² (Taking k = b + c)

                     = a² + 2ak + k²

                     = a² + 2a(b + c) + (b + c)² (Substituting the value of k)

                     = a² + 2ab + 2ac + b² + c² + 2bc = a² + b² + c² + 2ab + 2bc + 2ca

Example 5: Expand the following:

a. (a – b + c)²                            b. (2x – 3y + z)²

Solution: a. (a – b + c)² = (a)² + (–b)² + c² + 2(a)(–b) + 2(a)(c) + 2(–b)(c)

                                          = a² + b² + c² – 2ab + 2ac – 2bc

b. (2x – 3y + z)² = (2x)² + (–3y)² + (z)² + 2(2x)(–3y) + 2(2x)(z) + 2(–3y)(z) = 4x² + 9y² + z² – 12xy + 4xz – 6yz

Cubes of Binomials

Consider (a + b)³.

We can evaluate the product of (a + b)³ as follows:

(a + b)³ = (a + b)(a + b)(a + b) = (a + b)(a + b)² = (a + b)(a² + 2ab + b²)

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

              = a³ + 2a² b + ab² + ba² + 2ab² + b³  

               = a³ + b³ + 3a² b + 3ab² = a³ + b³ + 3ab(a + b)

From the above product, we have

a³ + b³ = (a + b)³ – 3ab(a + b)

             = (a + b) [(a + b)² – 3ab]  

             = (a + b)(a² + 2ab + b² – 3ab)

             = (a + b)(a² + b² – ab)

We thus get the formula: (a + b)³ = a³ + b³ + 3a²b + 3ab²

or (a + b)³ = a³ + b³ + 3ab(a + b)

Consider (a – b)³.

The product (a – b)³ can be calculated in the same way as above.

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

= (a – b) (a² + b² – 2ab)

= a (a² + b² – 2ab) – b(a² + b² – 2ab)

= a³ + ab² – 2a²b – ba² – b³ + 2ab²

= a³ – b³ – 3a²b + 3ab² = a³ – b³ – 3ab(a – b) (Rearranging and grouping the terms)

From the above product, we have

a³ – b³ = (a – b)³ + 3ab(a – b)

            = (a – b) [(a – b)² + 3ab]

            = (a – b) [a² + b² – 2ab + 3ab]

Thus, we get (a – b)³ = a³ – b³ – 3a²b + 3ab² or (a – b)³ = a³ – b³ – 3ab(a – b)

Example 6: Write the expansion of the following:

a. (2x + y)³                 b. (a – 3b)³

Solution: a. (2x + y)³ = (2x)³ + 3(2x)²(y) + 3(2x)(y)² + (y)³

                                     = 8x³ + 12x²y + 6xy² + y³

b. (a – 3b)³ = (a)³ – 3(a)²(3b) + 3(a)(3b)² – (3b)³

                   = a³ – 9a²b + 27ab² – 27b³