Special Products and Expansions

# 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