**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) = x

^{2}+ bx + ax + ab = x^{2}+ x (a + b) + ab
Thus, (x +
a) (x + b) = x

^{2}+ (a + b) x + ab
2. (x – a)
and (x + b)

(x – a) (x +
b) = x (x + b) – a (x + b) = x

^{2}+ bx – ax – ab = x^{2}+ x (b – a) – ab
Thus, (x –
a) (x + b) = x

^{2}+ (b – a) x – ab
3. (x + a)
and (x – b)

(x + a) (x –
b) = x (x – b) + a (x – b) = x

^{2}– bx + ax – ab = x^{2}– x (b – a) – ab
Thus, (x +
a) (x – b) = x

^{2}– (b – a) x – ab
4. (x – a)
and (x – b)

(x – a) (x –
b) = x (x – b) – a (x – b) = x

^{2}– bx – ax + ab = x^{2}– x (b + a) + ab
Thus, (x –
a) (x – b) = x

^{2}– (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) = x

^{2}+ (a + b) x + ab
Replacing
the values of a and b we get, (x + 4) (x + 6) = x

^{2}+ (6 + 4) x + (6 × 4)
Thus, (x +
4) (x + 6) = x

^{2}+ 10x + 24
b. (x + 3)
(x – 2)

Here, a = 3
and b = 2

We know that
(x + a) (x – b) = x

^{2}– (b – a) x – ab
Replacing
the values of a and b we get, (x + 3) (x – 2) = x

^{2}– (2 – 3) x – (3 × 2) = x^{2}+ 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) = x

^{2}+ (b – a) x – ab
Replacing
the values, we get (x – 6.5) (x + 2.5) = x

^{2}+ (2.5 – 6.5) x – (2.5 × 6.5) = x^{2}– 4x – 16.25
Thus, (x –
6.5) (x + 2.5) = x

^{2}– 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) = x

^{2}– xy + xy − y^{2}
Thus, (x +
y) (x – y) = x

^{2}– y^{2}^{}

**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) = x

^{2}– y^{2}
(x + 6) (x –
6) = x

^{2}– 6^{2}= x^{2}– 36
Thus, (x +
6) (x – 6) = x

^{2}– 36
b. (5x + 4y)
(5x – 4y) = (5x)

^{2}– (4y)^{2}
Thus, (5x +
4y) (5x – 4y) = 25x

^{2}– 16y^{2}^{}

**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) = x

^{2}+ xy + xy + y^{2}= x^{2}+ 2xy + y^{2}
Thus, (x + y)

^{2}= x^{2}+ 2xy + y^{2}
2.
Now,
consider (x – y)

^{2}= (x – y) (x – y).
(x – y) (x – y) = x (x – y) – y (x – y) = x

^{2}– xy – xy + y^{2}= x^{2}– 2xy + y^{2}
Thus, (x – y) (x – y) = x

^{2}– 2xy + y^{2}^{}

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

a. (x + 5)

^{2}b. (2x – 3y)^{2}^{}

**Solution:**a. (x + 5)

^{2}= x

^{2}+ 2 × x × 5 + 5

^{2}= x

^{2}+ 10x + 25

b. (2x – 3y)

^{2 }= (2x)^{2}– 2 × 2x × 3y + (3y)^{2}= 4x^{2}– 12xy + 9y^{2}**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)
= a

^{2}+ 2ak + k^{2}
= a

^{2}+ 2a(b + c) + (b + c)^{2}(Substituting the value of k) = a

^{2}+ 2ab + 2ac + b^{2}+ c^{2}+ 2bc = a^{2}+ b^{2}+ c^{2}+ 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}+ c

^{2}+ 2(a)(–b) + 2(a)(c) + 2(–b)(c)

= a

^{2}+ b^{2}+ c^{2}– 2ab + 2ac – 2bc
b. (2x – 3y
+ z)

^{2}**= (2x)**^{2}+ (–3y)^{2}+ (z)^{2}+ 2(2x)(–3y) + 2(2x)(z) + 2(–3y)(z) = 4x^{2}+ 9y^{2}+ z^{2}– 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)(a^{2}+ 2ab + b^{2})
= a(a

^{2}+ 2ab + b^{2}) + b(a^{2}+ 2ab + b^{2})
= a

^{3}+ 2a^{2}b + ab^{2}+ ba^{2}+ 2ab^{2}+ b^{3}
= a

^{3}+ b^{3}+ 3a^{2}b + 3ab^{2}= a^{3}+ b^{3}+ 3ab(a + b)
From the
above product, we have

a

^{3}+ b^{3}= (a + b)^{3}– 3ab(a + b)
= (a + b) [(a + b)

^{2}– 3ab]
= (a + b)(a

^{2 }+ 2ab + b^{2}– 3ab)
= (a + b)(a

^{2 }+ b^{2}– ab)
We thus get
the formula: (a + b)

^{3}= a^{3 }+ b^{3}+ 3a^{2}b + 3ab^{2}
or (a + b)

^{3}= a^{3}+ b^{3 }+ 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) (a

^{2}+ b^{2}– 2ab)
= a (a

^{2}+ b^{2}– 2ab) – b(a^{2}+ b^{2}– 2ab)
= a

^{3}+ ab^{2}– 2a^{2}b – ba^{2}– b^{3}+ 2ab^{2}
= a

^{3}– b^{3}– 3a^{2}b + 3ab^{2}= a^{3}– b^{3}– 3ab(a – b) (Rearranging and grouping the terms)
From the
above product, we have

a

^{3}– b^{3}= (a – b)^{3}+ 3ab(a – b)
= (a – b) [(a – b)

^{2}+ 3ab]
= (a – b) [a

^{2}+ b^{2}– 2ab + 3ab]
Thus, we get
(a – b)

^{3}= a^{3}– b^{3}– 3a^{2}b + 3ab^{2}or (a – b)^{3}= a^{3}– b^{3}– 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}

= 8x

^{3}+ 12x^{2}y + 6xy^{2}+ y^{3}
b. (a – 3b)

^{3}= (a)^{3}– 3(a)^{2}(3b) + 3(a)(3b)^{2}– (3b)^{3}
= a

^{3}– 9a^{2}b + 27ab^{2}– 27b^{3}
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