To understand the multiplication of binomials, we should know what is meant by Distributive Law of Multiplication. Suppose that we are to multiply (a + b) and m. We treat (a + b) as a compound expression and m as a simple expression. Therefore, (a + b)m by definition will be:
= m + m + m + m + ....... taken a + b times
= (m + m + m + .... taken a times) + (m + m + m + ..... taken b times)
= am + bm
Similarly (a  b)m = am  bm and (a  b + c)m = am  bm + cm. This is referred to as Distributive Law of Multiplication and it says that the product of a compound expression by a simple expression is the algebraic sum of the partial products of each term of the compound expression by that simple expression.
In the above, if we write (c + d) in place of m we will have
(a + b)(c + d) = a(c + d) + b(c + d)
= ac + ad + bc + bd
1. Multiply (3a + d) and (b + c).
We employ (a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd. Therefore, (3a + d)(b + c) = 3a(b + c) + d(b + c) = 3ab + 3ac + bd + cd. (This procedure can be extended to trinomials and polynomials also.)
2. Multiply 2a + 5c and 3d + 2b.
One way of doing this is to employ (a + b)(c + d) = ac + ad + bc + bd
That is,
(2a + 5c)(3d + 2b) = 2a(3d + 2b) + 5c(3d + 2b)
= 6ad + 4ab + 15cd + 10bc
In the second method, we position the binomials as we did in addition or subtraction and do the multiplication operation. That is,
2a + 5c

(x)

3d + 2b



6ad + 15cd



+ 4ab + 10bc


6ad + 15cd + 4ab + 10bc

This product is the same as one obtained earlier.
Multiply 

and 

That is, we have to compute
We write this as
(Note: While multiplying fractions, numerators and denominators of given fractions are multiplied respectively and the product also being expressed as a fraction.)

Add 3ac + 5bd  7cd and ac  5bd  4cd

3ac + 5bd  7cd 
(+)

ac  5bd  4cd 

4ac + 0  11cd 

Multiply 3a + 5b  7d and c  4e  5
That is, we require (3a + 5b  7d) x (c  4e  5)
= 3a (c  4e  5) + 5b (c  4e  5)  7d(c  4e  5)
= 3ac  12ae  15a + 5bc  20be  25b  7cd + 28de + 35d