Multiplication of binomials, Mathematics

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 1180_multiplication of binomials.png and 37_multiplication of binomials2.png

That is, we have to compute

1242_multiplication of binomials3.png

We write this as

1089_multiplication of binomials4.png 

(Note: While multiplying fractions, numerators and denominators of given fractions are multiplied respectively and the product also being expressed as a fraction.)

  1. Add 3ac + 5bd - 7cd and ac - 5bd - 4cd

  3ac + 5bd - 7cd

(+)

ac - 5bd - 4cd
  4ac +  0  - 11cd
  1. 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

Posted Date: 9/13/2012 3:07:15 AM | Location : United States







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