Page 77 - Maths Skills - 8
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Algebraic Expressions and Identities                                                                    75

          3.  What must be added to –7xy + 3y  – 13x y  to get –x y  + 5y  – 3xy?
                                                    2 2
                                                                2 2
                                                                       2
                                              2
          4.  The sides of a rectangle are 2x + y unit and 3y – 2x unit. Find its perimeter.
          5.  The perimeter of a triangle is 5a  + 3a –1 and two of its sides are 2a – 5 + 7a  and 3a  + 4 – a. Find the
                                                                                                  2
                                                                                          2
                                             2
            third side of the triangle.
        MULTIPLICATION OF ALGEBRAIC EXPRESSIONS
        Rules
           (i) The product of two factors with like signs is positive and the product of two factors with unlike signs is
            negative.
            Thus, we have

             (a)  (+) × (+) = +       (b)  (–) × (–) = +      (c) (+) × (–) = –       (d) (–) × (+) = –
          (ii) If p is any variable and m, n are positive integers, then p  × p  = p m + n , i.e., p  × p  = p 3 + 4  = p .
                                                                  m
                                                                                         4
                                                                                                    7
                                                                       n
                                                                                     3
        Multiplication of Two Monomials
        Product of two monomials = (Product of their numerical coefficients) × (Product of their literal factors)
        For example, let us multiply 3x y and – 4xy .
                                      2
                                                  2
               (3x y) ×(– 4xy ) = 3 × (– 4) × (x y × xy ) = –12(x  × y ) = –12 x y
                             2
                                                    2
                                              2
                  2
                                                                              3 3
                                                              2+1
                                                                    1+2
        Multiplication of a Monomial and a Binomial
        To multiply a monomial by a binomial, we use distributive law, i.e.,
          a × (b + c) = (a × b) + (a × c), [where a, b, c are three monomials.]
        For example, let us multiply 3x y by (– 2x  + 4).
                                      2
                                                2
          3x y × (– 2x  + 4) = [3x y × (– 2x )] + [3x y × 4] = – 6x y + 12x y
                     2
                                                  2
            2
                                                              4
                                          2
                                                                      2
                                2
              Let’s Attempt
        Example 1:  Multiply each of the following.
                                                                                                3
                      (i)  3xy and 7x y           (ii)  4  x y z  and   7  xy z      (iii)  – 8a bc ,  ab c and  15  a b c
                                                                                                               2 2 3
                                                                                                    2
                                                                       3 3
                                                                                              2
                                                                                           2
                                    2 3
                                                         3 2 3
                                                      5           10                            5          16
        Solution:      (i)  3xy × 7x y  = (3 × 7) × (x × x  × y × y ) = 21 × x 1 + 2  × y 1 + 3  = 21x y
                                                      2
                                                              3
                                                                                        3 4
                                   2 3
                       (ii)  4  x y z  ×   7  xy z
                                          3 3
                              3 2 3
                           5         10
                                4  7                               28                         14
                          =            × (x  × x × y  × y  × z  × z ) =    × (x 3 + 1  × y 2 + 3  × z 3 + 3 ) =    x y z
                                                               3
                                                                                                   4 5 6
                                                  2
                                                      3
                                                           3
                                          3
                               5 10                                50                         25
                                    3
                      (iii)  –8a bc ,  ab c and   15 a b c
                                                   2 2 3
                                        2
                               2
                                  2
                                    5          16

                                   3 15
                          =                 × (a  × a × a  × b × b  × b  × c  × c × c )
                                                                2
                                                                    2
                                               2
                                                                        2
                                                                                3
                                                        2


                                8
                                   5 16
                          = −  9   × a 2 + 1 + 2  × b 1 + 2 + 2  × c 2 + 1 + 3  =  −  9  a b c
                                                                   5 5 6
                               2                                2
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