While doing the product of algebraic expressions, we should follow the steps given below.

**Multiplication
of Algebraic Expressions**

While doing
the product of algebraic expressions, we should follow the steps given below.

**Step 1: **Multiply the signs of the
terms. That is, the product of two** **like signs** **are** **positive and the product of two unlike signs are negative.

**Step 2: **Multiply the corresponding
co-efficients of the terms.

**Step 3: **Multiply the variable factors
by using laws of exponents.

If ‘*x*’ is a variable and m, n are positive integers
then,

*x ^{m} *×

For
example, *x*^{3} ×
*x*^{4} =
*x*^{3 }^{+}^{4} = *x*^{7}

**Think**

Every algebraic expression is a polynomial. Is this statement true?
Why?

**Solution:**

No, This statement is not true. Because Polynomials contain only
whole numbers as the powers of their variables. But an algebraic expression may
contains fractions and negative powers on their variables.

Eg. 2*y*^{2} + 5*y*^{−1} − 3 is a an
algebraic expression. But not a polynomial.

A polynomial is a special kind of algebraic expression. The difference
between an algebraic expression and a polynomial is,

**Algebraic Expression**

May contains whole numbers, fractions, negative numbers as the power
of their variables.

Example: 4*x*^{3/2}
− 3*x* + 9

2 *y*^{2} + 5/*y* − 3 , 3*x*^{2} − 4*x* + 1

**Polynomial**

contains only whole numbers as the power of their variables.

**Example: **4*x*** **^{2}** **−** **3*x*** **+** **9

2 *y* ^{6} + 5*y*^{3}
− 3

**Note**

Product of two terms is represented by the symbols ( ), dot (.) or
× .

For example,

multiplying 4*x*^{2}
and *xy* can be written in any one of the
following ways.

__1. Multiplication
of two or more monomials__

Consider
that, Geetha buys 3 pens each @ ₹5, how much she has to pay to the shopkeeper?

Geetha has
to pay to the shopkeeper = 3 × ₹5

= ₹15

If there
are ‘*x*’ pens and the cost of each pen
is ₹
‘*y*’, then the cost of (3*x*^{2} ) pens bought by Geetha @ ₹
5*y*

@ - at the rate of

= (3*x* ^{2} ) × 5*y*

= (3 × 5)(*x* ^{2} × *y*)

= ₹
15*x*^{2}*y*

**Example 3.1**

If the length
and breadth of a rectangular painting are 4*xy*^{3}
and 3*x* ^{2} *y* . Find its area.

*Solution:*

Area of the
rectangular painting, *A* =
(*l* × *b*)
*sq.units*

= (4*xy*^{3} ) × (3*x* ^{2} *y*)

= (4 × 3)(*x* × *x*^{2} )( *y* ^{3} × *y*)

*A *=* *12*x *^{3}* y*^{4}* sq.units *

**Example 3.2**

Find the
product of 2*x* ^{2} *y*^{2} , 3 *y* ^{2}*z* and – *z* ^{2} *x*^{3}^{}

*Solution:*

We have,
(2*x* ^{2} *y*^{2} ) × (3 *y* ^{2} *z*) ×
(−*z* ^{2} *x*^{3}
)

= ( +) × ( +) × ( −)(2 × 3 ×1)(*x* ^{2} × *x*^{3}
)( *y* ^{2} ×
*y*^{2} )(*z* × *z*^{2} )

= −6*x* ^{5} *y*^{4}*z*^{3}

**Try these **

Find the product of

(i) 3*ab*^{2} , −2*a*^{2}*b*^{3}

(ii) 4*xy*, 5 *y*^{2} *x*,(−*x*^{2} )

(iii) 2*m*, −5*n*, −3*p*

**Solution:**

**(i) (3 ab^{2})
× (−2a^{2}b^{3}) **= (+) × (−) × (3 × 2) × (

**(ii) (4 xy) ×
(5y^{2}x) × (−x^{2})** = (+) × (+) × (−) × (4 × 5 × 1) × (

= −20 *x*^{4}*y*^{3}

**(iii) (2 m) ×
(−5n) × (−3p)** = (+) × (−) × (−) × (2 × 5 × 3)
×

= + 30 *mnp* = 30 *mnp*

__2. Multiplication
of a polynomial by a monomial__

If there
are ‘*a*’ shops and each shop has ‘*x*’ apples in 8 baskets and ‘*y*’ oranges in 3 baskets and ‘*z*’ bananas in 5 baskets, then the total number
of apples, oranges and bananas are

*= a *×* *(8*x *+* *3*y *+* *5*z*)

*= a*(8*x*)* *+* a*(3*y*)* *+* a*(5*z*)

(using distributive
law).

= 8*ax*
+
3*ay* + 5*az*

monomial × monomial = monomial

binomial × monomial = binomial

binomial × binomial = binomial/polynomial

polynomial × monomial = polynomial

**Note**

**Distributive law **

If *a* is a constant, *x* and *y* are variables then

*a*(*x* + *y*) = *ax* + *ay*

For example, 5(*x* + *y*) = 5*x* + 5 *y*

**Think**

Why 3+(4*x*–7*y*) ≠ 12*x*−21*y* ?

**Solution:**

Addition and multiplication are different 3 + (4*x* − 7*y*)
= 3 + 4*x* − 7*y*

We can add only like terms.

**Example 3.3**

Multiply
3*x* ^{2} *y* and (2*x* ^{3} *y*^{3} −
5*x* ^{2} *y* + 9*xy*)

*Solution:*

Now, (3*x* ^{2} *y*) × (2*x* ^{3} *y*^{3}
−
5*x* ^{2} *y* + 9*xy*)

= 3*x* ^{2} *y*(2*x* ^{3} *y*^{3} ) −
3*x* ^{2} *y* (5*x* ^{2} *y* ) + 3*x* ^{2} *y*(9*xy*)

multiplying
each term of the polynomial by the monomial

= (3 ×
2)(*x* ^{2} ×
*x*^{3} )( *y* × *y*^{3} ) −
(3 ×
5)(*x* ^{2} ×
*x*^{2} )( *y* × *y*) +
(3 ×
9)(*x* ^{2} ×
*x*)( *y* × *y*)

= 6*x*^{5}*y*^{4} − 15*x*^{4} *y*^{2}
+
27*x* ^{3} *y*^{2}

**Example 3.4**

Ram deposited
‘*x*’ number of ₹2000
notes, ‘*y*’ number of ₹500
notes, ‘*z*’ number of ₹100
notes in a bank and Velan deposited ‘3*xy*’
times of amount of what Ram had deposited. How much amount did Velan deposit in
the bank?

*Solution:*

Amount deposited
by Ram

= ( *x* ×₹ 2000 + *y* ×₹ 500 + *z* ×₹100)

= ₹( 2000*x* + 500 *y* +100*z* )

Amount deposited
by Velan =
3*xy* times ×
Amount deposited by Ram

= ₹ 3*xy* × (2000*x* + 500 *y* +
100*z*)

= (3 × 2000)(*x* × *x* ×
*y* ) + (3 × 500)(*x* × *y* ×
*y* ) + (3 ×100)(*x* × *y* ×
*z*)

= ₹ (6000*x* ^{2} *y* + 1500*xy* ^{2} + 300*xyz*)

**Try these **

Multiply

(i) (5 *x*^{2} +
7*x *– 3) by –4*x*^{2}

(ii) (10*x *– 7*y *+ 5*z*)
by 6*xy*z

(ii) (*ab*+3*bc* –5*ca*)
by 3*a*^{2}*bc*

(iv) (4*m*^{2} –
3*m* + 7) by –5*m*^{3}

**Solution:**

**(i)**** (5 x^{2} + 7x − 3) by – 4x^{2}**

(5*x*^{2} + 7*x* − 3) × (– 4*x*^{2})
= 5*x*^{2} (− 4*x*^{2}) + 7*x* (− 4*x*^{2})
− 3(− 4*x*^{2})

= − 20 *x*^{4} − 28*x*^{3} + 12*x*^{2}

**(ii)**** (10 x −7y + 5z) by 6xyz**

(10*x* −7*y* + 5*z*) × 6*xyz* = 6*xyz*
(10*x* − 7*y* + 5*z*) [∵ Multiplication is commutative]

= 6*xyz*(10*x*) + 6*xyz*(−7*y*) + 6*xyz*(5*z*)

= (6 × 10) (*x* × *x* × *y* × *z*) + (6 × −7)
+ (*x* × *y* × *y* × *z* ) + (6 × 5)(*x* × *y* × *z*
× *z*)

= 60*x*^{2}*yz* + (−42*xy*^{2}*z*)
+ 30*xyz*^{2}

= 60*x*^{2}*yz* − 42*xy*^{2}*z*
+ 30*xyz*^{2}

**(iii) ( ab +
3bc − 5ca) by 3a^{2} bc**

(*ab* + 3*bc* −
5*ca*) × (3*a*^{2}*bc*) = *ab*(3*a*^{2}*bc*)
+ 3*bc* (3*a*^{2}*bc*) − 5*ca* (3*a*^{2}*bc*)

= 3*a*^{2}*b*^{2}*c* + 9*a*^{2}*b*^{2}*c*^{2}
− 15*a*^{3}*bc*^{2}

**(iv) (4 m^{2}
− 3m + 7) by − 5m^{3} **

(4*m*^{2} − 3*m*
+ 7) × (− 5*m*^{3}) = 4*m*^{2} (− 5*m*^{3})
− (3*m*) (− 5*m*^{3}) + 7(− 5*m*^{3})

= − 20*m*^{5} + 15*m*^{4} − 35*m*^{3}

__3. Multiplication
of two binomials__

Consider
that a rectangular flower bed whose length is decreased by 5 *units* from the original length and whose
breadth is increased by 3 *units* to the
original breadth. What is the area of the rectangular flower bed?

Area of the
rectangle = *l* ×
*b*

Here, area
of the rectangular flower bed *A* =
(*l* − 5) × (*b* + 3) *sq.units*

How do we
multiply this?

Now, let us learn how to multiply two binomials

If (*x* + *y*)
and ( *p* +
*q*) are two binomials, we can find their
product as given below,

So, the above
area of the rectangle = (*l* − 5) × (*b* + 3)

(By horizontal
distributive approach) = *l*
(*b* + 3) − 5(*b* + 3)

*A *=* *(*lb *+* *3*l
*−* *5*b *−15)* sq.u*

Let us consider
one more example. Consider the given figure , In the square *OABC,*

*OA*= 4* units *;* OC *=4* units*

The area
of the square *OABC* = 4 × 4

*A *= 16* sq.units*

If the sides
of the square are increased by ‘*x’ units *and
‘*y*’ *units* respectively,then we get the rectangle *ODEF* whose sides are *OD*=(4+*x*) *units*
and *OF*=(4+*y*) *units*.

Now, the area of the rectangle ODEF = (4+x) (4+y) (by FOIL approach)

A = 16+ 4y
+ 4*x *+* x*y *sq. units.*

(all are unlike terms andso, we can’t add)

**Example 3.5**

Multiply
(2*x* + 5 *y*) and (3*x* −
4 *y*)

*Solution:*

By horizontal distributive approach,

(2*x *+ 5y) (3*x *− 4* y*) = 2* x *(*x
*− 4* y*) + 5* y *(* x *− 4* y*)

= 6*x*^{2} − 8*xy* + 15*xy* − 20 *y*^{2}

=
6*x*^{2} +
7*xy* − 20 *y*^{2} (simplify the like terms)

**Try these **

**Multiply**

(i) (a − 5) and (a + 4)

(ii) (*a + b*) and (a − b)

(iii) (*m*^{4} +
*n*^{4} ) and (*m* − *n*)

(iv) (2*x *+ 3)(*x *+ 4)

(v) (3*x *+ 7)(* x *−5)

(vi) (*x *− 2)(6*x *− 3)

**Solution:**

**(i) ( a − 5) and (a + 4)**

(*a − *5) (*a*
+ 4) = *a*(*a* + 4) − 5(*a* + 4)

= (*a* × *a*) + (*a* × 4) + (−5 × *a*) + (−5
× 4)

= *a*^{2} + 4*a* − 5*a* − 20 = *a*^{2}
− *a* − 20

**(ii) ( a + b) and (a − b)**

(*a + b*) (*a − b*)
= *a*(*a − b*) + *b*(*a − b*)

= (*a* × *a*) + (*a* × −*b*) + (*b* × *a*
) + *b*(− *b*)

= *a*^{2} − *ab* + *ab* − *b*^{2}
= *a*^{2} − *b*^{2}

**(iii) ( m^{4 }+ n^{4}) and (m −
n)**

(*m*^{4} + *n*^{4}) (*m − n*) = *m*^{4}
(*m − n*) + *n*^{4} (*m − n*)

= (*m*^{4} × *m*) + (*m*^{4} × (−
*n*)) + (*n*^{4} × *m*) + (*n*^{4} × (−*n*))

= *m*^{5} − *m*^{4}* n* + *mn*^{4}
– *n*^{5}

**(iv) (2 x + 3) (x + 4)**

(2*x* + 3) (*x*
+ 4) = 2*x*(*x* + 4) + 3(*x*
+ 4)

= (2*x*^{2} × *x*) + (2*x* × 4) + (3 × *x*)
+ (3 × 4)

= 2*x*^{2} + 8*x* + 3*x* + 12 = 2*x*^{2}
+ 11*x* + 12

**(v) (3 x + 7) (x − 5)**

(3*x* + 7) (*x*
− 5) = *x*(3*x* + 7) − 5(3*x*
+ 7)

= (*x* × 3*x*) + (*x* × 7) + (−5 × 3*x*) + (−5
× 7)

= 3*x*^{2} + 7*x* − 15*x* − 35

= 3*x*^{2} − 8*x* – 35

**(vi) ( x − 2) (6x − 3)**

(*x* − 2) (6*x*
− 3) = *x*(6*x* − 3) − 2(6*x*
− 3)

= (*x* × 6*x*) + (*x* × (−3) − (2 × 6*x*) − (2
× 3)

= 6*x*^{2} − 3*x* − 12*x* + 6

= 6*x*^{2} − 15*x* + 6

Think

(i) In 3*x*^{2} (*x*^{4}–7*x*^{3}+2), what is the highest power
in the expression?

**Solution:**

3*x*^{2} (*x*^{4} − 7*x*^{3}
+ 2) = (3*x*^{2}) (*x*^{4})
+ 3*x*^{2 }(− 7*x*^{3}) + (3*x*^{2})2

= 3*x*^{2 } −
21*x*^{5} + 6*x*^{2}

Highest power is 6 in *x*^{6}.

(ii) Is –5*y*^{2}
+2*y*–6 = –(5*y*^{2} +2*y*–6)? If not,
correct the mistake.

**Solution:**

No, – 5*y*^{2} + 2*y* – 6 = − (5*y*^{2}
− 2*y* + 6 )

Tags : Algebra | Chapter 3 | 8th Maths , 8th Maths : Chapter 3 : Algebra

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