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9. Algebric Expressions and Identities
1 What are Expressions?
In earlier classes, we have already become familiar with what algebraic expressions (or simply expressions) are. Examples of expressions are:
x + 3, 2y – 5, 3x^{2}, 4xy + 7 etc.
You can form many more expressions. As you know expressions are formed from variables and constants. The expression 2y – 5 is formed from the variable y and constants 2 and 5. The expression 4xy + 7 is formed from variables x and y and constants 4 and 7.
We know that, the value of y in the expression, 2y – 5, may be anything. It can be 2, 5, –3, 0,5/2 ,7/3 etc.; actually countless different values.
The value of an expression changes with the value chosen for the variables it contains. Thus as y takes on different values, the value of 2y – 5 goes on changing. When y = 2, 2y – 5 = 2(2) – 5 = –1; when y = 0, 2y – 5 = 2 × 0 –5 = –5, etc. Find the value of the expression 2y – 5 for the other given values of y.
Number line and an expression:Consider the expression x + 5. Let us say the variable x has a position X on the number line;
X may be anywhere on the number line, but it is definite that the value of x + 5 is given by a point P, 5 units to the right of X. Similarly, the value of x – 4 will be 4 units to the left of X and so on. What about the position of 4x and 4x + 5?
The position of 4x will be point C; the distance of C from the origin will be four times the distance of X from the origin. The position D of 4x + 5 will be 5 units to the right of C.
2 Terms, Factors and Coefficients
Take the expression 4x + 5. This expression is made up of two terms, 4x and 5. Terms are added to form expressions. Terms themselves can be formed as the product of factors. The term 4x is the product of its factors 4 and x. The term 5 is made up of just one factor, i.e., 5.
The expression 7xy – 5x has two terms 7xy and –5x. The term 7xy is a product of factors 7, x and y. The numerical factor of a term is called its coefficient. The coefficient in the term 7xy is 7 and the coefficient in the term –5x is –5.
3 Monomials, Binomials and Polynomials
Expression that contains only one term is called a monomial. Expression that contains two terms is called a binomial. An expression containing three terms is a trinomial and so on. In general, an expression containing, one or more terms with nonzero coefficient (with variables having non negative exponents) is called a polynomial. A polynomial may contain any number of terms, one or more than one.
Examples of monomials: 4x^{2}, 3xy, –7z, 5xy^{2}, 10y, –9, 82mnp, etc.
Examples of binomials: a + b, 4l + 5m, a + 4, 5 –3xy, z^{2} – 4y^{2}, etc.
Examples of trinomials: a + b + c, 2x + 3y – 5, x^{2}y – xy^{2} + y^{2}, etc.
Examples of polynomials: a + b + c + d, 3xy, 7xyz – 10, 2x + 3y + 7z, etc.
4 Like and Unlike Terms
Look at the following expressions:
7x, 14x, –13x, 5x^{2}, 7y, 7xy, –9y^{2}, –9x^{2}, –5yx
Like terms from these are:
(i) 7x, 14x, –13x are like terms.
(ii) 5x2 and –9x2 are like terms.
(iii) 7xy and –5yx are like terms.
Why are 7x and 7y not like?
Why are 7x and 7xy not like?
Why are 7x and 5x2 not like?
5 Addition and Substraction of Algebraic Expressions
In the earlier classes, we have also learnt how to add and subtract algebraic expressions.
For example, to add 7x^{2} – 4x + 5 and 9x – 10, we do
7x^{2} – 4x + 5
+ 9x – 10
7x^{2} + 5x – 5
Observe how we do the addition. We write each expression to be added in a separate row. While doing so we write like terms one below the other, and add them, as shown.
Thus 5 + (–10) = 5 –10 = –5. Similarly, – 4x + 9x = (– 4 + 9)x = 5x. Let us take some more examples.
Add: 7xy + 5yz – 3zx, 4yz + 9zx – 4y , –3xz + 5x – 2xy.
Writing the three expressions in separate rows, with like terms one below the other, we have
7xy + 5yz –3zx
+ 4yz + 9zx – 4y
+ –2xy– 3zx + 5x (Note xz is same as zx)
5xy + 9yz + 3zx + 5x – 4y
Thus, the sum of the expressions is 5xy + 9yz + 3zx + 5x – 4y. Note how the terms, – 4y in the second expression and 5x in the third expression, are carried over as they are, since they have no like terms in the other expressions.
Subtract 5x^{2} – 4y^{2} + 6y – 3 from 7x^{2} – 4xy + 8y^{2} + 5x – 3y.
7x^{2} – 4xy + 8y^{2} + 5x – 3y
5x^{2} – 4y^{2}+ 6y – 3
(–) (+) (–) (+)
+ –2xy– 3zx + 5x (Note xz is same as zx)
2x^{2} – 4xy +12y^{2} + 5x – 9y + 3
6 Multiplication of Algebraic Expressions: Introduction
 Look at the following patterns of dots.
 Can you now think of similar other situations in which
two algebraic expressions have to be multiplied?
Ameena gets up. She says, “We can think of area of a rectangle.” The area of a rectangle is l × b, where l is the length, and b is breadth. If the length of the rectangle is increased by 5 units, i.e., (l + 5) and breadth is decreased by 3 units , i.e., (b – 3) units, the area of the new rectangle will be (l + 5) × (b – 3).
 Can you think about volume? (The volume of a rectangular box is given by the product of its length, breadth and height).
 Sarita points out that when we buy things, we have to carry out multiplication. For example, if price of bananas per dozen = Rs p and for the school picnic bananas needed = z dozens, then we have to pay = Rs p × z
Pattern of dots  Total number of dots 
4 × 9  
5 × 7  
m × n  
(m + 2) × (n + 3) 
expressions have to be multiplied?
Suppose, the price per dozen was less by Rs 2 and the bananas needed were less by 4 dozens.
Then, price of bananas per dozen = Rs (p – 2) and bananas needed = (z – 4) dozens, Therefore, we would have to pay = Rs (p – 2) × (z – 4)
In all the above examples, we had to carry out multiplication of two or more quantities. If the quantities are given by algebraic expressions, we need to find their product. This means that we should know how to obtain this product. Let us do this systematically. To begin with we shall look at the multiplication of two monomials.
7 Multiplying a Monomial by a Monomial
7.1 Multiplying two monomials
We begin with 4 × x = x + x + x + x = 4x as seen earlier.
Similarly, 4 × (3x) = 3x + 3x + 3x + 3x = 12x
Now, observe the following products.
 x × 3y = x × 3 × y = 3 × x × y = 3xy
 5x × 3y = 5 × x × 3 × y = 5 × 3 × x × y = 15xy
 5x × (–3y) = 5 × x × (–3) × y = 5 × (–3) × x × y = –15xy
 5x × 4x^{2} = (5 × 4) × (x × x^{2})= 20 × x^{3} = 20x^{3}
 5x × (– 4xyz) = (5 × – 4) × (x × xyz)= –20 × (x × x × yz) = –20x^{2}yz
Notice that all the three products of monomials, 3xy, 15xy, –15xy, are also monomials.
Some more useful examples follow.
Observe how we collect the powers of different variables in the algebraic parts of the two monomials. While doing so, we use the rules of exponents and powers.
7.2 Multiplying three or more monomials
Observe the following examples.
 2x × 5y × 7z = (2x × 5y) × 7z = 10xy × 7z = 70xyz
 4xy × 5x2y2 × 6x3y3 = (4xy × 5x^{2}y^{2}) × 6x^{3}y^{3} = 20x^{3}y^{3} × 6x^{3}y^{3}= 120x^{3}y^{3} × x^{3}y^{3} = 120 (x^{3} × x^{3}) × (y^{3} × y^{3}) = 120x^{6} × y^{6} = 120x^{6}y^{6}
It is clear that we first multiply the first two monomials and then multiply the resulting monomial by the third monomial. This method can be extended to the product of any number of monomials.
Complete the table for area of a rectangle with given length and breadth.
length  breath  area 
3x  5y  3x × 5y = 15xy 
9y  4y^{2}  .............. 
4ab  5bc  .............. 
2l^{2}m  3lm^{2}  .............. 
Find the volume of each rectangular box with given length, breadth and height.
length  breath  area  
(i)  2ax  3by  5cz 
(ii)  m^{2}n  n^{2}p  p^{2}m 
(iii)  2q  4q^{2}  8q^{3} 
Volume = length × breadth × height
Hence, for
 volume = (2ax) × (3by) × (5cz)
= 2 × 3 × 5 × (ax) × (by) × (cz)
= 30abcxyz  volume = m^{2}n × n^{2}p × p^{2}m
= (m^{2} × m) × (n × n^{2}) × (p × p^{2})
= m^{3}n^{3}p^{3}  volume = 2q × 4q^{2} × 8q^{3}
= 2 × 4 × 8 × q × q^{2} × q^{3}
= 64q^{6}
8 Multiplying a Monomial by a Polynomial
8.1 Multiplying a monomial by a binomial
Let us multiply the monomial 3x by the binomial 5y + 2, i.e., find 3x × (5y + 2) = ? Recall that 3x and (5y + 2) represent numbers. Therefore, using the distributive law, 3x × (5y + 2) = (3x × 5y) + (3x × 2) = 15xy + 6x
8.2 Multiplying a monomial by a trinomial
Consider 3p × (4p2 + 5p + 7). As in the earlier case, we use distributive law;
3p × (4p2 + 5p + 7) = (3p × 4p2) + (3p × 5p) + (3p × 7)
= 12p3 + 15p2 + 21p
Multiply each term of the trinomial by the monomial and add products.
Simplify the expressions and evaluate them as directed:
 x (x – 3) + 2 for x = 1,
 3y (2y – 7) – 3 (y – 4) – 63 for y = –2
 x (x – 3) + 2 = x^{2} – 3x + 2
For x = 1, x2 – 3x + 2 = (1)2 – 3 (1) + 2
= 1 – 3 + 2 = 3 – 3 = 0  3y (2y – 7) – 3 (y – 4) – 63 = 6y2 – 21y – 3y + 12 – 63
= 6y2 – 24y – 51
For y = –2, 6y2 – 24y – 51 = 6 (–2)2 – 24(–2) – 51
= 6 × 4 + 24 × 2 – 51
= 24 + 48 – 51 = 72 – 51 = 21
Add
 5m (3 – m) and 6m^{2} – 13m
 4y (3y^{2} + 5y – 7) and 2 (y^{3} – 4y^{2} + 5)
 First expression = 5m (3  m) = (5m × 3)  (5m × m) = 15m  5m^{2}
Now adding the second expression to it,15m  5m^{2} + 6m^{2}  13m = m^{2} + 2m  The first expression = 4y (3y2 + 5y  7) = (4y × 3y2) + (4y × 5y) + (4y × (7))
= 12y^{3} + 20y228y
The second expression = 2 (y^{3}  4y^{2} + 5) = 2y^{3} + 2 × (4y^{2}) + 2 × 5 = 2y38y2 + 10
Adding the two expressions,
12y^{3}+20y^{2}28y
+2y^{3}8y^{2}+10
14y^{3}+ 12y^{2}28y+ 10
Subtract 3pq (p – q) from 2pq (p + q).
We have 3pq (p – q) = 3p^{2}q – 3pq^{2} and 2pq (p + q) = 2p^{2}q + 2pq^{2}
Subtracting, 2p^{2}q + 2pq^{2}
3p^{2}q – 3pq^{2}
 +
 p^{2}q + 5pq^{2}
9 Multiplying a Polynomial by a Polynomial
9.1 Multiplying a binomial by a binomial
Let us multiply one binomial (2a + 3b) by another binomial, say (3a + 4b). We do this stepbystep, as we did in earlier cases, following the distributive law of multiplication,
3a + 4b) × (2a + 3b) = 3a × (2a + 3b) + 4b × (2a + 3b)
= (3a × 2a) + (3a × 3b) + (4b × 2a) + (4b × 3b)
= 6a2 + 9ab + 8ba + 12b2
= 6a2 + 17ab + 12b2
Observe, every term in one binomial multiplies every term in the other binomial.
When we carry out term by term multiplication, we expect 2 × 2 = 4 terms to be present. But two of these are like terms, which are combined, and hence we get 3 terms.In multiplication of polynomials with polynomials, we should always look for like terms, if any, and combine them.
Multiply
 (x – 4) and (2x + 3)
 (x – y) and (3x + 5y)
 (x – 4) × (2x + 3) = x × (2x + 3) – 4 × (2x + 3)
= (x × 2x) + (x × 3) – (4 × 2x) – (4 × 3) = 2x^{2} + 3x – 8x – 12
= 2x^{2} – 5x – 12 (Adding like terms)  (x – y) × (3x + 5y) = x × (3x + 5y) – y × (3x + 5y)
= (x × 3x) + (x × 5y) – (y × 3x) – ( y × 5y)
= 3x^{2} + 5xy – 3yx – 5y^{2} = 3x2 + 2xy – 5y^{2} (Adding like terms)
Multiply
 (a + 7) and (b – 5)
 (a2 + 2b2) and (5a – 3b)
 (a + 7) × (b – 5) = a × (b – 5) + 7 × (b – 5)
= ab – 5a + 7b – 35  (a^{2} + 2b^{2}) × (5a – 3b) = a^{2} (5a – 3b) + 2b^{2} × (5a – 3b)
= 5a^{3} – 3a^{2}b + 10ab^{2}  6b^{3}
Note that there are no like terms involved in this multiplication.
9.2 Multiplying a binomial by a trinomial
n this multiplication, we shall have to multiply each of the three terms in the trinomial by each of the two terms in the binomial. We shall get in all 3 × 2 = 6 terms, which may reduce to 5 or less, if the term by term multiplication results in like terms. Consider
(a +7) × (a^{2} +3a+5) = a × (a^{2} + 3a + 5) + 7 × (a^{2} + 3a + 5)
binomial trinomial [using the distributive law]
= a3 + 3a2 + 5a + 7a2 + 21a + 35
= a3 + (3a2 + 7a2) + (5a + 21a) + 35
= a3 + 10a2 + 26a + 35 (Why are there only 4 terms in the final result?)
Simplify (a + b) (2a – 3b + c) – (2a – 3b) c.
We have
(a + b) (2a – 3b + c) = a (2a – 3b + c) + b (2a – 3b + c)
= 2a2 – 3ab + ac + 2ab – 3b2 + bc
= 2a2 – ab – 3b2 + bc + ac (Note, –3ab and 2ab are like terms)
and (2a – 3b) c = 2ac – 3bc
Therefore,
(a + b) (2a – 3b + c) – (2a – 3b) c = 2a2 – ab – 3b2 + bc + ac – (2ac – 3bc)
= 2a2 – ab – 3b2 + bc + ac – 2ac + 3bc
= 2a2 – ab – 3b2 + (bc + 3bc) + (ac – 2ac)
= 2a2 – 3b2 – ab + 4bc – ac
10 What is an Identity?
Consider the equality (a + 1) (a +2) = a^{2} + 3a + 2 We shall evaluate both sides of this equality for some value of a, say a = 10. For a = 10, LHS = (a + 1) (a + 2) = (10 + 1) (10 + 2) = 11 × 12 = 132 RHS = a^{2} + 3a + 2 = 10^{2} + 3 × 10 + 2 = 100 + 30 + 2 = 132
Thus, the values of the two sides of the equality are equal for a = 10.
Let us now take a = –5
LHS = (a + 1) (a + 2) = (–5 + 1) (–5 + 2) = (– 4) × (–3) = 12
RHS = a2 + 3a + 2 = (–5)2 + 3 (–5) + 2
= 25 – 15 + 2 = 10 + 2 = 12
Thus, for a = –5, also LHS = RHS.
We shall find that for any value of a, LHS = RHS Such an equality, true for every value of the variable in it, is called an identity. Thus,
(a + 1) (a + 2) = a2 + 3a + 2 is an identity. An equation is true for only certain values of the variable in it. It is not true for all values of the variable. For example, consider the equation
a^{2} + 3a + 2 = 132
It is true for a = 10, as seen above, but it is not true for a = –5 or for a = 0 etc. Try it: Show that a^{2} + 3a + 2 = 132 is not true for a = –5 and for a = 0.
11 Standard Identities
We shall now study three identities which are very useful in our work. These identities are obtained by multiplying a binomial by another binomial.
Let us first consider the product (a + b) (a + b) or (a + b)^{2}.
(a + b)^{2} = (a + b) (a + b)
= a(a + b) + b (a + b)
= a^{2} + ab + ba + b^{2}
= a^{2} + 2ab + b^{2} (since ab = ba)
Thus (a + b)^{2} = a^{2} + 2ab + b^{2} (I)
Clearly, this is an identity, since the expression on the RHS is obtained from the LHS by actual multiplication. One may verify that for any value of a and any value of b, the values of the two sides are equal.

Next we consider (a – b)^{2} = (a – b) (a – b) = a (a – b) – b (a – b)
We have = a^{2} – ab – ba + b^{2} = a^{2} – 2ab + b^{2}
Or a^{2} – 2ab + b^{2} (II)

Finally, consider (a + b) (a – b). We have (a + b) (a – b) = a (a – b) + b (a – b)
= a2 – ab + ba – b2 = a2 – b2(since ab = ba)
Or (a + b) (a – b) = a^{2}b^{2} (III)
The identities (I), (II) and (III) are known as standard identities.
 We shall now work out one more useful identity.
(x + a) (x + b) = x (x + b) + a (x + b)
= x2 + bx + ax + ab
Or (x + a) (x + b) = x^{2} + (a + b) x + ab (IV)
We can see that Identity (IV) is the general form of the other three identities also.
12 Applying Identities
We shall now see how, for many problems on multiplication of binomial expressions and also of numbers, use of the identities gives a simple alternative method of solving them.
Using the Identity (I), find
 (2x + 3y)^{2}
 1032

(2x + 3y)^{2} = (2x)^{2} + 2(2x) (3y) + (3y)^{2} [Using the Identity (I)]
= 4x^{2} + 12xy + 9y^{2}
We may work out (2x + 3y)^{2} directly.
(2x + 3y)^{2} = (2x + 3y) (2x + 3y)
= (2x) (2x) + (2x) (3y) + (3y) (2x) + (3y) (3y)
= 4x^{2} + 6xy + 6 yx + 9y^{2} (as xy = yx)
= 4x^{2}+ 12xy + 9y^{2}
Using Identity (I) gave us an alternative method of squaring (2x + 3y). Do you notice that the Identity method required fewer steps than the above direct method? You will realize the simplicity of this method even more if you try to square more complicated binomial expressions than (2x + 3y).

(103)^{2} = (100 + 3)2
= 100^{2} + 2 × 100 × 3 + 3^{2} (Using Identity I)
= 10000 + 600 + 9 = 10609
We may also directly multiply 103 by 103 and get the answer. Do you see that Identity (I) has given us a less tedious method than the direct method of squaring 103? Try squaring 1013. You will find in this case, the method of using identities even more attractive than the direct multiplication method.
Using Identity (II), find
 (4p – 3q)^{2}
 (4.9)^{2}

(4p – 3q)^{2} =(4p)^{2} – 2 (4p) (3q) + (3q)^{2} [Using the Identity (II)]
= 16p^{2} – 24pq + 9q^{2}
Do you agree that for squaring (4p – 3q)^{2} the method of identities is quicker than the direct method?

(4.9)^{2} =(5.0 – 0.1)^{2} = (5.0)^{2} – 2 (5.0) (0.1) + (0.1)^{2}
= 25.00 – 1.00 + 0.01 = 24.01
Is it not that, squaring 4.9 using Identity (II) is much less tedious than squaring it by direct multiplication?
Using Identity (III), find
 (3/2 m + 2/3 n)(3/2 m  2/3 n)
 983^{2} – 17^{2}
 194 × 206
 (3/2 m + 2/3 n)(3/2 m  2/3 n) = (3/2 m)^{2} (2/3 n)^{2}
=(9/4 m^{2})  4/9 n^{2} 983^{2}  17^{2} = (983 + 17) (983 – 17)
[Here a = 983, b =17, a2 – b2 = (a + b) (a – b)]
Therefore, 9832 – 172 = 1000 × 966 = 966000
Try doing this directly. You will realise how easy our method of using Identity (III) is.
194 × 206 = (200 – 6) × (200 + 6) = 2002 – 62
= 40000 – 36 = 39964
Use the Identity (x + a) (x + b) = x^{2} + (a + b) x + ab to find the following:
 501 × 502
 95 × 103

501 × 502 = (500 + 1) × (500 + 2) = 5002 + (1 + 2) × 500 + 1 × 2
= 250000 + 1500 + 2 = 251502

5 × 103 = (100 – 5) × (100 + 3) = 100^{2} + (–5 + 3) × 100 + (–5) × 3
= 10000 – 200 – 15 = 9785