9.1 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, 3x2, 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, 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.

TRY THESE

9.2 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 numerical coefficient or simply coefficient. The coefficient in the term 7xy is 7 and the coefficient in the term –5x is –5.

9.3 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 non-zero coefficient (with variables having non negative integers as exponents) is called a polynomial. A polynomial may contain any number of terms, one or more than one.

Examples of monomials: 4x2, 3xy, –7z, 5xy2, 10y, –9, 82mnp, etc.

Examples of binomials: a + b, 4l + 5m, a + 4, 5 –3xy, z2 – 4y2, etc.

Examples of trinomials: a + b + c, 2x + 3y – 5, x2y – xy2 + y2, etc.

Examples of polynomials: a + b + c + d, 3xy, 7xyz – 10, 2x + 3y + 7z, etc.

9.4 Like and Unlike Terms

Look at the following expressions:

7x, 14x, –13x, 5x2, 7y, 7xy, –9y2, –9x2, –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?

9.5 Addition and Subtraction of Algebraic Expressions

In the earlier classes, we have also learnt how to add and subtract algebraic expressions. For example, to add 7x2 – 4x + 5 and 9x – 10, we do

7x2 – 4x + 5

+ 9x – 10

7x2 + 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.

Example 1: Add: 7xy + 5yz – 3zx, 4yz + 9zx – 4y , –3xz + 5x – 2xy.

Solution: 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.

Example 2: Subtract 5x2 – 4y2 + 6y – 3 from 7x2 – 4xy + 8y2 + 5x – 3y.

Solution:

7x2 – 4xy + 8y2 + 5x – 3y

5x2 – 4y2 + 6y – 3

(–) (+) (–) (+)

2x2 – 4xy + 12y2 + 5x – 9y + 3

Note that subtraction of a number is the same as addition of its additive inverse. Thus subtracting –3 is the same as adding +3. Similarly, subtracting 6y is the same as adding – 6y; subtracting – 4y2 is the same as adding 4y2 and so on. The signs in the third row written below each term in the second row help us in knowing which operation has to be performed.

1. Identify the terms, their coefficients for each of the following expressions.

(i) 5xyz2 – 3zy (ii) 1 + x + x2    (iii) 4x2y2 – 4x2y2z2 + z2

(iv) 3 – pq + qr – rp (v)     (vi) 0.3a – 0.6ab + 0.5b

2. Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

x + y, 1000, x + x2 + x3 + x4, 7 + y + 5x, 2y – 3y2, 2y – 3y2 + 4y3, 5x – 4y + 3xy, 4z – 15z2, ab + bc + cd + da, pqr, p2q + pq2, 2p + 2q

3. Add the following.

(i) ab – bc, bc – ca, ca – ab  (ii) a – b + ab, b – c + bc, c – a + ac

(iii) 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2  (iv) l2 + m2, m2 + n2, n2 + l2,2lm + 2mn + 2nl

4. (a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3

(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz

(c) Subtract 4p2q – 3pq + 5pq2 – 8p + 7q – 10 from

18 – 3p – 11q + 5pq – 2pq2 + 5p2q

9.6 Multiplication of Algebraic Expressions: Introduction

(i) Look at the following patterns of dots.

(ii) 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).

(iv) Sarita points out that when we buy things, we have to carry out multiplication. For example, if

price of bananas per dozen = ` p

and for the school picnic bananas needed = z dozens,

then we have to pay = ` p × z

Suppose, the price per dozen was less by ` 2 and the bananas needed were less by 4 dozens.

Then, price of bananas per dozen = ` (p – 2)

and bananas needed = (z – 4) dozens,

Therefore, we would have to pay = ` (p – 2) × (z – 4)

9.7 Multiplying a Monomial by a Monomial

9.7.1 Multiplying two monomials

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.

We begin with

Similarly, 4 × (3x) = 3x + 3x + 3x + 3x = 12x

Now, observe the following products.

(i) x × 3y = x × 3 × y = 3 × x × y = 3xy

(ii) 5x × 3y = 5 × x × 3 × y = 5 × 3 × x × y = 15xy

(iii) 5x × (–3y) = 5 × x × (–3) × y

= 5 × (–3) × x × y = –15xy

Some more useful examples follow.

(iv) 5x × 4x2 = (5 × 4) × (x × x2)

= 20 × x³ = 20x³

(v) 5x × (– 4xyz) = (5 × – 4) × (x × xyz)

= –20 × (x × x × yz) = –20x2yz

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.

9.7.2 Multiplying three or more monomials

Observe the following examples.

(i) 2x × 5y × 7z = (2x × 5y) × 7z = 10xy × 7z = 70xyz

(ii) 4xy × 5x2y2 × 6x3y3 = (4xy × 5x2y2) × 6x3y3 = 20x3y3 × 6x3y3 = 120x3y3 × x3y3

= 120 (x3 × x3) × (y3 × y3) = 120x6 × y6 = 120x6y6

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.

TRY THESE

Example 3: Complete the table for area of a rectangle with given length and breadth.

Solution:

Example 4: Find the volume of each rectangular box with given length, breadth and height.

Solution: Volume = length × breadth × height

Hence, for (i) volume = (2ax) × (3by) × (5cz)

= 2 × 3 × 5 × (ax) × (by) × (cz) = 30abcxyz

for (ii) volume = m2n × n2p × p2m

= (m2 × m) × (n × n2) × (p × p2) = m3n3p3

for (iii) volume = 2q × 4q2 × 8q3

= 2 × 4 × 8 × q × q2 × q3 = 64q6

Exercise 9.2

1. Find the product of the following pairs of monomials.

(i) 4, 7p (ii) – 4p, 7p (iii) – 4p, 7pq (iv) 4p3, – 3p

(v) 4p, 0

(p, q); (10m, 5n); (20x2, 5y2); (4x, 3x2); (3mn, 4np)

3. Complete the table of products.

4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

(i) 5a, 3a2, 7a4 (ii) 2p, 4q, 8r (iii) xy, 2x2y, 2xy2 (iv) a, 2b, 3c

5. Obtain the product of

(i) xy, yz, zx (ii) a, – a2, a3 (iii) 2, 4y, 8y2, 16y3

(iv) a, 2b, 3c, 6abc (v) m, – mn, mnp

9.8 Multiplying a Monomial by a Polynomial

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

Similarly, (–3x) × (–5y + 2) = (–3x) × (–5y) + (–3x) × (2) = 15xy – 6x

and 5xy × (y2 + 3) = (5xy × y2) + (5xy × 3) = 5xy3 + 15xy.

What about a binomial × monomial? For example, (5y + 2) × 3x = ?

We may use commutative law as : 7 × 3 = 3 × 7; or in general a × b = b × a

Similarly, (5y + 2) × 3x = 3x × (5y + 2) = 15xy + 6x as before.

Find the product (i) 2x (3x + 5xy) (ii) a2 (2ab – 5c)

9.8.2 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.

Observe, by using the distributive law, we are able to carry out the multiplication term by term.

TRY THESE

Find the product: (4p2 + 5p + 7) × 3p

Example 5: Simplify the expressions and evaluate them as directed:

(i) x (x – 3) + 2 for x = 1, (ii) 3y (2y – 7) – 3 (y – 4) – 63 for y = –2

Solution:

(i) x (x – 3) + 2 = x2 – 3x + 2

For x = 1, x2 – 3x + 2 = (1)2 – 3 (1) + 2

= 1 – 3 + 2 = 3 – 3 = 0

(ii) 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

Example 6: Add

(i) 5m (3 – m) and 6m2 – 13m (ii) 4y (3y2 + 5y – 7) and 2 (y3 – 4y2 + 5)

Solution:

(i) First expression = 5m (3 – m) = (5m × 3) – (5m × m) = 15m – 5m2

Now adding the second expression to it,15m – 5m2 + 6m2 – 13m = m2 + 2m

(ii) The first expression = 4y (3y2 + 5y – 7) = (4y × 3y2) + (4y × 5y) + (4y × (–7))

= 12y3 + 20y2 – 28y

The second expression = 2 (y3 – 4y2 + 5) = 2y3 + 2 × (– 4y2) + 2 × 5

= 2y3 – 8y2 + 10

Adding the two expressions, 12y3 + 20y2 – 28y

+ 2y3 – 8y2 + 10

14y3 + 12y2 – 28y + 10

Example 7: Subtract 3pq (p – q) from 2pq (p + q).

Solution: We have 3pq (p – q) = 3p2q – 3pq2 and

Subtracting, 2p2q + 2pq2

3p2q – 3pq2

–          +               

– p2q + 5pq2

Exercise 9.3

1. Carry out the multiplication of the expressions in each of the following pairs.

(i) 4p, q + r (ii) ab, a – b (iii) a + b, 7a2b2 (iv) a2 – 9, 4a

(v) pq + qr + rp, 0

2. Complete the table.

First expression Second expression Product

3. Find the product.

(i) (a2) × (2a22) × (4a26) (ii)

(iii) (iv) x × x2 × x3 × x4

4. (a) Simplify 3x (4x – 5) + 3 and find its values for (i) x = 3 (ii) x = .

(b) Simplify a (a2 + a + 1) + 5 and find its value for (i) a = 0, (ii) a = 1

(iii) a = – 1.

5. (a) Add: p ( p – q), q ( q – r) and r ( r – p)

(b) Add: 2x (z – x – y) and 2y (z – y – x)

(c) Subtract: 3l (l – 4 m + 5 n) from 4l ( 10 n – 3 m + 2 l )

(d) Subtract: 3a (a + b + c ) – 2 b (a – b + c) from 4c ( – a + b + c )

9.9 Multiplying a Polynomial by a Polynomial

9.9.1 Multiplying a binomial by a binomial

Let us multiply one binomial (2a + 3b) by another binomial, say (3a + 4b). We do this step-by-step, as we did in earlier cases, following the distributive law of multiplication,

(3a + 4b) × (2a + 3b) = 3a × (2a + 3b) + 4b × (2a + 3b)

Observe, every term in one binomial multiplies every term in the other binomial

= (3a × 2a) + (3a × 3b) + (4b × 2a) + (4b × 3b)

= 6a2 + 9ab + 8ba + 12b2

= 6a2 + 17ab + 12b2 (Since ba = ab)

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.

Example 8: Multiply

(i) (x – 4) and (2x + 3) (ii) (x – y) and (3x + 5y)

Solution:

(i) (x – 4) × (2x + 3) = x × (2x + 3) – 4 × (2x + 3)

= (x × 2x) + (x × 3) – (4 × 2x) – (4 × 3) = 2x2 + 3x – 8x – 12

= 2x2 – 5x – 12 (Adding like terms)

(ii) (x – y) × (3x + 5y) = x × (3x + 5y) – y × (3x + 5y)

= (x × 3x) + (x × 5y) – (y × 3x) – ( y × 5y)

= 3x2 + 5xy – 3yx – 5y2 = 3x2 + 2xy – 5y2 (Adding like terms)

Example 9: Multiply

(i) (a + 7) and (b – 5) (ii) (a2 + 2b2) and (5a – 3b)

Solution:

(i) (a + 7) × (b – 5) = a × (b – 5) + 7 × (b – 5)

= ab – 5a + 7b – 35

Note that there are no like terms involved in this multiplication.

(ii) (a2 + 2b2) × (5a – 3b) = a2 (5a – 3b) + 2b2 × (5a – 3b)

= 5a3 – 3a2b + 10ab2 – 6b3

9.9.2 Multiplying a binomial by a trinomial

In 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 × (a2 + 3a + 5) + 7 × (a2 + 3a + 5)

= 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?)

Example 10: Simplify (a + b) (2a – 3b + c) – (2a – 3b) c.

Solution: We have

= 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

Exercise 9.4

1. Multiply the binomials.

(i) (2x + 5) and (4x – 3) (ii) (y – 8) and (3y – 4)

(iii) (2.5l – 0.5m) and (2.5l + 0.5m) (iv) (a + 3b) and (x + 5)

(v) (2pq + 3q2) and (3pq – 2q2)

(vi)

2. Find the product.

(i) (5 – 2x) (3 + x) (ii) (x + 7y) (7x – y)

(iii) (a2 + b) (a + b2) (iv) (p2 – q2) (2p + q)

3. Simplify.

(i) (x2 – 5) (x + 5) + 25 (ii) (a2 + 5) (b3 + 3) + 5

(iii) (t + s2) (t2 – s)

(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)

(v) (x + y)(2x + y) + (x + 2y)(x – y) (vi) (x + y)(x2 – xy + y2)

(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y

(viii) (a + b + c)(a + b – c)

9.10 What is an Identity?

Consider the equality (a + 1) (a +2) = a2 + 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 = a2 + 3a + 2 = 102 + 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

a2 + 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 a2 + 3a + 2 = 132 is not true for a = –5 and for a = 0.

9.11 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)

= a2 + ab + ba + b2

= a2 + 2ab + b2 (since ab = ba)

Thus (a + b)2 = a2 + 2ab + b2 (I)

• Next we consider (a – b)2 = (a – b) (a – b) = a (a – b) – b (a – b)

We have = a2 – ab – ba + b2 = a2 – 2ab + b2

or (a – b)2 = a2 – 2ab + b2 (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) = a2 – b2 (III)

The identities (I), (II) and (III) are known as standard identities.

1. Put – b in place of b in Identity (I). Do you get Identity (II)?

(x + a) (x + b) = x (x + b) + a (x + b)

= x2 + bx + ax + ab

or (x + a) (x + b) = x2 + (a + b) x + ab (IV)

TRY THESE

We can see that Identity (IV) is the general form of the other three identities also.

9.12 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.

Example 11: Using the Identity (I), find (i) (2x + 3y)2 (ii) 1032

Solution:

(i) (2x + 3y)2 = (2x)2 + 2(2x) (3y) + (3y)2 [Using the Identity (I)]

= 4x2 + 12xy + 9y2

We may work out (2x + 3y)2 directly.

(2x + 3y)2 = (2x + 3y) (2x + 3y)

= (2x) (2x) + (2x) (3y) + (3y) (2x) + (3y) (3y)

= 4x2 + 6xy + 6 yx + 9y2 (as xy = yx)

= 4x2 + 12xy + 9y2

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 realise the simplicity of this method even more if you try to square more complicated binomial expressions than (2x + 3y).

(ii) (103)2 = (100 + 3)2

= 1002 + 2 × 100 × 3 + 32 (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.

Example 12: Using Identity (II), find (i) (4p – 3q)2 (ii) (4.9)2

Solution:

(i) (4p – 3q)2 = (4p)2 – 2 (4p) (3q) + (3q)2 [Using the Identity (II)]

= 16p2 – 24pq + 9q2

(ii) (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?

Example 13: Using Identity (III), find

(i) (ii) 9832 – 172 (iii) 194 × 206

Solution:

(i) =

=

Try doing this directly. You will realise how easy our method of using Identity (III) is.

[Here a = 983, b = 17, a2 – b2 = (a + b) (a – b)]

(iii) 194 × 206 = (200 – 6) × (200 + 6) = 2002 – 62

= 40000 – 36 = 39964

Example 14: Use the Identity (x + a) (x + b) = x2 + (a + b) x + ab to find the following:

(i) 501 × 502 (ii) 95 × 103

Solution:

(i) 501 × 502 = (500 + 1) × (500 + 2) = 5002 + (1 + 2) × 500 + 1 × 2

= 250000 + 1500 + 2 = 251502

(ii) 95 × 103 = (100 – 5) × (100 + 3) = 1002 + (–5 + 3) × 100 + (–5) × 3

= 10000 – 200 – 15 = 9785

Exercise 9.5

1. Use a suitable identity to get each of the following products.

(i) (x + 3) (x + 3) (ii) (2y + 5) (2y + 5) (iii) (2a – 7) (2a – 7)

(iv) (3a – ) (3a – ) (v) (1.1m – 0.4) (1.1m + 0.4)

(vi) (a2 + b2) (– a2 + b2) (vii) (6x – 7) (6x + 7) (viii) (– a + c) (– a + c)

(ix) (x) (7a – 9b) (7a – 9b)

2. Use the identity (x + a) (x + b) = x2 + (a + b) x + ab to find the following products.

(i) (x + 3) (x + 7) (ii) (4x + 5) (4x + 1)

(iii) (4x – 5) (4x – 1) (iv) (4x + 5) (4x – 1)

(v) (2x + 5y) (2x + 3y) (vi) (2a2 + 9) (2a2 + 5)

(vii) (xyz – 4) (xyz – 2)

3. Find the following squares by using the identities.

(i) (b – 7)2 (ii) (xy + 3z)2 (iii) (6x2 – 5y)2

(iv) (v) (0.4p – 0.5q)2 (vi) (2xy + 5y)2

4. Simplify.

(i) (a2 – b2)2 (ii) (2x + 5)2 – (2x – 5)2

(iii) (7m – 8n)2 + (7m + 8n)2 (iv) (4m + 5n)2 + (5m + 4n)2

(v) (2.5p – 1.5q)2 – (1.5p – 2.5q)2

(vi) (ab + bc)2 – 2ab2c (vii) (m2 – n2m)2 + 2m3n2

5. Show that.

(i) (3x + 7)2 – 84x = (3x – 7)2 (ii) (9p – 5q)2 + 180pq = (9p + 5q)2

(iii) + 2mn =

(iv) (4pq + 3q)2 – (4pq – 3q)2 = 48pq2

(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0

6. Using identities, evaluate.

(i) 712 (ii) 992 (iii) 1022 (iv) 9982

(v) 5.22 (vi) 297 × 303 (vii) 78 × 82 (viii) 8.92

(ix) 10.5 × 9.5

7. Using a2 – b2 = (a + b) (a – b), find

(i) 512 – 492 (ii) (1.02)2 – (0.98)2 (iii) 1532 – 1472

(iv) 12.12 – 7.92

8. Using (x + a) (x + b) = x2 + (a + b) x + ab, find

(i) 103 × 104 (ii) 5.1 × 5.2 (iii) 103 × 98 (iv) 9.7 × 9.8

What have we Discussed?