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ALGEBRAIC EXPRESSIONS, , AND I DENTITIES, , Algebraic Expressions, and Identities, , 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, 5 7, , – etc.; actually countless different values. The value of an expression, 2 3, 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., , 2, 5, –3, 0,, , 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., 2021–22, , 137, , CHAPTER, , 9

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138, , MATHEMATICS, , TRY THESE, 1. Give five examples of expressions containing one variable and five examples of, expressions containing two variables., 2. Show on the number line x, x – 4, 2x + 1, 3x – 2., , 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., TRY THESE, The expression 7xy – 5x has two terms 7xy and –5x. The term, Identify the coefficient of each, 7xy is a product of factors 7, x and y. The numerical factor of a term, term in the expression, is called its numerical coefficient or simply coefficient. The coefficient, x2y2 – 10x2y + 5xy2 – 20., 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., , TRY THESE, 1. Classify the following polynomials as monomials, binomials, trinomials., – z + 5, x + y + z, y + z + 100, ab – ac, 17, 2. Construct, (a) 3 binomials with only x as a variable;, (b) 3 binomials with x and y as variables;, (c) 3 monomials with x and y as variables;, (d) 2 polynomials with 4 or more terms., , 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., , 2021–22

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ALGEBRAIC EXPRESSIONS, , AND I DENTITIES, , To find the number of, dots we have to multiply, the expression for the, number of rows by the, expression for the, number of columns., , m×n, , (m + 2) × (n + 3), , Here the number of rows, is increased by, 2, i.e., m + 2 and number, of columns increased by, 3, i.e., n + 3., , (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)., (iii) Can you think about volume? (The volume of a, rectangular box is given by the product of its length,, breadth and height)., , To find the area of a rectangle, we, have to multiply algebraic, expressions like l × b or, (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), , 2021–22, , 141

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142, , MATHEMATICS, , TRY THESE, Can you think of two more such situations, where we may need to multiply algebraic, expressions?, [Hint: • Think of speed and time;, • Think of interest to be paid, the principal and the rate of simple interest; etc.], 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., , 9.7 Multiplying a Monomial by a Monomial, 9.7.1 Multiplying two monomials, We begin with, 4 × x = x + x + x + x = 4x as seen earlier., Notice that all the three, Similarly, 4 × (3x) = 3x + 3x + 3x + 3x = 12x, products of monomials, 3xy,, Now, observe the following products., 15xy, –15xy, are also, monomials., (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), , 2, , 2, , 5x × 4x = (5 × 4) × (x × x ), = 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., , Note that 5 × 4 = 20, i.e., coefficient of product = coefficient of, first monomial × coefficient of second, monomial;, and, x × x2 = x 3, i.e.,, algebraic factor of product, = algebraic factor of first monomial, × algebraic factor of second monomial., , 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., 2021–22

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146, , MATHEMATICS, , 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, , (i), , a, , b+c+d, , ..., , (ii), , x+y–5, , 5xy, , ..., , (iii), , p, , 6p2 – 7p + 5, , ..., , (iv), , 4p2q2, , p2 – q2, , ..., , (v), , a+b+c, , abc, , ..., , 3. Find the product., (i) (a2) × (2a22) × (4a26), ,  10 3   6 3 , pq  ×  p q, (iii)  −,  5, , 3, ,  2   −9 2 2 , x y , (ii)  xy × ,  3   10, , (iv) x × x2 × x3 × x4, , 1, ., 2, (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 ), , 4. (a) Simplify 3x (4x – 5) + 3 and find its values for (i) x = 3 (ii) x =, , 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), = (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., Observe, every term in one, binomial multiplies every, term in the other binomial., , 2021–22

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148, , MATHEMATICS, , EXERCISE 9.4, 1. Multiply the binomials., (i) (2x + 5) and (4x – 3), (iii) (2.5l – 0.5m) and (2.5l + 0.5m), (v) (2pq + 3q2) and (3pq – 2q2), , (ii) (y – 8) and (3y – 4), (iv) (a + 3b) and (x + 5), , (vi), 2. Find the product., (i) (5 – 2x) (3 + x), (ii) (x + 7y) (7x – y), 2, 2, (iii) (a + b) (a + b ), (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., , 2021–22

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ALGEBRAIC EXPRESSIONS, , 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, , AND I DENTITIES, , (since ab = ba), , (a + b)2 = a2 + 2ab + b2, , Thus, , (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, or, , = a2 – ab – ba + b2 = a2 – 2ab + b2, (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), or, , = a2 – ab + ba – b2 = a2 – b2(since ab = ba), (a + b) (a – b) = a2 – b2, (III), , The identities (I), (II) and (III) are known as standard identities., , TRY THESE, 1. Put – b in place of b in Identity (I). Do you get Identity (II)?, , • We shall now work out one more useful identity., , or, , (x + a) (x + b) = x (x + b) + a (x + b), = x2 + bx + ax + ab, (x + a) (x + b) = x2 + (a + b) x + ab, , (IV), , TRY THESE, 1. Verify Identity (IV), for a = 2, b = 3, x = 5., 2. Consider, the special case of Identity (IV) with a = b, what do you get? Is it, related to Identity (I)?, 3. Consider, the special case of Identity (IV) with a = – c and b = – c. What do you, get? Is it related to Identity (II)?, 4. Consider the special case of Identity (IV) with b = – a. What do you get? Is it, related to Identity (III)?, 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., 2021–22, , 149

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150, , MATHEMATICS, , 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)], 2, 2, = 4x + 12xy + 9y, 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), 2, 2, = 4x + 12xy + 9y, 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, Do you agree that for squaring (4p – 3q)2 the method of identities is quicker than the, direct method?, (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, , 2  3, 3, (i)  m + n  m −, 2, 3, 2, Solution:, 2  3, 3, (i)  m + n  m −, 2, 3, 2, , 2 , n, 3 , , (ii), , 9832 – 172, , (iii), , 194 × 206, , 2   3 2  2 2, Try doing this directly., n =, m −  n, 3 , You will realise how easy, 3   2 , our method of using, 9 2 4 2, Identity (III) is., m, −, n, =, 4, 9, (ii) 9832 – 172 = (983 + 17) (983 – 17), [Here a = 983, b =17, a2 – b2 = (a + b) (a – b)], Therefore,, 9832 – 172 = 1000 × 966 = 966000, 2021–22

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152, , MATHEMATICS, , 6. Using identities, evaluate., (i) 712, (ii) 992, (iii), 2, (v) 5.2, (vi) 297 × 303, (vii), (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), 2, 2, (iv) 12.1 – 7.9, 8. Using (x + a) (x + b) = x2 + (a + b) x + ab, find, (i) 103 × 104, (ii) 5.1 × 5.2, (iii), , 1022, 78 × 82, , (iv) 9982, (viii) 8.92, , 1532 – 1472, , 103 × 98, , (iv) 9.7 × 9.8, , WHAT HAVE WE DISCUSSED?, 1. Expressions are formed from variables and constants., 2. Terms are added to form expressions. Terms themselves are formed as product of factors., 3. Expressions that contain exactly one, two and three terms are called monomials, binomials and, trinomials respectively. In general, any expression containing one or more terms with non-zero, coefficients (and with variables having non- negative integers as exponents) is called a polynomial., 4. Like terms are formed from the same variables and the powers of these variables are the same,, too. Coefficients of like terms need not be the same., 5. While adding (or subtracting) polynomials, first look for like terms and add (or subtract) them;, then handle the unlike terms., 6. There are number of situations in which we need to multiply algebraic expressions: for example, in, finding area of a rectangle, the sides of which are given as expressions., 7. A monomial multiplied by a monomial always gives a monomial., 8. While multiplying a polynomial by a monomial, we multiply every term in the polynomial by the, monomial., 9. In carrying out the multiplication of a polynomial by a binomial (or trinomial), we multiply term by, term, i.e., every term of the polynomial is multiplied by every term in the binomial (or trinomial)., Note that in such multiplication, we may get terms in the product which are like and have to be, combined., 10. An identity is an equality, which is true for all values of the variables in the equality., On the other hand, an equation is true only for certain values of its variables. An equation is not an, identity., 11. The following are the standard identities:, (a + b)2 = a2 + 2ab + b2, (I), (II), (a – b)2 = a2 – 2ab + b2, (a + b) (a – b) = a2 – b2, (III), 12. Another useful identity is (x + a) (x + b) = x2 + (a + b) x + ab, (IV), 13. The above four identities are useful in carrying out squares and products of algebraic expressions., They also allow easy alternative methods to calculate products of numbers and so on., , 2021–22