9.1. INTRODUCTION
a) What are expressions
In earlier classes, we have already become familiar with what algebraic expressions (or simply expressions) are.
Examples of expressions are: $x+3,2y\u20135,3{x}^{2},4xy+7$ etc.
We can form many more expressions. As we 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 $2,5,\u20133,0,\frac{5}{2},\u2013\frac{7}{3}$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.
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.
Let’s see 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.
b) 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.
c) 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 exponents) is called a polynomial. A polynomial may contain any number of terms, one or more than one.
Examples of monomials: $4{x}^{2},3xy,\u20137z,5x{y}^{2},10y,\u20139,82mnp$, etc.
Examples of binomials: $\mathrm{a}+\mathrm{b},4l+5\mathrm{m},\mathrm{a}+4,5\u20133\mathrm{xy},{\mathrm{z}}^{2}\u20134{\mathrm{y}}^{2}$, etc.
Examples of trinomials: $a+b+c,2x+3y\u20135,{x}^{2}y\u2013x{y}^{2}+{y}^{2}$
Examples of polynomials: a + b + c + d, 3xy, 7xyz – 10, 2x + 3y + 7z, etc.
d) Like and Unlike terms
Look at the following expressions:
$7x,14x,\u201313x,5{x}^{2},7y,7xy,\u20139{y}^{2},\u20139{x}^{2},\u20135yx$
Like terms from these are:
(i) 7x, 14x, –13x are like terms.
(ii) $5{x}^{2}$ and $\u20139{x}^{2}$ are like terms.
(iii) 7xy and –5yx are like terms.
Unlike terms from these are:
(i) 7x and 7y are unlike terms.
(ii) 7x and 7xy are unlike terms.
(iii) 7x and $5{x}^{2}$ are unlike terms.
9.2. ADDITION AND SUBTRACTION OF ALGEBRAIC EXPRESSIONS
Let’s add $7{x}^{2}\u20134x+5$and 9x – 10, we do
$7{x}^{2}\u20134x+5\phantom{\rule{0ex}{0ex}}+9x\u201310\phantom{\rule{0ex}{0ex}}\overline{)7{x}^{2}+5x\u20135}$
During addition of algebraic expression, 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 above
Thus 5 + (–10) = 5 –10 = –5.
Similarly, – 4x + 9x = (– 4 + 9)x = 5x.
Let us take some more examples.
9.3. MULTIPLICATION OF MONOMIALS WITH ALGEBRAIC EXPRESSIONS
I. Multiplying two monomials
We know that
4 × x = x + x + x + x = 4x as seen earlier.
Similarly, 4 × (3x) = 3x + 3x + 3x + 3x = 12x
Let’s observe the multiplication of few 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
Note: We observe that all the three products of monomials, 3xy, 15xy, –15xy, are also monomials.
(iv) 5x × 4x^{2} = (5 × 4) × (x × x^{2})
= 20 × x^{3} = 20x^{3}
Note: 5 × 4 = 20
i.e., coefficient of product = coefficient of first monomial × coefficient of second monomial; and x × x^{2} = x^{3}
algebraic factor of product = algebraic factor of first monomial × algebraic factor of second monomial.
(v) 5x × (– 4xyz) = (5 × – 4) × (x × xyz)
= –20 × (x × x × yz) = –20x^{2} yz
Let see few more examples.
Suppose we have two monomials 4x and 5y. By multiplying them, we get
4x × 5y = (4 × x) × (5 × y)
= (4 × 5) × (x × y) = 20xy
Hence, we can say that 4x × 5y = 20xy
Now, 10x × 5x^{2}z = (10 × x) × (5 × x^{2} × z)
This becomes (10 × 5) × (x × x^{2} × z) = 50x^{3}z
II. Multiplying three or more monomials
Observe the following examples.
(i) 2x × 5y × 7z = (2x × 5y) × 7z = 10xy × 7z = 70xyz
(ii) 4xy × 5x^{2} y^{2} × 6x^{3}y^{3} = (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.
Example: Complete the table for area of a rectangle with given length and breadth.
Solution
Length |
Breadth |
Area |
3x 9y 4ab 2l^{2}m |
5y 4y^{2} 5bc 3lm^{2} |
3x×5y = 15xy 9y×4y^{2} = 36y^{3} 4ab×5bc = 20ab^{2}c 2l^{2}m×3lm^{2} = 6l^{3}m^{3} |
III. Multiplication of a monomial with a polynomial
a) 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
We commonly use distributive law in our calculations.
For example
7 × 106 = 7 × (100 + 6) = 7 × 100 + 7 × 6
(Here, we used distributive law)
= 700 + 42 = 742
7 × 38 = 7 × (40 – 2)
= 7 × 40 – 7 × 2
(Here, we used distributive law)
= 280 – 14 = 266
Similarly, (–3x) × (–5y + 2)
= (–3x) × (–5y) + (–3x) × (2) = 15xy – 6x
and 5xy × (y^{2} + 3) = (5xy × y^{2} ) + (5xy × 3)
= 5xy^{3} + 15xy.
Multiplication of 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.
b) Multiplying a monomial by a trinomial
Consider 3p × (4p^{2} + 5p + 7). As in the earlier case, we use distributive law;
3p × (4p^{2} + 5p + 7) = (3p × 4p^{2} ) + (3p × 5p) + (3p × 7)
= 12p^{3} + 15p^{2} + 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.
Example: 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 = x^{2} – 3x + 2
For x = 1, x^{2} – 3x + 2 = (1)^{2} – 3 (1) + 2
= 1 – 3 + 2 = 3 – 3 = 0
(ii) 3y (2y – 7) – 3 (y – 4) – 63 = 6y^{2} – 21y – 3y + 12 – 63
= 6y^{2} – 24y – 51
For y = –2, 6y^{2} – 24y – 51 = 6 (–2)^{2} – 24(–2) – 51
= 6 × 4 + 24 × 2 – 51
= 24 + 48 – 51 = 72 – 51 = 21
9.4. MULTIPLICATION OF BINOMIALS WITH BINOMIALS AND TRINOMIALS
I. Multiplication of a binomial by a binomial
Let us multiply one binomial (2a + 3b) by another binomial, say (3a + 4b).
Observe, every term in one binomial multiplies every term in the other binomial.
(3a + 4b) × (2a + 3b) = 3a × (2a + 3b) + 4b × (2a + 3b)
= (3a × 2a) + (3a × 3b) + (4b × 2a) + (4b × 3b)
= 6a^{2} + 9ab + 8ba + 12b^{2}
= 6a^{2} + 17ab + 12b^{2}
(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.
In multiplication of polynomials with polynomials, we sholud always look for like terms, if any, and combine them.
II. Multiplication of a binomial with 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
$\underset{\text{binomial}}{\underset{\u23df}{(a+7)}}\times \underset{\text{trinomial}}{\underset{\u23df}{\left({a}^{2}+3a+5\right)}}$
= a × (a + 3a + 5) + 7 × (a + 3a + 5)
[using the distributive law]
= a^{3} + 3a^{2} + 5a + 7a2 + 21a + 35
= a^{3} + (3a^{2} + 7a^{2} ) + (5a + 21a) + 35
= a^{3} + 10a^{2} + 26a + 35
9.5. 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 = a^{2} + 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) = a^{2} + 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.
9.6. STANDARD IDENTITIES
I. Use of the Identities (a + b)^{2} and (a – b)^{2}
Let us try to find the square of the number 102. The square of a number, as we know, is the product of the number with itself. One way to do this is by writing the numbers one below the other, and then multiplying them as we normally do. The other way is to break the numbers and then apply distributive property. This will make our work much easier.
Let us see how.
102^{2} = 102 × 102
= (100 + 2) (100 + 2)
= 100 (100 + 2) + 2 (100 + 2)
= 100 × 100 + 100 × 2 + 2 × 100 + 2 × 2
= 10000 + 200 + 200 + 4
= 10404
Let us try to find the square of the expression (a + b).
(a + b)^{2} = (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} + 2ab + b^{2}
$\overline{)\therefore {\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}}$
This is, in fact, an important identity. It is known as an identity because for any value of a and b, the LHS is always equal to the RHS. The difference between an identity and an equation is that for an equation, its LHS and RHS are equal only for some values of the variable.
On the other hand, as we discussed, for an identity, the LHS equals the RHS for any value of the variable.
There is another important identity for the square of the expression (a – b). Let us try and simplify it.
(a – b)^{2} = (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} – 2ab + b^{2}
$\overline{)\therefore {\left(a\u2013b\right)}^{2}={a}^{2}\u20132ab+{b}^{2}}$
Many a times, these identities help in shortening our calculations.
II. Use of the Identity (x + a) (x + b)
A very important identity that we have to learn is regarding the expression (x + a) (x + b).
We know how to multiply binomials. By finding the product of these binomials, we can find the identity.
Let’s understand this identity from the following examples:
Example: Use the appropriate identity to simplify the following expressions.
(a) (3p + 5q) (3p – 7z)
(b) $\left({z}^{2}\u2013\frac{4}{3}\right)\left({z}^{2}\u2013\frac{3}{2}\right)$
Solution
(a) The given expression is (3p + 5q)(3p – 7z).
This expression is of the form (x + a) (x + b).
Thus, we can use the identity (x + a) (x + b) = x^{2} + (a + b) x + ab
Here, x = 3p, a = 5q, b = –7z
$\therefore $ (3p + 5q)(3p – 7z)
= (3p)^{2} + (5q – 7z)(3p) + (5q)(–7z)
= 9p^{2} + (5q)(3p) – (7z)(3p) – 35qz
= 9p^{2} + 15pq – 21pz – 35qz
(b)The given expression is $\left({z}^{2}\u2013\frac{4}{3}\right)\left({z}^{2}\u2013\frac{3}{2}\right)$
This expression is of the form (x + a) (x + b).
Thus, we can use the identity (x + a) (x + b) = x^{2} + (a + b) x + ab
Here, x = z^{2}, $a=\u2013\frac{4}{3},b=\u2013\frac{3}{2}$
$\therefore \left({z}^{2}\u2013\frac{4}{3}\right)\left({z}^{2}\u2013\frac{3}{2}\right)$
$={\left({z}^{2}\right)}^{2}+\left\{\left(\u2013\frac{4}{3}\right)+\left(\u2013\frac{3}{2}\right)\right\}{z}^{2}+\left(\u2013\frac{4}{3}\right)\times \left(\u2013\frac{3}{2}\right)$
$={z}^{4}+\left(\frac{\u20138\u20139}{6}\right){z}^{2}+2={z}^{4}\u2013\frac{17}{6}{z}^{2}+2$
III. Use of the Identity (a + b) (a – b)
Suppose we need to find the product of the numbers 79 and 81. Instead of multiplying these two numbers, we can use the identity (a + b) (a – b). This identity is very important and is applicable in various situations.
Let us first understand this identity.
(a + b) (a – b) = a (a – b) + b (a – b)
(By distributive property)
= a^{2} – ab + ba – b^{2}
= a^{2} – ab + ab – b^{2} (ab = ba)
= a^{2} – b^{2}
$\overline{)\therefore (a+b)(a\u2013b)={a}^{2}\u2013{b}^{2}}$
Now, let us solve some examples in which the above identity can be applied.
Example: Simplify the following expressions.
(a) $\left(\frac{3}{7}l+\frac{4}{5}m\right)\left(\frac{3}{7}l\u2013\frac{4}{5}m\right)$
(b) (x^{2} – y^{3})(x^{2} + y^{3}) + (y^{3} – z^{4})(y^{3} + z^{4}) + (z^{4} – x^{2})(z^{4} + x^{2})
Solution
(a) The given expression is $\left(\frac{3}{7}l+\frac{4}{5}m\right)\left(\frac{3}{7}l\u2013\frac{4}{5}m\right)$.
This expression is of the form (a + b) (a – b).
Hence, we can use the identity (a + b) (a – b) = a^{2} – b^{2}
$\therefore \left(\frac{3}{7}l+\frac{4}{5}\mathrm{m}\right)\left(\frac{3}{7}l\u2013\frac{4}{5}\mathrm{m}\right)={\left(\frac{3}{7}l\right)}^{2}\u2013{\left(\frac{4}{5}\mathrm{m}\right)}^{2}$
$=\frac{9}{49}{l}^{2}\u2013\frac{16}{25}{\mathrm{m}}^{2}$
(b) (x^{2} – y^{3})(x^{2} + y^{3}) + (y^{3} – z^{4})(y^{3} + z^{4}) + (z^{4} – x^{2})(z^{4} + x^{2})
On using the identity (a + b) (a – b) = a^{2} – b^{2}, we get
$\left\{{\left({x}^{2}\right)}^{2}\u2013{\left({y}^{3}\right)}^{2}\right\}+\left\{{\left({y}^{3}\right)}^{2}\u2013{\left({z}^{4}\right)}^{2}\right\}+\left\{{\left({z}^{4}\right)}^{2}\u2013{\left({x}^{2}\right)}^{2}\right\}$
$={x}^{4}\u2013{y}^{6}+{y}^{6}\u2013{z}^{8}+{z}^{8}\u2013{x}^{4}$
$=\left({x}^{4}\u2013{x}^{4}\right)+\left(\u2013{y}^{6}+{y}^{6}\right)+\left(\u2013{z}^{8}+{z}^{8}\right)$
= 0.