{"id":2659,"date":"2021-01-08T14:29:18","date_gmt":"2021-01-08T14:29:18","guid":{"rendered":"https:\/\/themindpalace.in\/?p=2659"},"modified":"2021-08-26T05:36:58","modified_gmt":"2021-08-26T05:36:58","slug":"algebraic-expression-and-identities","status":"publish","type":"post","link":"https:\/\/themindpalace.in\/index.php\/2021\/01\/08\/algebraic-expression-and-identities\/","title":{"rendered":"Algebraic Expression and Identities"},"content":{"rendered":"\n

Summary of algebraic expression and Identities<\/a><\/p>\n\n\n\n

Solved exercise of algebraic expression and Identities<\/a><\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Summary<\/h1>\n\n\n\n

A combination of constants and variables connected by the signs of fundamental operations of addition, subtraction, multiplication and division is called an algebraic expression.<\/p>\n\n\n\n

Algebraic Expression<\/span><\/strong><\/p>\n\n\n\n

A symbol which takes various numerical values is called a variable.Various parts of an algebraic expression which are separated by the signs of \u2018+\u2019 or \u2018-\u2018 are called the terms of the expression.<\/p>\n\n\n\n

What are Expressions?<\/strong><\/span><\/p>\n\n\n\n

We know that a constant is a symbol having fixed numerical value whereas a variable is a symbol assuming various numerical values.
An algebraic expression is formed from variables and constants. A combination of variables and constants connected by the signs +, -, \u00d7 and \u00f7 is called an algebraic expression. The variable\/variables in an algebraic expression can assume countless different values. The value of algebraic expression changes with the value (s) assumed by the variable (s) it contains.<\/p>\n\n\n\n

Algebraic Expression<\/span><\/strong><\/p>\n\n\n\n

Any mathematical expression which consists of numbers, variables and operations are called Algebraic Expression<\/strong>.<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

1. Terms<\/span><\/strong><\/p>\n\n\n\n

Every expression is separated by an operation which is called Terms<\/strong>. Like 7n and 2 are the two terms in the above figure.<\/p>\n\n\n\n

2. Factors<\/span><\/strong><\/p>\n\n\n\n

Every term is formed by the product of the factors<\/strong>.7n is the product of 7 and n which are the factors of 7n.<\/p>\n\n\n\n

3. Coefficient<\/span><\/strong><\/p>\n\n\n\n

The number placed before the variable or the numerical factor of the term is called Coefficient of that variable<\/strong>.7 is the numerical factor of 7n so 7 is coefficient here.<\/p>\n\n\n\n

4. Variable<\/span><\/strong><\/p>\n\n\n\n

Any letter like x, y etc. are called Variables.<\/strong> The variable in the above figure is n.<\/p>\n\n\n\n

5. Operations<\/span><\/strong><\/p>\n\n\n\n

Addition, subtraction etc. are the operations <\/strong>which separate each term.<\/p>\n\n\n\n

6. Constant<\/span><\/strong><\/p>\n\n\n\n

The number without any variable is constant<\/strong>. 2 is constant here.<\/p>\n\n\n\n

Number Line and an Expression<\/span><\/strong><\/p>\n\n\n\n

An expression can be represented on the number line.<\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

How to represent x + 5 and x \u2013 5 on the number line?<\/p>\n\n\n\n

Solution:<\/span><\/strong><\/p>\n\n\n\n

First, mark the distance x and then x + 5 will be 5 unit to the right of x<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

In the case of x \u2013 5 we will start from the right and move towards the negative side. x \u2013 5 will be 5 units to the left of x.<\/p>\n\n\n\n

Monomials, Binomials and Polynomials<\/span><\/strong><\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

An expression that contains exactly one, two or three terms is called a monomial, binomial or trinomial, respectively. In general, an expression containing, one or more terms with non-zero coefficients and with variable having non-negative exponents is called a polynomial.<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Like and Unlike Terms<\/span><\/strong><\/p>\n\n\n\n

Like (or similar) terms are formed from the same variables and the powers of these variables are also the same. Coefficients of like terms need not be the same. In case otherwise, they are called, unlike (or dissimilar) terms.<\/p>\n\n\n\n

Terms having the same variable are called Like Terms<\/strong>.<\/p>\n\n\n\n

Examples of Like Terms<\/strong> \u20222x and -9x \u202224xy and 5yx \u20226x2<\/sup> and 12x2<\/sup><\/p>\n\n\n\n

The terms having different variable are called, Unlike Terms<\/strong>.<\/p>\n\n\n\n

Examples of Unlike Terms<\/strong> \u20222x and – 9y \u202224xy and 5pq \u20226x2<\/sup> and 12y2<\/sup><\/p>\n\n\n\n

Addition and Subtraction of Algebraic Expressions<\/span><\/strong><\/p>\n\n\n\n

Steps to add or Subtract Algebraic Expression<\/p>\n\n\n\n

\u2022First of all, we have to write the algebraic expressions in different rows in such a way that the like terms come in the same column. \u2022 <\/p>\n\n\n\n

\u2022Add them as we add other numbers. <\/p>\n\n\n\n

\u2022If any term of the same variable is not there in another expression then write is as it is in the solution.<\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

Add 15p2<\/sup> \u2013 4p + 5 and 9p \u2013 11<\/p>\n\n\n\n

Solution:<\/span><\/strong><\/p>\n\n\n\n

Write down the expressions in separate rows with like terms in the same column and add.<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

Subtract 5a2<\/sup> \u2013 4b2 <\/sup>+ 6b \u2013 3 from 7a2<\/sup> \u2013 4ab + 8b2<\/sup> + 5a \u2013 3b.<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

For subtraction also write the expressions in different rows. But to subtract we have to change their signs from negative to positive and vice versa.<\/p>\n\n\n\n

7a2<\/sup> – 4ab  + 8b2    <\/sup>+ 5a \u2013 3b<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Multiplication of Algebraic Expressions<\/span><\/strong><\/p>\n\n\n\n

While multiplying we need to take care of some points about the multiplication of like and unlike terms.<\/p>\n\n\n\n

1. Multiplication of Like Terms<\/span><\/strong> The coefficients will get multiplied. \u2022The power will not get multiplied but the resultant variable will be the addition of the individual powers.<\/p>\n\n\n\n

Example<\/strong> \u2022The product of 4x and 3x will be 12x2<\/sup>. \u2022The product of 5x, 3x and 4x will be 60x3<\/sup>.<\/p>\n\n\n\n

2. Multiplication of Unlike Terms<\/strong> <\/span>The coefficients will get multiplied. \u2022The power will remain the same if the variable is different. \u2022If some of the variables are the same then their powers will be added.<\/p>\n\n\n\n

Example<\/strong> \u2022The product of 2p and 3q will be 6pq \u2022The product of 2x2<\/sup>y, 3x and 9 will  be 54x3<\/sup>y<\/p>\n\n\n\n

Multiplying a Monomial by a Monomial<\/span><\/strong><\/p>\n\n\n\n

1. Multiplying Two Monomials<\/span><\/strong><\/p>\n\n\n\n

While multiplying two polynomials the resultant variable will come by \u2022The coefficient of product = Coefficient of the first monomial \u00d7 Coefficient of the second monomial \u2022The algebraic factor of product = Algebraic factor of the first monomial \u00d7 Algebraic factor of the second monomial.<\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

25y \u00d7 3xy = 125xy2<\/sup><\/p>\n\n\n\n

2. Multiplying Three or More Monomials<\/span><\/strong><\/p>\n\n\n\n

While multiplying three or more monomial the criterion will remain the same.<\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

4xy \u00d7 5x2<\/sup>y2<\/sup> \u00d7 6x3 <\/sup>y3<\/sup><\/p>\n\n\n\n

    = (4 \u00d7 5 \u00d7 6) \u00d7 (x \u00d7 x2<\/sup> \u00d7 x3<\/sup>) \u00d7 (y \u00d7 y2<\/sup> \u00d7 y3<\/sup>)<\/p>\n\n\n\n

    = 120 x6<\/sup>y6<\/sup><\/p>\n\n\n\n

Multiplying a Monomial by a Polynomial<\/span><\/strong><\/p>\n\n\n\n

1. Multiplying a Monomial by a Binomial<\/span><\/strong><\/p>\n\n\n\n

To multiply a monomial with a binomial we have to multiply the monomial with each <\/strong>term of the binomial.<\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

\u2022Multiplication of 8 and (x + y) will be (8x + 8y). <\/p>\n\n\n\n

\u2022Multiplication of 3x and (4y + 7) will be (12xy + 21x).<\/p>\n\n\n\n

\u2022Multiplication of 7x3<\/sup> and (2x4 <\/sup>+ y4<\/sup>) will be (14x7<\/sup>+ 7x3<\/sup>y4<\/sup>).<\/p>\n\n\n\n

\u2022Multiplication of 7x3<\/sup> and (2x4 <\/sup>+ y4<\/sup>) will be (14x7<\/sup>+ 7x3<\/sup>y4<\/sup>).<\/p>\n\n\n\n

2. Multiplication of Monomial by a trinomial<\/span><\/strong><\/p>\n\n\n\n

This is also the same as above<\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

\u2022Multiplication of 8 and (x + y + z) will be (8x + 8y + 8z). <\/p>\n\n\n\n

\u2022Multiplication of 4x and (2x + y + z) will be (8x2<\/sup> + 4xy + 4xz). <\/p>\n\n\n\n

\u2022Multiplication of 7x3<\/sup> and (2x4<\/sup>+ y4<\/sup>+ 2) will be (14x7 <\/sup>+ 7x3<\/sup>y4 <\/sup>+ 14x3<\/sup>).<\/p>\n\n\n\n

Multiplying a Polynomial by a Polynomial<\/span><\/strong><\/p>\n\n\n\n

1. Multiplying a Binomial by a Binomial<\/span><\/strong><\/p>\n\n\n\n

We use the distributive law of multiplication in this case. Multiply each term of a binomial with every term of another binomial. After multiplying the polynomials we have to look for the like terms and combine them.<\/p>\n\n\n\n

Example<\/strong>:Simplify (3a + 4b) \u00d7 (2a + 3b)<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(3a + 4b) \u00d7 (2a + 3b)<\/p>\n\n\n\n

= 3a \u00d7 (2a + 3b) + 4b \u00d7 (2a + 3b)    [distributive law]<\/p>\n\n\n\n

= (3a \u00d7 2a) + (3a \u00d7 3b) + (4b \u00d7 2a) + (4b \u00d7 3b)<\/p>\n\n\n\n

= 6 a2 <\/sup>+ 9ab + 8ba + 12b2<\/sup><\/p>\n\n\n\n

= 6 a2<\/sup> + 17ab + 12b2<\/sup>     [Since ba = ab]<\/p>\n\n\n\n

2. Multiplying a Binomial by a Trinomial<\/span><\/strong><\/p>\n\n\n\n

In this also we have to multiply each term of the binomial with every term of trinomial.<\/p>\n\n\n\n

Example<\/strong>:Simplify (p + q) (2p \u2013 3q + r) \u2013 (2p \u2013 3q) r.<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have a binomial (p + q) and one trinomial (2p \u2013 3q + r)<\/p>\n\n\n\n

(p + q) (2p \u2013 3q + r)<\/p>\n\n\n\n

= p(2p \u2013 3q + r) + q (2p \u2013 3q + r)<\/p>\n\n\n\n

= 2p2<\/sup> \u2013 3pq + pr + 2pq \u2013 3q2<\/sup> + qr<\/p>\n\n\n\n

= 2p2<\/sup> \u2013 pq \u2013 3q2<\/sup> + qr + pr    (\u20133pq and 2pq are like terms)<\/p>\n\n\n\n

(2p \u2013 3q) r = 2pr \u2013 3qr<\/p>\n\n\n\n

(p + q) (2p \u2013 3q + r) \u2013 (2p \u2013 3q) r<\/p>\n\n\n\n

= 2p2<\/sup> \u2013 pq \u2013 3q2<\/sup> + qr + pr \u2013 (2pr \u2013 3qr)<\/p>\n\n\n\n

= 2p2<\/sup> \u2013 pq \u2013 3q2<\/sup> + qr + pr \u2013 2pr + 3qr<\/p>\n\n\n\n

= 2p2<\/sup> \u2013 pq \u2013 3q2<\/sup> + (qr + 3qr) + (pr \u2013 2pr)<\/p>\n\n\n\n

= 2p2<\/sup> \u2013 3q2<\/sup> \u2013 pq + 4qr \u2013 pr<\/p>\n\n\n\n

Identities<\/span><\/strong><\/p>\n\n\n\n

An identity is an equality which is true for every value of the variable but an equation is true for only some of the values of the variables.<\/p>\n\n\n\n

So an equation is not an identity.<\/p>\n\n\n\n

Like, x2<\/sup> = 1, is valid if x is 1 but is not true if x is 2.so it is an equation but not an identity.<\/p>\n\n\n\n

Some of the Standard Identities<\/span><\/strong><\/p>\n\n\n\n

(a + b)2 <\/sup>= a2<\/sup> + 2ab + b2<\/sup><\/p>\n\n\n\n

(a – b)2<\/sup> = a2<\/sup> \u2013 2ab + b2<\/sup><\/p>\n\n\n\n

a2<\/sup> \u2013 b2<\/sup> = (a + b) (a – b)<\/p>\n\n\n\n

(x + a) (x + b) = x2<\/sup> + (a + b)x + ab<\/p>\n\n\n\n

These identities are useful in carrying out squares and products of algebraic expressions. They give alternative methods to calculate products of numbers and so on.<\/p>\n\n\n\n

Exercise 6.1 , 6.2 and 6.3<\/strong><\/span><\/h1>\n\n\n\n

1.Identify the terms, their coefficients for each of the following expressions.<\/p>\n\n\n\n

(i)5xyz2<\/sup> \u2013 3zy           (ii) 1 + x + x2                     <\/sup>(iii) 4x2<\/sup>y2<\/sup> \u2013 4x2<\/sup>y2<\/sup>z2<\/sup> + z2<\/sup><\/p>\n\n\n\n


(iv) 3 \u2013 pq + qr \u2013 rp    (v) x2 + y2 \u2013 xy       (vi) 0.3a \u2013 0.6ab + 0.5b<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

2. Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?<\/p>\n\n\n\n

x + y, 1000, x + x2<\/sup> + x3<\/sup> + x4<\/sup>, 7 + y + 5x, 2y \u2013 3y2<\/sup>, 2y \u2013 3y2<\/sup> + 4y3<\/sup>, 5x \u2013 4y + 3xy, 4z \u2013 15z2<\/sup>,<\/p>\n\n\n\n

 ab + bc + cd + da, pqr, p2<\/sup>q + pq2<\/sup>, 2p + 2q<\/p>\n\n\n\n

Solution:<\/span><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

3. Add the following:<\/span><\/p>\n\n\n\n

i)  ab \u2013 bc, bc \u2013 ca, ca \u2013 ab<\/p>\n\n\n\n

(ab \u2013 bc) + (bc \u2013 ca) + (ca \u2013 ab) (Adding all the terms)
= ab \u2013 bc + bc \u2013 ca + ca \u2013 ab
= (ab \u2013 ab) + (bc \u2013 bc) + (ca \u2013 ca) (Collecting the like terms together)
= 0 + 0 + 0
= 0<\/p>\n\n\n\n

(ii)  a \u2013 b + ab, b \u2013 c + bc, c \u2013 a + ac<\/p>\n\n\n\n

We have (a \u2013 b + ab) + (b \u2013 c + bc) + (c \u2013 a + ac) (Adding all the terms)
= a \u2013 b + ab + b \u2013 c + bc + c \u2013 a + ac
= (a \u2013 a) + (b \u2013 b) + (c \u2013 c) + ab + bc + ac (Collecting all the like terms together)
= 0 + 0 + 0 + ab + bc + ac
= ab + bc + ac<\/p>\n\n\n\n

(iii)  2p2<\/sup>q2<\/sup> \u2013 3pq + 4, 5 + 7pq \u2013 3p2<\/sup>q2<\/sup><\/p>\n\n\n\n

(Adding columnwise)<\/p>\n\n\n\n

By arranging the like terms in the same column, we have<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

iv)  l2<\/sup> + m2<\/sup>, m2<\/sup> + n2<\/sup>, n2<\/sup> + l2<\/sup>, 2lm + 2mn + nl  By arranging the like terms in the same column, we have <\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Thus, the sum of the given expressions is 2(l2<\/sup> + m2<\/sup> + n2<\/sup> + lm + mn + nl)<\/p>\n\n\n\n

 4. Subtract <\/p>\n\n\n\n

(a) Subtract 4a \u2013 7ab + 3b + 12 from 12a \u2013 9ab + 5b \u2013 3<\/p>\n\n\n\n

Arranging the like terms column-wise, we have<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

[Change the signs of all the terms of lower expressions and then add]<\/p>\n\n\n\n

(b) Arranging the like terms column-wise, we have<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

[Change the signs of all the terms of lower expressions and then add]<\/p>\n\n\n\n

(c) Arranging the like terms column-wise, we have<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

[Change the signs of all the terms of lower expressions and then add]<\/p>\n\n\n\n

The terms are p2<\/sup>q \u2013 7pq2<\/sup> + 8pq \u2013 18q + 5p + 20<\/p>\n\n\n\n

Exercise 9.2<\/span><\/strong><\/p>\n\n\n\n

1. Find the product of the following pairs of monomials.<\/span><\/p>\n\n\n\n

(i) 4, 7p  (ii) -4p, 7p      (iii) -4p, 7pq   (iv) 4p3<\/sup>, -3p     (v) 4p, 0<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

(i)4 \u00d7 7p = (4 \u00d7 7) \u00d7 p = 28p<\/p>\n\n\n\n


(ii) -4p \u00d7 7p = (-4 \u00d7 7) \u00d7 p \u00d7 p = -28p2<\/sup><\/p>\n\n\n\n


(iii) -4p \u00d7 7pq = (-4 \u00d7 7) \u00d7 p \u00d7 pq = -28p2<\/sup>q<\/p>\n\n\n\n


(iv) 4p3<\/sup> \u00d7 -3p = (4 \u00d7 -3) \u00d7 p3<\/sup> \u00d7 p = -12p4<\/sup><\/p>\n\n\n\n


(v) 4p x 0 = (4 \u00d7 0) \u00d7 p = 0 \u00d7 p = 0<\/p>\n\n\n\n

2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.<\/p>\n\n\n\n

(p, q); (10m, 5n); (20x2<\/sup>, 5y2<\/sup>); (4x, 3x2<\/sup>); (3mn, 4np)<\/p>\n\n\n\n

Solution:(i) Length = p units and breadth = q units<\/p>\n\n\n\n

Area of the rectangle = length \u00d7 breadth = p \u00d7 q = pq sq units<\/p>\n\n\n\n

(ii) Length = 10 m units, breadth = 5n units<\/p>\n\n\n\n

Area of the rectangle = length \u00d7 breadth = 10 m \u00d7 5 n = (10 \u00d7 5) \u00d7 m \u00d7 n = 50 mn sq units<\/p>\n\n\n\n

iii) Length = 20x2<\/sup> units, breadth = 5y2<\/sup> units<\/p>\n\n\n\n

Area of the rectangle = length \u00d7 breadth = 20x2<\/sup> \u00d7 5y2<\/sup> = (20 \u00d7 5) \u00d7 x2<\/sup> \u00d7 y2<\/sup> = 100x2<\/sup>y2<\/sup> sq units<\/p>\n\n\n\n

(iv) Length = 4x units, breadth = 3x2<\/sup> units<\/p>\n\n\n\n

Area of the rectangle = length \u00d7 breadth = 4x \u00d7 3x2<\/sup> = (4 \u00d7 3) \u00d7 x \u00d7 x2<\/sup> = 12x3<\/sup> sq units<\/p>\n\n\n\n

(v) Length = 3mn units, breadth = 4np units<\/p>\n\n\n\n

Area of the rectangle = length \u00d7 breadth = 3mn \u00d7 4np = (3 \u00d7 4) \u00d7 mn \u00d7 np = 12mn2<\/sup>p sq units<\/p>\n\n\n\n

3. Complete the table of Products.<\/span><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\n\n\n\n

4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively.<\/p>\n\n\n\n

i) 5a, 3a2<\/sup>, 7a4         <\/sup>(ii) 2p, 4q, 8r          (iii) xy, 2x2<\/sup>y, 2xy2                <\/sup>(iv) a, 2b, 3c<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

(i) Here, length = 5a, breadth = 3a2<\/sup>, height = 7a4<\/sup><\/p>\n\n\n\n

Volume of the box = l \u00d7 b \u00d7 h = 5a \u00d7 3a2<\/sup> \u00d7 7a4<\/sup> = 105 a7<\/sup> cu. units<\/p>\n\n\n\n

(ii) Here, length = 2p, breadth = 4q, height = 8r<\/p>\n\n\n\n

Volume of the box = l \u00d7 b \u00d7 h = 2p \u00d7 4q \u00d7 8r = 64pqr cu. units<\/p>\n\n\n\n

iii) Here, length = xy, breadth = 2x2<\/sup>y, height = 2xy2<\/sup><\/p>\n\n\n\n

Volume of the box = l \u00d7 b \u00d7 h = xy \u00d7 2x2<\/sup>y \u00d7 2xy2<\/sup> = (1 \u00d7 2 \u00d7 2) \u00d7 xy \u00d7 x2<\/sup>y \u00d7 xy2<\/sup> = 4x4<\/sup>y4<\/sup> cu. units<\/p>\n\n\n\n

iv) Here, length = a, breadth = 2b, height = 3c<\/p>\n\n\n\n

Volume of the box = length \u00d7 breadth \u00d7 height = a \u00d7 2b \u00d7 3c = (1 \u00d7 2 \u00d7 3)abc = 6 abc cu. units<\/p>\n\n\n\n

\n\n\n\n

5. Obtain the product of<\/span><\/p>\n\n\n\n

i) xy, yz, zx    (ii) a, -a2<\/sup>, a3    <\/sup>(iii) 2, 4y, 8y2<\/sup>, 16y3    <\/sup>(iv) a, 2b, 3c, 6abc     (v) m, -mn, mnp<\/p>\n\n\n\n

Solution:(i) xy \u00d7 yz \u00d7 zx = x2<\/sup>y2<\/sup>z2<\/sup><\/p>\n\n\n\n

(ii) a \u00d7 (-a2<\/sup>) \u00d7 a3<\/sup> = -a6<\/sup>(iii) 2 \u00d7 4y \u00d7 8y2<\/sup> \u00d7 16y3<\/sup> = (2 \u00d7 4 \u00d7 8 \u00d7 16) \u00d7 y \u00d7 y2<\/sup> \u00d7 y3<\/sup> = 1024y6<\/sup><\/p>\n\n\n\n

(iv) a \u00d7 2b \u00d7 3c \u00d7 6abc = (1 \u00d7 2 \u00d7 3 \u00d7 6) \u00d7 a \u00d7 b \u00d7 c \u00d7 abc = 36 a2<\/sup>b2<\/sup>c2<\/sup><\/p>\n\n\n\n

(iv) a \u00d7 2b \u00d7 3c \u00d7 6abc = (1 \u00d7 2 \u00d7 3 \u00d7 6) \u00d7 a \u00d7 b \u00d7 c \u00d7 abc = 36 a2<\/sup>b2<\/sup>c2<\/sup><\/p>\n\n\n\n

(v) m \u00d7 (-mn) \u00d7 mnp = [1 \u00d7 (-1) \u00d7 1 ]m \u00d7 mn \u00d7 mnp = -m3<\/sup>n2<\/sup>p<\/p>\n\n\n\n

\n\n\n\n

Exercise 9.3<\/span><\/strong><\/p>\n\n\n\n

1. Carry out the multiplication of the expressions in each of the following pairs:     <\/p>\n\n\n\n

(i) 4p, q + r  (ii) ab, a \u2013 b  (iii) a + b, 7a2<\/sup>b2  <\/sup>(iv) a2<\/sup> \u2013 9, 4a     (v) pq + qr + rp, 0<\/p>\n\n\n\n

Solution:i) 4p \u00d7 (q + r) = (4p \u00d7 q) + (4p \u00d7 r) = 4pq + 4pr<\/p>\n\n\n\n

ii) ab, a \u2013 b = ab \u00d7 (a \u2013 b) = (ab \u00d7 a) \u2013 (ab \u00d7 b) = a2<\/sup>b \u2013 ab2<\/sup><\/p>\n\n\n\n

iii) (a + b) \u00d7 7a2<\/sup>b2<\/sup> = (a \u00d7 7a2<\/sup>b2<\/sup>) + (b \u00d7 7a2<\/sup>b2<\/sup>) = 7a3<\/sup>b2<\/sup> + 7a2<\/sup>b3<\/sup>
(iv) (a2<\/sup> \u2013 9) \u00d7 4a = (a2<\/sup> \u00d7 4a) \u2013 (9 \u00d7 4a) = 4a3<\/sup> \u2013 36a
(v) (pq + qr + rp) \u00d7 0 = 0    [\u2235 Any number multiplied by 0 is = 0]<\/p>\n\n\n\n

\n\n\n\n

2. Complete the table.<\/span><\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Solution:
(i) a \u00d7 (b + c + d) = (a \u00d7 b) + (a \u00d7 c) + (a \u00d7 d) = ab + ac + ad<\/p>\n\n\n\n


(ii) (x + y \u2013 5) (5xy) = (x \u00d7 5xy) + (y \u00d7 5xy) \u2013 (5 \u00d7 5xy) = 5x2<\/sup>y + 5xy2<\/sup> \u2013 25xy<\/p>\n\n\n\n


(iii) p \u00d7 (6p2<\/sup> \u2013 7p + 5) = (p \u00d7 6p2<\/sup>) \u2013 (p \u00d7 7p) + (p \u00d7 5) = 6p3<\/sup> \u2013 7p2<\/sup> + 5p<\/p>\n\n\n\n


(iv) 4p2<\/sup>q2<\/sup> \u00d7 (p2<\/sup> \u2013 q2<\/sup>) = 4p2<\/sup>q2<\/sup> \u00d7 p2<\/sup> \u2013 4p2<\/sup>q2<\/sup> \u00d7 q2<\/sup> = 4p4<\/sup>q2<\/sup> \u2013 4p2<\/sup>q4<\/sup><\/p>\n\n\n\n

\n\n\n\n


(v) (a + b + c) \u00d7 (abc) = (a \u00d7 abc) + (b \u00d7 abc) + (c \u00d7 abc) = a2<\/sup>bc + ab2<\/sup>c + abc2<\/sup><\/p>\n\n\n\n

3. Find the products.<\/span><\/p>\n\n\n\n

\n\n\n\n

4. (a) Simplify: 3x(4x \u2013 5) + 3 and find its values for (i) x = 3 (ii) x = 1\/2.<\/p>\n\n\n\n


(b) Simplify: a(a2<\/sup> + a + 1) + 5 and find its value for (i) a = 0 (ii) a = 1 (iii) a = -1<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

(a) We have 3x(4x \u2013 5) + 3 = 4x \u00d7 3x \u2013 5 \u00d7 3x + 3 = 12x2<\/sup> \u2013 15x + 3<\/p>\n\n\n\n

\n\n\n\n

i) For x = 3, we have
12 \u00d7 (3)2<\/sup> \u2013 15 \u00d7 3 + 3 = 12 \u00d7 9 \u2013 45 + 3 = 108 \u2013 42 = 66<\/p>\n\n\n\n\n\n

(b) We have a(a2<\/sup> + a + 1) + 5
                        = (a2<\/sup> \u00d7 a) + (a \u00d7 a) + (1 \u00d7 a) + 5
= a3<\/sup> + a2<\/sup> + a + 5<\/p>\n\n\n\n

\n\n\n\n


(i) For a = 0, we have
             = (0)3<\/sup> + (0)2<\/sup> + (0) + 5 = 5<\/p>\n\n\n\n

\n\n\n\n


(ii) For a = 1, we have
        = (1)3<\/sup> + (1)2<\/sup> + (1) + 5<\/p>\n\n\n\n

     = 1 + 1 + 1 + 5 = 8<\/p>\n\n\n\n

\n\n\n\n


(iii) For a = -1, we have
           = (-1)3<\/sup> + (-1)2<\/sup> + (-1) + 5<\/p>\n\n\n\n

\u00a0\u00a0\u00a0\u00a0 = -1 + 1 \u2013 1 + 5 = 4.<\/p>\n\n\n\n

\n\n\n\n

5. (a) Add: p(p \u2013 q), q(q \u2013 r) and r(r \u2013 p)
(b) Add: 2x(z \u2013 x \u2013 y) and 2y(z \u2013 y \u2013 x)<\/p>\n\n\n\n

(c) Subtract: 3l(l \u2013 4m + 5n) from 4l(10n \u2013 3m + 2l)
(d) Subtract: 3a(a + b + c) \u2013 2b(a \u2013 b + c) from 4c(-a + b + c)<\/p>\n\n\n\n

Solution:<\/span><\/strong><\/p>\n\n\n\n

(a) p(p \u2013 q) + q(q \u2013 r) + r(r \u2013 p)
= (p \u00d7 p) \u2013 (p \u00d7 q) + (q \u00d7 q) \u2013 (q \u00d7 r) + (r \u00d7 r) \u2013 (r \u00d7 p)
= p2<\/sup>\u00a0\u2013 pq + q2<\/sup>\u00a0\u2013 qr + r2<\/sup>\u00a0\u2013 rp
= p2<\/sup>\u00a0+ q2<\/sup>\u00a0+ r2<\/sup>\u00a0\u2013 pq \u2013 qr \u2013 rp<\/p>\n\n\n\n

\n\n\n\n

(b) 2x(z \u2013 x \u2013 y) + 2y(z \u2013 y \u2013 x)
= (2x \u00d7 z) \u2013 (2x \u00d7 x) \u2013 (2x \u00d7 y) + (2y \u00d7 z) \u2013 (2y \u00d7 y) \u2013 (2y \u00d7 x)
= 2xz \u2013 2x2<\/sup>\u00a0\u2013 2xy + 2yz \u2013 2y2<\/sup>\u00a0\u2013 2xy
= -2x2<\/sup>\u00a0\u2013 2y2<\/sup>\u00a0+ 2xz + 2yz \u2013 4xy
= -2x2<\/sup>\u00a0\u2013 2y2<\/sup>\u00a0\u2013 4xy + 2yz + 2xz<\/p>\n\n\n\n

\n\n\n\n

(c) 4l(10n \u2013 3m + 2l) \u2013 3l(l \u2013 4m + 5n)
= (4l \u00d7 10n) \u2013 (4l \u00d7 3m) + (4l \u00d7 2l) \u2013 (3l \u00d7 l) \u2013 (3l \u00d7 -4m) \u2013 (3l \u00d7 5n)
= 40ln \u2013 12lm + 8l2<\/sup>\u00a0\u2013 3l2<\/sup>\u00a0+ 12lm \u2013 15ln
= (40ln \u2013 15ln) + (-12lm + 12lm) + (8l2<\/sup>\u00a0\u2013 3l2<\/sup>)
= 25ln + 0 + 5l2<\/sup>
= 25ln + 5l2<\/sup>
= 5l2<\/sup>\u00a0+ 25ln<\/p>\n\n\n\n

\n\n\n\n

(d) [4c(-a + b + c)] \u2013 [3a(a + b + c) \u2013 2b(a \u2013 b + c)]
= (-4ac + 4bc + 4c2<\/sup>) \u2013 (3a2<\/sup> + 3ab + 3ac \u2013 2ab + 2b2<\/sup> \u2013 2bc)
= -4ac + 4bc + 4c2<\/sup> \u2013 3a2<\/sup> \u2013 3ab \u2013 3ac + 2ab \u2013 2b2<\/sup> + 2bc
= -3a2<\/sup> \u2013 2b2<\/sup> + 4c2<\/sup> \u2013 ab + 6bc \u2013 7ac<\/p>\n\n\n\n

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