class 8 algebraic expressions and identities …...class 8 algebraic expressions and identities...

Class 8 A lgebraic Expressions And Identities Exercise 6.1

RD Sharma Solutions Class 8 Chapter 6 Exercise 6.1

Q1: Identify the terms, their coefficients for each of the following expressions:

(i) 7x2yz — 5 xy

(ii) x2 + x + 1

(iii) 3a;21/2 - 5x2y2z2 + z2

(iv) 9 - ab + be - ca

(v ) f + | - a 6

(vi) 0.2x - 0.3xy + 0.5y

Solution:

Definitions:

A term in an algebraic expression can be a constant, a variable or a product o f constants and variables separated by the signs o f addition (+) or subtraction (-). Examples: 27, x, xyz, ^ x2yz etc.

The number factor of the term is called its coefficient.

(i) The expression 7x2yz — 5xy consists o f two terms, i.e., 7x2yz and — 5xy.

The coefficient o f 7x2yz is 7 and the coefficient of -5xy. is -5.

(ii) The expression x2 + x +1 consists o f three terms, i.e., x2, x and 1.

The coefficient o f each term is 1.

(iii) The expression ZuPy2 - 5x2y2z2 + z2 consists of three terms, i.e., 3x2y2, - 5x2y2z2 and z2.

The coefficient o f ZuPy2 is 3.

The coefficient o f - 5x2y2z2 is -5 and the coefficient o f z2 is 1.

(iv) The expression 9 - ab + be - ca consists of four terms -9, -ab, be and - ca.

The coefficient o f the term 9 is 9.

The coefficient o f -ab is -1.

The coefficient o f be is 1, and the coefficient o f -ca is -1.

(v) The expression -| + -| — ab consists of three term s, i.e., and — ab.

The coefficient o f j is j .

The coefficient o f j is and the coefficient o f-ab is -1.

(vi) The expression 0.2x - 0.3xy + 0.5y consists o f three terms, i.e., 0.2x, -0.3xy and 0.5y.

The coefficient o f 0.2x is 0.2.

The coefficient o f -0.3xy is -0.3, and the coefficient o f 0.5y is 0.5.

Q2) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any category?

( i) x + y

(ii) 1000

(iii) x + x2 + x3 + x4

(iv) 7 + a + 5b

(v) 26 - 362

(vi) 2y - 3y2 + 4y3

(vii) 5x - 4y + 3x

(viii) 4a — 15a2

(ix) xy + yz + zt + tx

(x) pqr

(xi )p2q + pq2

(xii) 2p + 2q

Solution:

Definitions:

A polynomial is monomial if it has exactly one term. It is called binomial if it has exactly two non-zero terms. A polynomial is a trinomial if it has exactly three non-zero terms.

(i) The polynomial x + y has exactly two non zero terms, i.e., x and y. Therefore, it is a binomial.

(ii) The polynomial 1000 has exactly one term, i.e., 1000. Therefore, it is a monomial.

(iii) The polynomial x + x2 + x3 + x4 has exactly four terms, i.e., x, x2, x3 and x4. Therefore, it doesn't belong to any of the categories.

(iv) The polynomial 7 + a + 5b has exactly three terms, i.e., 7, a and 5b. Therefore, it is a trinomial.

(v) The polynomial 2b - 3b2 has exactly two terms, i.e., 2b and -3b2. Therefore, it is a binomial.

(vi) The polynomial 2y - 3y2 + 4y3 has exactly three terms, i.e., 2y, 3y2 and 4y3.. Therefore, it is a trinomial.

(vii) The polynomial 5x - 4y + 3x has exactly three terms, i.e., 5x, -4y and 3x. Therefore, it is a trinomial.

(viii) The polynomial 4a - 15a2 has exactly two terms, i.e., 4a and -15a2. Therefore, it is a binomial.

(ix) The polynomial xy + yz + zt + tx has exactly four terms xy, yz, zt and tx. Therefore, it doesn't belong to any of the categories.

(x) The polynomial pqr has exactly one term, i.e., pqr. Therefore, it is a monomial.

(xi) The polynomial p2q + pq2 has exactly two terms, i.e., p2q and pq2. Therefore, it is a binomial.

(xii) The polynomial 2p+ 2q has two terms, i.e., 2p and 2q. Therefore, it is a binomial.



Q.1: Add the following algebraic expressions:

(i) 3o26, —4<j2&, 9a26

00 3 °) 5 5 a

(iii) Axy2- 7x2y, 12x2y-6xy2, —3x2y + 5xy2

+ - f y - f x - f x y

+ fa;3 + fa;2-* + f , f*2-f*-2Solution:

(i) To add the like terms, we proceed as follows:

3a2b + (-4a2b) + 9a2b

= 3a2b - 4a2b + 9a2b (Distributive Law)

= 8a2b

(ii) To add like terms, we proceed as follows:

2 „ . 3 , / 6 \3a + 5 “ + ( — 5 a)

2 , 3 6- 3 O + j a - gO

= (- | + | - f ) a (Distributive Law)

= 15 a

(iii) To add, we proceed as follows:

(4xy2- 7x2y) + (12 x2y) + (—6xy2) + (—3x2y + 5xy2)

= 4 xy2- 7x2y + 12 x2y - Qxy2- 3x2y + 5a:y2

= 4xy2- 6xy2 + 5xy2- 7x2y + 12x2y - 3x2y (Collecting like terms)

= 3xy2 + 2x2y (Combining like terms)

(iv) To add, we proceed as follows:

( t « - lb + f c ) + ( f a - lb + f c) + ( f a + f b- | c )

= f a - f 6 + f c + | o - | 6 + | c + | o + f ^ f c

= | a + | a + | a_ | ^ I 6 + | 6 + | c + l ^ | c

(Collecting like terms)

= f f a - \b + | f c (Combining like terms)

(v) To add, we proceed as follows:

( f x y + f y + f x ) + ( - f y - f * ~ f * » )

= t xv + f y + T x~ t v - T X~ T XV

= f x y - f x y + f y - f y + f x- f x

= f f a;y- f f y - f f a; (Combining like terms)


(vi) To add, we proceed as follows:

(la:3- fa:2 + f ) + (fa:3 + \x2- x + f ) + (fa:2- fa:-2)

= I * 3- i * 2 + ! + f * 3 + i x2~ x + 1 + ! * 2- f * - 2

= l x 3 + fa :3- fa ;2 + fa :2 + f a:2- a : - fa : + f + f - 2


= 5a;3 + f x 2- fa : (Combining like terms)

Q2) Subtract:

( i) -5xyfrom 12xy

(ii) 2a2 from -7a 2

(iii) 2a - b from 3a - 5b

(iv) 2a;3- 4a;2 + 3a: + 5 from 4a:3 + x2 + x + 6

(v) f y3~ f y2~5 from f y3 + j y 2 + y-2

(vi) \x~ f y~ \z from \x + f y -\ z

(vii) x2y - f a :y2 + f xy from f x2y + f xy2- \xy

(viii) y - fb c + f ac from f be- f ac

Solution:

( i ) 1 2 x y -(-5 x y )

= 12xy + 5xy = 17xy

(ii) -7 a 2 - (2a2)

= -7 a 2 - 2a2 = -9 a 2

(iii) (3a - 5b) - (2a - b)

= (3a - 5b) - 2a + b

= 3a - 5b - 2a + b

= 3a - 2a - 5b + b = a - 4b

(iv) (4a;3 + x2 + x + 6) - (2a;3- 4a;2 + 3x + 5)

= 4a;3 + a:2 + x + 6- 2a;3 + 4a;2- 3a;- 5

= 4a;3- 2a;3 + a;2 + 4a:2 + x - 3x + 6- 5 (Collecting like terms)

= 2a;3 + 5a;2-2 a : + 1 (Combining like terms)

(v ) ( l 2/3 + f y 2 + ^ 2 ) - ( f 2 / 3- f 2 / 2- 5 )

= \y3 + j V 2 + y -2 - \ y 3+ 2j y 2 + b

= j y 3- 1 y3 + j y 2 + j y 2 + y-2 + 5 (Collecting like terms)

= — \y3 + y2 + y + 3 (Combining like terms)

W i h + \ y - h ) - { \ x - \ y - \ z )

= l x + \ y - ± z - \ x + \ y + \ z

= f a;- f a: + f y + f y - f z + j z (Collecting like terms)

= —f a: + y j / + y - z (Combining like terms)

(vii) ( f x2y + |xy2- \xy)~ (x2y - \xy2 + f xy)

= \x2y + jxy2- \ x y - x 2y + ±xy2- ±xy

= | x2y -x2y + | xy2 + \xy2- \ xy- jxy (Collecting like terms)

= — j x 2y + | | xy2- | xy (Combining like terms)

(viii) ( f 6^ | ac ) - ( f - f 6c + | aC)

_ 3 7 4 ab - 35 7 6- g- 0C~ -g-QC— Tj—r yOC- jflC

= f 6c + y be- \ac- f a c - y (Collecting like terms)

= be- 2ac- y (Combining like terms)

Q3) Take away:

(i) fa :2- f a:3 + f + \x from f a:2 + fa ; + f

(ii) f a :3 + f a :2 + i a : + f

m ^ + \y2 + \ y + \ f r o m \ - \ y 2

(iv) f ac- jab + f be from jab- f ac- f 6c

Solution:

(i) The difference is given by:

/x3 5 2 _i_ 3 . 1 \ / 6 2 4 3 , 5 , 3 \( 3 2 X 5®"*" 4 ) ( 5 * 5 X 6 “*” 2® )

_ x3 5 2 1 3 . 1 6 2 1 4 3 5 3- 3 235 531 4 5 * 5 * 6 2 X

= — + —a:3— - x 2— —x2 4- - x — —x + - — — 3 - r 5a. 5 a , - r 5 a, 2<c t 4 6

1 ( ^ i i r ' ) a ;3 + (■ 1037 2_ 9 _ J_

15 * 10 x 10 12


)a;2 + ( ^ a : ) + ( ^ )

(Combining like terms)

(ii) The difference is given by:


( M - t )- (^ 3 + ! * 2 + t + !)- 1 _ * I _ I 7.3- - T 2- - - -“ 2 3 5 4 5 A 2 2

- Z _ £ _ £ _ £ _ 3x2 7a:3” 2 2 3 2 5 5 4

. _5x_ 4a;2 7a;36 5 4

(iii) The difference is given by:

( i - j v 2) - ( T + j y 2 + \ y + \ )

(Combining like terms)

I - V - -3 3 " 37 2 1 1» r - i V - a

■ 1 _ 1 _ » _ 5 2_ 7 2_tr3 2 2 3 y 3 » 3 (Collecting like terms)

r -5 -7 ,

-4y2- - (Combining like terms)

(iv) The difference is given by:

( |a 6 - ^ a c - |6 c ) - ( |a c - y a & + |-6c)

— y db y (ic 16c - g'nc + yrz6 y be

= | a 6 + - |a 6- -j-ac- - |a c - | - 6c - | 6c (Collecting like terms)

■ -2 1 -8 ) a c + ( ^ l ) 6c= ( ^ ) a & + ( ^ 12 v 6

= f j-o b - | | oc_ 6c (Combining like terms)

Q4: Subtract 3x - 4y - 7z from the sum o f x - 3y + 2z and -4 x + 9y -1 1 z

Solution:

First add the expressions x - 3y + 2z and -4 x + 9y - 11 z we get:

(x — 3y + 2 z ) + (-4x + 9y - 11z)

= x - 3y + 2z - 4x + 9y - 11z

= x - 4x - 3y + 9y + 2z - 11 z (Collecting like terms)

= -3 x + 6y - 9z (Combining like terms)

Now, Subtracting the expression 3x - 4y - 7z from the above sum, we get:

(-3x + 6y - 9z) - (3x - 4y - 7z)

= -3 x + 6y - 9z - 3x + 4y + 7z

= -3 x - 3x + 6y + 4y - 9z + 7z (Collecting like terms)

= - 6x + 1 0 y -2 z (Combining like terms)

Thus, the answer is - 6x + 10y - 2z.

Q5) Subtract the sum o f 31 - 4m - 7n2 and 21 + 3m - 4n2 from the sum o f 91 + 2m - 3n2 and -31 + m + 4n2.

Solution:

We have to subtract the sum of (31 - 4m - 7n2) and (21 + 3m - 4n2) from the sum of (91 + 2m - 3n2) and (-31 + m + 4n2)

{(91 + 2m - 3n2) + (-31 + m + 4n2)} - {(31 - 4m - 7n2) + (21 + 3m - 4n2)}

= (91 - 31 + 2m + m - 3n2 + 4n2) - (31 + 21 - 4m + 3m - 7n2 - 4n2)

= (61 + 3m + n2) - (51 - m - 11 n2) (Combining like terms inside the parenthesis)

= 61 + 3m + n2 - 51 + m + 11 n2

= 61 - 51 + 3m + m + n2 + 11 n2 (Collecting like terms)

= I + 4m + 12n2 (Combining like terms)

Thus, the required solution is I + 4m + 12n2.

Q6) Subtract the sum 2x - x2 + 5 and -4 x - 3 + 7x2 from 5.

Solution:

We have to subtract the sum of (2x - x2 + 5) and (-4x - 3 + 7x2) from 5.

5 - {(2x - x2 + 5) + (-4x - 3 + 7x2)}

= 5 - (2x - 4x - x2 + 7x2 + 5 - 3)

= 5 - 2x + 4x + x2 - 7x2 - 5 + 3

= 5 - 5 + 3 - 2x + 4x + x2 - 7x2 (Collecting like terms)

= 3 + 2x - 6x2 (Combining like terms)

Thus, the answer is 3 + 2x - 6x2.

Q7) S im plify each o f the follow ing:

(i) x2- 3a; + 5 - f (3a;2- 5a; + 7)

(ii) [5 - 3x + 2y - (2x - y)] - (3x - 7y + 9)

(in) Y x2y~ i xy2 + \xy~ i j y 2* + i j y * 2 + 7 *y

(iv) ( f y2- j y + 11)- i j y - 3 + 2 y2) - ( f y - f y2 + 2)

(v) — f a 2&2c + f a 62c - f abc2- fc & 2a2 + f c 62a - ^<?ab + jca2b.

Solution:

(I) a;2- 3a; + 5 - 1 (3a;2- 5a; + 7)

= a:2-3 a ; + 5 - ^ + f - f

= a;2- — 3a; + -y- + 5 - f (Collecting like terms)

- ~xl £ j _ i 2 2 2

Thus, the answer is ^ — 7 + f •

(ii) [5 - 3x + 2y - (2x - y)] - (3x - 7y + 9)

= [5 - 3x + 2y - 2x + y] - (3x - 7y + 9)

= [5 - 5x + 3y] - (3x - 7y + 9)

= 5 - 5x + 3y - 3x + 7y - 9

= 5 - 9 - 5x - 3x + 3y + 7y = -4 - 8x + lOy

(iii) Y x2y~ \ xy2 + \xy~ T iy2x + T5 yx2 + 1* 2/

= Y x2y + t * ? /2 - ^ 2/2* + l xy + 7 * 2/

= ( i ^ ) a ; 2J/ + ( ^ ) a ;2,2 + ( ^ ) x ?/

= f j j f x2y- 1 | xy2 + |xy (Combining like terms)

(iv) ( i t / 2- f t , + 1 1 )- ( i j r 3 + 2a,2) - ( f y - f t ,2 + 2)

= \l/2- y 2/ + 1 1 - y f / + 3 -2 t /2- f t , + f t , 2- 2

= I y2 + \y2- 2 y 2- f y - i y - f y + l l + 3 - 2



= ( i z | ± a )y 2 + ( ^ i )y + i 2

= -y 2 - 7y + 12 (Combining like terms)

(v) — f a ^ c + f o 62c - fafec2- f c 62a2 + f c 62o - j(?ab + f c a 26

= —f a 262c - f cb2a 2 + f ob2c + f cb2a- f o 6c2- f c2a6 + f ca2b (Collecting like terms)

= ( ^ lr - W b2c + ( n r 1 )ab2c + ( ^ ) ( ? a b + f c a 26

= —j^a2b2c + i-a 62c - f f a&c2 + f a2bc (Combining like terms)



Find each o f the follow ing products: (1-8)

Q1) 5a;2 x 4a;3

Solution:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices. However, use of these laws subject to their applicability in the given expressions.

In the present problem, to perform the multiplication, we can proceed as follows:

5a:2 x 4a:3

= (5 x 4) x (a:2 x a:3)

= 20a:5 (•_■ am xan = am+n)

Thus, the answer is 20a:5.

Q2) - 3 a 2 x 464

Solution:

To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, am x a " applicable.

We have:

- 3 a 2 x 464

= ( - 3 x 4) x (a2 x 64)

= —12a264

Thus, the answer is — 12a264.

Q3) ( —5* 3/) x (~3x2yz)

Solution:

To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, am x an applicable.

We have:

( —5* 3/) x (—3x2yz)

= [(—5) x ( —3)] x (x x x2) x (y x y) x z

= 15 x ( * 1+2) x ( i/1+1) x z

=15x3y2z

Thus, the answer is 15x3y2z.

Q4) \xy x | x2yz2

Solution:

To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, that is, an

We have:

\xy x | x2yz2

= ( | x f ) x ( * x a;2) x {y x y) x z2

= ( - j x | ) x ( * 1+2) x (3/1+1) x z2

= \x3y2z2

Thus, the answer is j x 3y2z2.

Q5) (—jxy2z) x ( f - x2yz2)

Solution:

To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, that is, an

We have:

{ - \ x y 2z) x ( f x2yz2)

= ( - } X f ) x ( * x X2) x ( y x y ) x ( z x z2)

= ( - I x f ) x ( * W ) x ( » w ) x ( * w )

= - % x 3y3z3

Thus, the answer is — j^x3y3z3.

Q6) ( - | \x3z ) x { - ^ x z 2y)

Solution:

To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, that is, a'

am+n, wherever

am+n, wherever

x an = am+n

x an = am+n

x an — am+n

We have:

( - 1 * * * ) x ( - f * * v )

^ ( - i x C - l D l x ^ x ^ x j /

= T 5 * V 3

Thus, the answer is -^xiyz3.

Q7) ( - ^ a 262) x ( f a W )

Solution:

To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, that is, am x an = am+n

We have:

( - ± a W ) x ( f a W )

= [ ( - ^ - x ( | ) ] x (a2 x a3) x (62 x 62) x c2

= [ ( - £ x (§ ) ] x (a2+3) x (62+2) x c2

= - ± a 564c2

Thus, the answer is — a ^ c 2.

Q8) ( - 7 xy) x (\x2yz)

Solution:


( - 7 xy) x ( j x 2yz)

= ( - 7 x j ) x (x x x2) x (y x y) x z

= ( - 7 x | ) x (a;1+2) x ( j /1+1) x z

= - { x * y 2z

Thus, the answer is —jx 3y2z.

Find each o f the follow ing products: (9-17)

Q9) (7ab) x ( —5ab2c) x (6abc2)

Solution:

To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, that is, am x an — am+n

We have:

(7ab) x ( —5a62c) x (6o6c2)

= [7 x ( - 5 ) x 6] x (a x a x a) x (b x b2 x b) x (c x c2)

= [7 x ( - 5 ) x 6] x (a 1+1+1) x (&1+2+1) x (c1+2)

= —210a364c3

Thus, the answer is —210o364c3.

Q10) ( - 5 a ) x ( - 1 0 a 2) x ( - 2 a 3)

Solution:


We have:

( —5a) x ( —10a2) x ( —2a3)

= [ ( - 5 ) x ( - 1 0 ) x ( - 2 ) ] x (a x a2 x a3)

= [ ( - 5 ) x ( - 1 0 ) x ( - 2 ) ] x (a 1+2+3)

= - 100a6

Thus, the answer is —100a6.

Q11) ( -4 a ;2) x ( -6xy2) x ( - 3 yz2)

Solution:


We have:

( —4a;2) x (—6xy2) x ( —3yz2)

= [ ( - 4 ) x ( - 6 ) x ( - 3 ) ] x (a;2 x a;) x (y2 x y) x z2

= [ ( - 4 ) x ( - 6 ) x ( - 3 ) ] x (x2+1) x (y2+1) x z2

= - 72x3i/3z2

Thus, the answer is —72x3y3z2.

Q 1 2 ) ( - f o 4) x ( —f o 26) x ( —^ f c 2)

Solution:


We have:

( - f a 4) x ( —-| a2b) x ( - f b2)

[ ( - f ) x ( - ! ) x ( —H ) ] x (a4 x a 2) x ( 6 x 62)

= H f * f x f )] x (a4+2) x (b4+2)

= - f a 6b35

Thus, the answer is — f a 6b3.5

Q 1 3 )( |a 6 2) x ( ^ a c 26 ) x ( - | a 2c)

Solution:


We have:

( f a b 2) x (y -a f^ b ) x ( — f a 2c)

= [ ( f ) x ( y - ) x ( - f )] x (a x a x a2) x (b2 x b) x (c2 x c)

= I? x y x ( - f )] x (a 1+1+2) x (b2+1) x (c2+1)

= - a 4b3c3

Thus, the answer is - a4b3c3.

Q14) ( f -u2vw) x ( —5uvw2) x (^v2wu)

Solution:


We have:

(^u2vw) x (—buvw2) x ( j v 2wu)

= [ ( f ) x ( —5) x ( f )] x ( it2 x u x u) x (v x v x v2) x (w x w2 x w)

= [ | X ( —5) X f ] x ( i t2+1+1) x (i>1+1+2) x ( i t ; ^ 1)

= - f t t V i a 4

Thus, the answer is - ™uiv4“wi.

Q15) (0.5a;) x (^xy2z4) x (24x2yz)

Solution:


We have:

(0.5a;) x ( f xy2z4) x (24x2yz)

= [0.5a; x | x 24] x ( i x i x x2) x (y2 xy ) x (z4 x z)

= [0.5a; x f x 24] x (a:1+1+2) x (y 2+1) x (z4+1)

= 4xiy3z5

Thus, the answer is 4xiy3z5.

Q16) ( f pq2) x ( ~ jP 2r) x (16p V r 2)

Solution:


We have:

( f M 2) x { - { i P r ) x (16p2q2r2)

= [ f x ( —f ) x 16] x {p x p 2 x p2) x (q2 x q2) x ( r x r 2)

= [ j X ( - i ) x 16] x (p1+2+2) x (g2+2) x ( r 1+2)

= - f p 5« V

Thus, the answer is — ™p5qir3.

Q17) (2.3xy) x (0.1a:) x (0.16)

Solution:

To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, that is, om x an = am+n

We have:

(2.3xy) x (0.1a:) x (0.16)

= (2.3 x 0.1 x 0.16) x (x x x) xy

= (2.3 x 0.1 x 0.16) x (a:1+1) x y

= 0.0368a;2?/

Thus, the answer is 0.0368a:2?/.

Express each o f the following products as a monomials and verify the result in each case for x = 1: (18-26)

Q18) (3a:) x (4a;) x ( —5a;)

Solution:

We have to find the product o f the expression in order to express it as a monomial.


We have:

(3a:) x (4a:) x ( —5a;)

= (3 x 4 x ( - 5 ) ) x (x xx x x)

= (3 x 4 x ( - 5 ) ) x (a:1+1+1)

= —60a:3

Substituting x = 1 in LHS, we get:

LHS = (3a:) x (4a:) x ( —5a:)

= (3 x 1) x (4 x 1) x ( - 5 x 1)

= -60

Putting x = 1 in RHS, we get:

RHS = -6 0 a :3

= —60(1)3

= - 6 0 x 1

= -60

Since, LHS = RHS for x =1; therefore, the result is correct.

Thus, the answer is —60a;3.

Q19) (4a:2) x ( -3 a :) x ( fa :3)

Solution:



We have:

(4a:2) x ( —3a;) x ( fa ;3)

= (4 x ( - 3 ) x f ) x ( x 2 x a: x x 3)

= (4 x ( - 3 ) x f ) x ( x 2+1+3)

_ 5 X

(4a;2) x ( —3a;) x ( fa ;3)

" 5 X


LHS = (4a;2) x (-3 a ;) x ( fa ;3)

= (4 x l 2) x ( - 3 x 1) x ( f x l 3)

= 4 x ( —3) x f

- _ i i” 5

Putting x = 1 in RHS, we get:

RHS = - y X 6

= - f X l 6

-" 5

Since, LHS = RHS for x = 1; therefore, the result is correct

Thus, the answer is — - x 6.

Q20) (5a:4) x (a:2) 3 x (2a:)2

Solution:


To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, that is, am x an = om+"

We have:

(5a;4) x (a;2)3 x (2a;)2

= (5a;4) x (a;6) x (22 x a:2)

= (5 x 22) x (a:4 x i s x x 2)

= (5 x 22) x ( x 4+6+2)

= 20x 12

(5 x4) x ( x 2) 3 x (2 x ) 2 = 20 x12


LHS = (5 x4) x (x 2) 3 x (2 x ) 2

= (5 x 1) x ( l 6) x (2) 2

= 5 x 1 x 4

= 20

Put x = 1 in RHS, we get:

RHS = 20 x12

= 20 x l 12

= 20 x 1

= 20


Thus, the answer is 20x12

Q21) (x 2) 3 x (2 x ) x ( - 4 x ) x (5)

Solution:



We have:

(a;2)3 x (2x) x ( —4 * ) x 5

= (a;6) x (2a;) x ( —4a;) x 5

= (2 x ( —4) x 5) x (a:6 x x x x)

= (2 x ( - 4 ) x 5) x (a:6+1+1)

= -4 0 a :8

(a:2) 3 x (2a:) x ( —4a;) x 5 = —40a;8


LHS = (a:2)3 x (2a:) x (-4 a :) x 5

= ( l 2)3 x (2 x 1) x ( - 4 x 1) x 5

= l 6 x 2 x ( - 4 ) x 5

= -40

Put x = 1 in RHS, we get:

RHS = -4 0 a ;8

= - 4 0 X l 8

= - 4 0 x 1

= -40


Thus, the answer is —40a;8

Q22) Write down the product o f -8 x2y6 and — 20xy. Verify the product for x = 2.5, y = 1.

Solution:


We have:

( —8x2y6) x ( —20xy)

= [ ( - 8) x ( - 20)] x (a:2 x * ) x (y6 x y)

= [ ( - 8) x ( - 20)] x (a;2+1) x (y6+1)

= —160 x3y7

( —8x2y6) x ( —20xy) = —160a:3j / 7

Substituting x = 2.5 and y = 1 in LHS, we get:

LHS = ( - 8x2y6) x ( - 2 0 xy)

= ( —8(2 .5 )2( l ) 6) x ( —20(2 .5 )(1 ))

= ( —8(6 .2 5 )(1 )) x ( 20 (2 .5 )(1 ))

= ( - 5 0 ) x ( - 5 0 )

= 2500

Substituting x = 2.5 and y = 1 in RHS, we get:

RHS = -160a ; V

= —160(2.5 )3(1 )7

= —160(15.625)(1)

= 2500

Because LHS is equal to RHS, the result is correct.

Thus, the answer is — 160x 3j/7

Q23) Evaluate (3.2x6y3) x (2.1 x2y2) when x = 1 and y = 0.5.

Solution:


We have:

(3.2x6y3) x (2.1a:2y 2)

= (3.2 x 2.1) x (x8 x x2) x (y3 x y2)

= (3.2 x 2.1) x (a:6+2) x ( j/3+2)

= 6.72a; V

.-. (3 .2 a :V ) x (2.1 x2y2) = 6.72a:8i/5

Substituting x = 1 and y = 0.5 in the result, we get:

6 .7 2 a :V

= 6 .72(1 )8(0.5 )s

= 6.72 x 1 x 0.03125

= 0.21Thus, the answer is 0.21.

Q24) Find the value o f (5a;6) x ( —1.5x2y3) x ( —12xy2) when x = 1 ,y = 0.5.

Solution:


We have:

(5a;6) x (—1.5x2y3) x ( —12xy2)

= [5 x ( —1.5) x ( —12)] x (x6 x x2 x x) x (y3 x y2)

= [5 x ( - 1 .5 ) x ( - 1 2 ) ] x (a;6+2+1) x (y3+2)

= 90x9y5

(5a:6) x ( —1.5x2y3) x ( —12xy2) = 90xsy5

Substituting x = 1 and y = 0.5 in the result, we get:

90a;9 j /5

= 90 (1 )9(0 .5 )5

= 90 x 1 x 0.03125

= 2.8125

Thus, the answer is 2.8125.

Q25) Evaluate when (2.3a562) x (1.2a262) when a = 1 and b = 0.5.

Solution:


We have:

(2.3a562) x (1.2o262)

= (2.3 x 1.2) x (a5 x o2) x {b2 x b2)

= (2.3 x 1.2) x (a5+2) x (62+2)

= 2.76a764

.-. (2.3a562) x (1.2a262) = 2.76a764

Substituting a = 1 and b = 0.5 in the result, we get:

2.76a764

= 2 .76(1 )7(0 .5 )4

= 2.76 x 1 x 0.0625

= 0.1725

Thus, the answer is 0.1725.

Q26) Evaluate fo r ( —8a;2y 6) x ( —20a-.y) x = 2.5 and y = 1.

Solution:


We have:

( - 8 x2y6) x ( —20xy)

= [ ( - 8) x ( - 20)] x (a:2 x x) x (y6 x y)

= [ ( - 8) x ( - 20)] x (x2+1) x (y 6+1)

= 160a;32/ 7

( - 8x2y6) x ( —20xy) = 160x3y*

Substituting x = 2.5 and y =1 in the result, we get:

160a:3 j /7

= 160(2.5 )3(1 )7

= 160 x 15.625

= 2500

Thus, the answer is 2500.

Express each o f the following products as a monomials and verify the result for x = 1, y =2: (27-31)

Q27) (—xy3) x (yx3) x (xy)

Solution:


We have:

( - x y 3) x (yx3) x (xy)

= ( —1) x (x x x3 x x) x (y3 x y xy )

= ( - 1) x (a;1+3+1) x (jj3+1+1)

= - x 5y5

To verify the result, we substitute x = 1 and y = 2 in LHS; we get:

LHS = (~xy3) x (yx3) x (xy)

= [ ( - 1) x 1 x 23] x (2 x l 3) x (1 x 2)

= [ ( - 1) x 1 x 8] x (2 x 1) x 2

= ( - 8) x 2 x 2

= -32

Substitute x = 1 and y = 2 in RHS, we get:

RHS = - x 5y5

= ( 1) ( 1) 5(2 )5

= ( - l ) x 1 x 32

= -32


Thus, the answer is - x3y5

Q28) ( j x 2yi) x (\xAy2) x (xy) x 5

Solution:


We have:

( | * 2y 4) x ( j * V ) x (xy) x 5

= ( i x i x 5) x ( * 2 x * 4 x * ) x (y4 x y 2 xy )

= (1 X i x 5) x ( x 2+4+1) x (t/4+2+1)

To verify the result, we substitute x = 1 and y = 2 in LHS; we get:

LHS = ( - | a : V ) x ( ^ * V ) x (*?/) x 5

= ( | x ( l )2 x (2 )4) x ( I x ( l )4 x (2 )2) x (1 x 2) x 5

= ( ^ x 1 x 16) x ( j x 1 x 4 ) x (1 x 2) x 5

= 2 x l x 2 x 5

= 20

Substituting x = 1 and y = 2 in RHS, we get:

RHS = ^ x 7y7

= | f ( l ) 7(2 )7

= ^ x 1 x 128

= 20


Thus, the answer is -^x7y7.

Q29) ( f a 26) x ( -1 5 6 2ac) x ( - ^ c 2)

Solution:


We have:

( | a 2&) x ( —1562ac) x ( — c 2)

= [■§■ x ( - 1 5 ) x ( — -|) ] x (a2 x a ) x (6 x f>2) x (c x c2)

= [ ( f x ( - 1 5 ) x ( - 1 ) ] x (a2+1) x (61+2) x (c4+2)

= 3a363c3

Since the expression doesn't consist o f the variables x and y.

Therefore, the result cannot be verified for x = 1 and y = 2

Thus, the answer is 3a363c3.

Q30) ( - f a26) x ( - f b2c) x ( - | c 2a)

Solution:


We have:

(—| o26) x (-■§ 62c) x ( - | c 2o)

= [ (—y ) x ( — f ) x ( “ f )] x («2 x o) x (b x b2) x (c x c2)

= K - f ) x ( - 1 ) x ( - 1 )] x (a2+1) x (61+2) x (c4+2)

= - | a 363c3



Thus, the answer is —^a3b3< .

Q31) ( f abc3) x ( - y a 362) x ( - 863c)

Solution:


We have:

( ^ abc3) x ( —-y-a362) x ( —863c)

= [ ( | ) x ( - ^ ) x ( - 8)] x (ax a3) x (b x b2 x b3) x (c3 x c)

= [ ( f ) X ( - f ) X ( - 8)] x (a 1+3) x (61+2+3) x (c3+4)

= 0



Thus, the answer is y 0466c4

Evaluate each o f the follow ing when x = 2, y = -1.

Q32) (2xy) x ( ^ r ) x (a:2) x (y2)

Solution:


We have:

(2xy) x x ( * 2) x (y2)

= (2 x j ) x ( i x i 2 x x2) x (y x y x y2)

= (2 x \ ) x (x1+2+2) x (y1+1+2)

= \ * VSubstituting x = 2 and y = -1 in the result, we get:

= I ( 2)5( - i )4

= \ x 32 x 1

= 16


m { \ x 2y ) ^ { ^ x y2) x { l x 2y2)

Solution:

To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, that is, om x an = am+n

We have:

( f x2y) x ( - ^ -x y 2) x ( \x2y2)

= ( | x ( ~ t ) x | ) x (i 2 x j x a:2) x {y x y2 x y 2)

= ( f x ( - f ) x f ) x (x2+1+2) x ( y 1+2+2)

= - l * v

Substituting x = 2 and y = -1 in the result, we get:

- l a y

= - { ( 2 ) 5( - l ) 5

= ( - t ) x 32 x ( - 1 )

= 56




Find the follow ing products: (1-15)

Q 1 .2 o 3(3a + 5b)

SOLUTION:

To find the product, we will use distributive law as follows:

2a3 (3a + 5b)

= 2a3 x 3a + 2a3 x 5b

= (2 x 3)(a3 x a) + (2 x 5)a3b

= ( 2 x 3 ) o3+1+ (2 x 5) a3b

= 6a4 +10o3b

Thus, the answer is 6a4 +10a3b.

Q 2.-11a(3a + 2b)

SOLUTION:


-11a(3a + 2b)

= (-11 a) x 3a + (-11a) x 2b

= (-11 x 3) x (a x a) + (-11 x 2) x (a x b)

= (-33) x (a1+1) + (-22) x (a x b)

= -33a2 - 22ab

Thus, the answer is -33a2 - 22ab.

Q 3. -5a(7a - 2b)

SOLUTION:


-5a(7a - 2b)

= (-5a) x 7a + (-5a) x (-2b)

= (-5 x 7) x (a x a) + (-5 x (-2)) x (a x b)

= (-35) x (a1+1) + (10) x (a x b)

= -35a2 + 1 0ab

Thus, the answer is -35a2 + 1 0ab.

Q 4. - I l y 2 7)

SOLUTION:


—11 y2(3y + 7)

= (-11 y2) x 3y + (-11 y2) x 7

= (-11 x 3)(y2 x y) + (-11 x 7) x (y2)

= ("33)(y2+1) + (-77) x (y2)

= -33y3-77y2

Thus, the answer is - 33y3- 77y2.

Q5.^-(x3+ y 3)

SOLUTION:


f ^ + s/3)

= £ x x 3 + f x y 3

= § x (a; x a;3) + § x (a; x y 3)

= | x (x x a;1"1"3) + | x (a: x y3)

— 4- 6x^3— 5 + 5

Thus, the answer is + jr~.

Q 6. xy(a:3- 3/3)

SOLUTION:

To find the product, we will use the distributive law in the following way:

xy(x3- y 3)

= xy x a;3 - xy x y3

= (x x a;3) x y - x x (y x j /3 )

= a:1+3y - x j/1+3

= a:4y - xy4

Thus, the answer is a;4y - xy4.

Q 7 .0.1y(0.1a:5 + 0.1y)

SOLUTION:


0.1y(0.1a;5 + 0.1 y)

= (O.ly) (0.1 a;5) + (0.ly)(0.ly)

= (0.1 x 0.1 )(y x a:5) + (0.1 x 0.1 )(y x y)

= (0.1 x 0.1) (a;5 x y) + (0.1 x 0.1) ( y 1+1)

= 0.01 a:5y + 0.01 y2

Thus, the answer is 0.01 a:5y + 0.01 y2

Q 8. ( -^ -a b 2c - f f a 2^ ( - 50 a2 b2c?)

SOLUTION:


(-^■ab2c - f j r a 2c2^ ( - 5 0 a ^ c 2)

= j ( ^ a b 2c^ ( -5 0 a 262c2) | - | ^ a 2c2^ ( —SOa2^2^ ) ^

= | | x x ( - 5 0 ) j (a x a2) (b2 x b2) x (ex c2) |

“ { ( f ) ( - 50) K x a2) x b2 x (c2 x c2) }

^ a W - ^ U a W )

= 111 a3b4c3 + 12a4b2c4

Thus, the answer is ^p -o3b4c3 + 12o4b2c4.

Q9. —§jxyz ^ x y z2- j x y 2z3

SOLUTION:


= { ( - W ^ z) ( I * * * 2) } " { ( - £ * » * ) ( f * 2 /^ 3) }

= { ( ^ F x f ) ( * x x) x (y x y) x (z x z2) | - 1 ( f £ x | j ( * x x) x (yx y2) x ( z x z3) }

SOLUTION:


= { ( - ± x y z ) ( f a : V ) } - { ( —^ a * / * ) ( fa r y z 2) }

---- g-a; y z, t y &

Thus, the answer is — f x2y2z3 + j x 2y3z‘

f a -3y2z2 + ±x2y2z-2 „3,,2_2 , I~ 2 ,,2 _3

Thus, the answer is — j x 3y2z2 + ^x2y2z'i « 2„ 2_3

Q 11. 1.5x(10x2y - lOOxy2)

SOLUTION:


1 .5x(10 x2y - lOOajy2)

= (1 .5 * x 10x2y) - (1 .5 * x 100* j / 2)

= (1 5 *1+2j/) - ( l5 0 * 1+1j/2)

= 1 5 *3y - 1 5 0 * 2j/2

Thus, the answer is 1 5 *3y - 150x2y2.

Q 12.4 .1 xy ( 1.1 x -y )

SOLUTION:


4 .1 *y ( l . i * - y)= (4.1*2/ x 1 -1 *) - (4 .1 *2 / x V)

= {(4 .1 x 1.1) x xy x * } - ( 4 .1* 2/ x y)= (4 .5 1 *1+12/) — (4.1*2/1+1)

= 4 .5 1 *22/— 4.1*2/2

Thus, the answer is 4 .5 1 *22/-4.1*2/2

Q 13.250.5*2/ ( xz + y(j-)

SOLUTION:


250.5*2/ + t j )

= 250.5*2/ x xz + 250.5*2/ x

= 250.5x1+1yz + 25.05xy1+1 — 250.5x2yz + 25.05xy2

Thus, the answer is 250.5x2yz + 25.05xy2.

SOLUTION:



Q 16. Find the product 24x2 (1 - 2 x ) and evaluate Its value fo r x = 3.

-ru . 21 3 3 | 14 3Thus, the answer is -^x y + -^x y

Q15. | o ( a 2 + 62- 3 c 2)

SOLUTION:

f a (a 2 + b2- 3c2)

= f a x a2 + f a x b2- f a x 3c2

SOLUTION:


24a:2 ( l- 2 a : )

= 24a:2 x l-2 4 a ;2 x 2a:= 2Ax2-48x1+2 = 24a:2-4 8 a :3

Substituting x = 3 in the result, we get

24a:2-4 8 a :3

24(3)2- 48 (3 )3

= 24 x 9 -4 8 x 27 = 216-1296 = -1 0 8 0

Thus, the product is 24a;2- 48a:3 and its value for x = 3 is -1080.

Q 17. Find the product —3y (xy + y2) and find its value fo r x = 4 and y = 5.

SOLUTION:


- 3 y (xy + y2)

= - 3 y x x y + ( - 3 y) x y2 = -3xy1+1-3y1+2 = —3xy2- 3y3

Substituting x = 4 and y = 5 in the result, we get

-3xy2-3y3

= - 3 ( 4 ) ( 5 ) 2- 3 ( 5 )3

= - 3 ( 4 ) ( 2 5 ) - 3 (125)

= -3 0 0 -3 7 5 = -6 7 5

Thus, the product is —3xy2- 3y3, and its value for x = 4 and y = 5 is -675.

Q 18. M ultiply - 1 x2y3 by (2x - y) and verify the answer fo r x = 1 and y = 2.

SOLUTION:


Substituting x = 1 and y = 2 in the result, we get

—3x3y3 + \x2yi

— —3(1 )3(2 )3 + - | ( 1) 2(2 )4

= —3 x l x 8 + | x l x l 6

= - 2 4 + 24 = 0Thus, the product is —3x3y3 + | ■x2yi, its value for x = 1 and y = 2 is 0.

Q 19. M ultiply the monomial by the binom ial and find the value o f each fo r x = -1, y = 0.25 and z = 0.05:

(I) 15 j/2 ( 2 - 3a:)

SOLUTION:


15y2 (2- 3a:)

= 15y2 x 2 -1 5 y2 x 3a:

= 30 j/2- Ahxy2

~^x2y3 x (2x -y )

= ( —3a;2+ y ) - ( —fa ;2^ 1)

= —3x3y3 + \x2yi

Substituting x = -1 and y = 0.25 in the result, we get:

30y2- 45 xy2= 30(0 .25)2- 4 5 ( —1) (0 .25)2

= 30 x 0 .0 6 2 5 -{4 5 x ( - 1 ) x 0 .0625}

= 1 .8 7 5 -(-2 .8 1 2 5 )

= 1.875 + 2.8125 = 4.6875

(ii) -3 a : (y2 + z2)

SOLUTION:


-3 a ; (y2 + z2)

— —3a: x y2 + ( —3a:) x z 2

= —3xy2-3xz2

Substituting x = -1, y = 0.25 and z = 0.05 in the result, we get:

-3xy2-3xz2

= - 3 ( - 1 ) (0 .25)2- 3 ( - 1 ) (0 .05 )2

= - 3 ( - 1 ) (0.0625) - 3 ( - 1 ) (0.0025)

= 0.1875 + 0.0075 = 0.195

(iii) z2 (x -y )

SOLUTION:


z2 (x -y )= z2 x x -z 2 x y = xz2-y z 2


xz2-y z2

= ( - 1 ) (0 .05 )2- (0.25) (0 .05 )2

= ( - 1 ) (0 .0 0 2 5 )-(0 .2 5 ) (0.0025)

= -0 .0 0 2 5 -0 .0 0 0 6 2 5 = -0 .00 31 25

(iv) xz (a:2 + y2)

SOLUTION:


xz (x2 + y2)

= xz x x2 + xz x. y2 = x3z + xy2z


x3z + xy2z

= ( - 1 )3 (0.05) + ( - 1 ) (0 .25)2 (0.05)

= ( - 1 ) (0.05) + ( - 1 ) (0.0625) (0.05)

= -0 .0 5 -0 .0 0 3 1 2 5 = -0 .05 31 25

Q 20. Simplify:

(i) 2a;2 (x3- x ) -3x (a;4 + 2a;) - 2 ( x 4- 3x2)

SOLUTION:

To simplify, we will use distributive law as follows:

2 x 2 (x3- x) - 3x (x4 + 2x) - 2 (a:4- 3a:2)

= 2 x5- 2x3- 3 x 5- 6x 2- 2x4 + 6x 2

= 2a;5-3 a ;5- 2a;4- 2a;3- 6a:2 + 6x 2 = —x 5- 2x4- 2x3

(ii) x 3y (x2- 2a:) + 2xy (a;3- a;4)

SOLUTION:


x 3y (x2- 2x) + 2xy (x3- x 4)

= x 5y - 2x4y + 2x4y - 2x 5y = x 5y - 2 x3y - 2x4y + 2xAy = ~ x 5y

(iii) 3a2 + 2 (o + 2) - 3a (2o + 1)

SOLUTION:


3a2 + 2 (a + 2) - 3a (2a + 1)

= 3a2 2a -1- 4— 6a2 - 3a = 3a2- 6a2- 3a + 4 = - 3a2- a + 4

(iv) x (x + 4) + 3a: (2a;2- 1) + 4a:2 + 4

SOLUTION:


a; (x + 4) + 3a: (2a;2- 1) + 4a:2 + 4

= a;2 + 4a; + 6a;3- 3a; + 4a;2 + 4 = a;2 + 4a;2 + 4a;- 3a; + 6a;3 + 4 = 5a:2 + x + 6x3 + 4

(v) a(b - c) - b(c - a) - c(a - b)

SOLUTION:


a(b - c) - b(c - a) - c(a - b)

= ab - ac - be + ba - ca + cb

= ab + ba - ac - ca - be - cb

= 0

(vi) a(b - c) + b(c - a) + c(a - b)

SOLUTION:


a(b - c) + b(c - a) + c(a - b)

= ab - ac + be - ba + ca - cb

= ab - ba - ac + ca + be - cb

= 0

(vii) 4a6 (a - b) - 6a2 (b- 62) - 362 (2a2- a) + 2a6 (6- a)

SOLUTION:


4a6 (a - 6) - 6a2 (6- b2) - 3b2 (2a2- a) + 2ab (b- a)

= 4a2b -4ab2- 6a2b + 6a2b2- 6b2a2 + 3b2 a + 2ab2- 2a2b = 4a?b-Ga2b -2a2b -4ab2 + 3 b2a + 2ab2 + 6a262- 662a2 = —4a2 6 + ab2

(viii) X2 (x2 + l ) - x 3 (x + l ) - x (x 3- x )

SOLUTION:


x 2 (x2 + l ) - a;3 (a; + 1) - * (x 3- x )

= X4 + x 2- x i - x 3- x i + X2 = X4— X4— X4- X3 + x2 + x2= - x4- x3 + 2x2

(ix) 2a2 + 3a ( l - 2a3) + a (a + 1)

SOLUTION:


2a2 3a ( l — 2a3) -1- a (a -1- 1)

= 2a2 + 3 a - 6a4 + a2 + a = 2a2 + a2 + 3a + a - 6a4

(x) a2 ( 2 a - 1) + 3a + a3- 8

SOLUTION:


a2 (2a-1) + 3a + a3- 8

= 2a3- a2 + 3a + a3- 8 = 2a3 + a3- a2 + 3 a - 8 = 3a3- a2 + 3 a - 8

(xO f a 2 (a:2- 1) + \x2 (x2 + x ) - \ x (x3- 1)

SOLUTION:


f a 2 (x2- l ) + \x2 (x2 + x ) - j x (x3- 1)

= | * 4- f * 2 + i * 4 + 1 * 3- f * 4 + 1 *3 4 , 1 4 3 4 , 1 3 3 2 , 3— X + X jX + - X 2 X + jX„ 4 1 1 _3 3 9 1 3 „= X + jX ° - -X + —X

(xii) a2b (a - fe2) + ab2 (4 a 6 - 2a2) - a3b ( 1 - 26)

SOLUTION:


a2b (a- 62) + ab2 (4 a 6 -2 a 2) - a 36 ( l - 2 6 )

= a3b-a2b3 + 4a263- 2 a 362- a 36 + 2a3 b2 = a3b-a3b-a2b3 + 4a263- 2a362 + 2a3 62

= 3a2 63

(x iii) a2b (a 3- a + l ) - ab (a4- 2a2 + 2a) - 6 (a3- a2- 1)

SOLUTION:


a2b (a 3- a + l ) - a6 (a4- 2a2 + 2a) - 6 (a3- a2- 1)

= a5b-a3b + a2b-a5b + 2a3b-2a2b-a3b + a2b + 6 = a5b-a5b — a3b + 2a3b-2a2b + a2b + b = b



M ultiply

Q1. (5x + 3) by (7x + 2)

SOLUTION:

To multiply, we will use distributive law as follows:

(5a; + 3) (7x + 2)

= 5a: (7a; + 2) + 3 (7a: + 2)

= (5a: x 7a: + 5a: x 2) + (3 x 7a: + 3 x 2)

= (35a:2 + 10a:) + (21a: + 6)

= 35a:2 + 10a: + 21a: + 6

= 35a:2 + 31a; + 6

Thus, the answer is 35a:2 + 31a: + 6

Q2. (2x + 8) by (x - 3)

SOLUTION:


(2a :+ 8) ( x - 3 )

= 2a: (a :-3) + 8 (a ;-3)

= (2 * x x - 2x x 3) + (8a:- 8 x 3 )

= (2a:2- 6a:) + (8 a :-24)

= 2a;2- 6a: + 8x- 24 = 2a;2 + 2a:- 24

Thus, the answer is 2a:2 + 2a:- 24.

Q3. (7x + y) by (x + 5y)

SOLUTION:


(7a; + y) {x + 5y)= 7x{x + 5 y) + y( x + by)= 7x2 + 3 5 xy + xy + 5 y2= 7x2 + 36 xy + by2

Thus, the answer is 7x2 + 36xy + by2.

Q4. (a - 1) by (0.1a2 + 3)

SOLUTION:


( a -1 ) (0.1a2 + 3)

= 0.1a2 ( a - 1 ) + 3 (a -1 )

= 0.1a3-0 .1 a 2 + 3 a -3

Thus, the answer is 0.1a3-0 .1 a 2 + 3 a -3 .

Q5. (3a:2 + y2) (2a:2 + 3y2)

SOLUTION:


(3a:2 + y2) (2a:2 + by2)

= bx2 (2a:2 + by2) + y2 (2a;2 + by2)

= 6x4 + 9x2y2 + 2a-2y2 + by4= 6a;4 + l l x 2y 2 + by4

Thus, the answer is 6a:4 + 11 x2y2 + by4.

Q6- ( f a : + i » ) ( f a: + 43/)

SOLUTION:


( ! * + * » ) ( I a:+ 4y)

= \ x ( I * 4j/) + \y ( f * + 4?/)

= i * 2 + T xy + ~kxv + 2y2

. ^ + ( a g » ) „ + v

Thus, the answer is ^x2 + XU + 2y2.

Q7. ( * 6- t / 6) ( * 2 + i / 2)

SOLUTION:


(x « -y « ) (a:2 + y2)

= x6 (x2 + y2) - y 6 (x2 + y2)

= (a:8 + x6y2) - (y6x2 + y8)

= a:8 + x6y2- y 6x2- y 8

Thus, the answer is a:8 + x6y2- y 6x2- y 8.

Q8. (x2 + y2) (3a + 26)

SOLUTION:


(x2 + y2) (3a + 26)

= x2 (3a + 26) + y2 (3a + 26)

= 3aa;2 + 26a:2 + 3a y2 + 2by2

Thus, the answer is 3aa:2 + 26a:2 + 3ay2 + 2by2.

Q 9 . [ - 3 d + ( - 7 / ) ] (5d + f )

SOLUTION:


[ -3 d + ( -7 f ) } (5 d + f )= (-3d) (5d + / ) + ( -7 /) (5d + / )= ( - 15d2- 3 d / ) + ( - 3 5 d / - 7 / 2)

= - 1 5 d 2- 3 d / - 3 5 d / - 7 / 2

= - 15d2- 3 8 d / - 7/ 2

Thus, the answer is - 15d2- 38 d f - 7 f 2.

Q10. (0 .8 a -0.56) (1 .5 a -36)

SOLUTION:


(0 .8 a -0.56) (1 .5 a -36)

= 0.8a (1 .5 a -36) -0 .5 6 (1 .5 a - 36)

= 1.2a2-2 .4 a 6 -0 .7 5 a 6 + 1.562 = 1.2a2-3 .1 5 a 6 + 1.562

Thus, the answer is 1.2a2- 3.15a& + 1.562.

Q11. (2* 2j/2- 5xy2) (x2- y 2)

SOLUTION:


(2x2y2-5xy2) (x2- y 2)

= 2x2y2 (x2- y 2) -5xy2 (x2- y 2)

= 2xiy2-2x2y4- 5x3y2 + hxyi

Thus, the answer is 2xiy2- 2x2yi- 5x3y2 + bxy4.

Q12- (f + t ) (f + f )SOLUTION:


( f + f ) (f + f )__ x f 2 , 9a; \ , x?_ ( 2 . 9x \“ 7 ^ 5 " h 4 / " h 2 \ 5 " h 4 >/__ 2x _|_ 9s2 | x2 ■ 9s3— 35 28 5 8

_ 2x , / 45+28 >\ _2 , 9*3 35 "h \ 140 ) X "h 8

__ 2x_ j_ 73 9 j_ 9a3— 35 140X 8

Thus, the answer is | | + -tj x2 35 1409x3

8

Q13. ( - f + i ) ( | - f )

SOLUTION:


( - t + t ) ( l - f )

__ / _ ab_ _|_ ab2 \ , f a2b a262 \— ^ 14 21 ) ^ 18 27 /

___ ab , ab2 i a2b a2b2— 14 21 18 27

TL .. . ab i ab2 i a2b a2b2Thus, the answer is - 4- ^ ^ r .

Q14. (3x2y-5xy2) (\ x 2 + | y 2)

SOLUTION:


(3x2y-5xy2) ^ a ;2 + | y 2)

= \x2 (3x2y-5xy2) + \y2 (3x2y-5xy2)

= \x4y -x3y2 + x2y3- |a ;y 4

Thus, the answer is j x 4y -x 3y2 + x2y3- jxy4.

SOLUTION:


(2a:2- 1 ) (4a:3 + 5a:2)

= 2a:2 (4a:3 + 5a:2) - 1 (4a:3 + 5a;2)

= 8a:5 + 10a:4- 4a:3- 5a:2

Thus, the answer is 8a:5 + 10a:4- 4a:3- 5a:2.

Q15. (2a;2- 1) (4a;3 + 5a;2)

Q16. (2xy + 3y2) (3y2- 2 )

SOLUTION:


(2xy + 3y2) (3a/2 2)

= 2xy(3y2- 2 ) + 3 y 2 (3a/2 2)

= 6 xy3- Axy + 9 a/4— 6 y2 = 9 yi + 6xy3—6y2—Axy

Thus, the answer is 9 j/4 + 6xy3- 6a/2— 4xy.

Find the follow ing products and verify the result fo r x = -1 and y = -2:

Q17. (3a:- 5a/) (x + y)

SOLUTION:


{3x-by){x + y)= 3a: (x + y ) - by ( x + y)= 3x2 + 3 xy- bxy- by2 = 3x2- 2 xy- by2=:. (3x-by) (x + y) = 3x2-2xy-by2Now, we put x = -1 and y = -2 on both sides to verify the result.

LHS =(3a;- by) (x + y)= { 3 ( - l ) - 5 ( - 2 ) } { - l + ( - 2 ) }

= ( - 3 + 1 0 ) ( - 3 )

= -2 1 RHS =3x2-2xy-by2

— 3 (—1)2- 2 ( —1) ( —2 ) - 5 ( —2) 2 = 3 x 1 - 4 - 5 x 4 = 3 - 4 - 2 0 = -2 1Because LHS is equal to RHS, the result is verified.

Q18. (x2y - 1) ( 3 - 2a:2y)

SOLUTION:


(x2y - l ) (3- 2x2y)

= x 2y (3- 2x 2y) -1 (3- 2x2y)

= 3x2y - 2 x i y2- 3 + 2 2y — 5x2y - 2 x i y2- 3

(x2y - 1) (3-2x2yj = bx2y - 2 x i y2- 3

Now, we put x = -1 and y = -2 on both sides to verify the result.

(x2y - l) (3-2x2y)

= [ ( —l ) 2 ( - 2 ) - l ] [3- 2 ( —l ) 2 ( —2 )j

= [1 x (—2 ) - l ] [3-2 x 1 x (-2)]= ( - 2 - 1 ) (3 + 4)= - 3 x 7 = -2 1

RHS =5x2y- 2xiy2- 3

= 5(—l)2 ( 2) — 2( 1)4( 2)2— 3 = [5 x 1 x ( - 2 ) ] - [ 2 x 1 x 4 ] -3 = - 1 0 - 8 - 3 = -2 1Because LHS is equal to RHS, the result is verified.

SOLUTION:


Now, we put x = -1 and y = -2 on both sides to verify the result.

LHS =

_ l 2 xr xv v9 X 15 15 2 5

LHS =

1 192 2 5

RHS =

9 25

— i l i2 2 5

Because LHS is equal to RHS, the result is verified.

Simplify:

Q20. x2 (x + 2y) (x-3y)

SOLUTION:


x2 (x + 2y) (x-3y)= [x2(x + 2y)\ (x-3y)

= (a:3 + 2x2y) (x-3y)

= x3 (a;- 3y) + 2x2y (x - 3y)= x4- 3x3 + 2x3-6x2y2= x4- x 3-6x2y2

Thus, the answer is x4- x3- 6x2y2.

Q21. (x2-2y2) (x + Ay) x2y2

SOLUTION:


(x2-2y2) (x + Ay) x2y2

= [ * 2 (x + Ay) - 2y2 (x + Ay)] x2y2

= {x3 + Ax2y-2xy2-8y3) x2y2

= x5y2 + Ax4y3-2x3y4-8x2y5

Thus, the answer is x5y2 + Ax4y3-2x3y4-8x2y5.

Q22. a262 (a + 26) (3a + 6)

SOLUTION:


a2b2 (a + 2b) (3a + b)= [a2b2 (a + 2b)] (3a + 6)

= (a362 + 2a263) (3a + 6)

= 3a (a3b2 + 2a2b3) + b (a3b2 + 2a2b3)

= 3a4b2 + 6a363 + a3b3 + 2 a2b4= 3a4b2 + la3b3 + 2a2b4Thus, the answer is 3a462 + 7a3b3 + 2a2b4.

Q23. x2 (x -y ) y2 (x + 2y)

SOLUTION:


z 2 (x -y )y 2 (x + 2 y)= [x2(x-y)\ \y2(x + 2y)]

= (x3- x 2y) (xy2 + 2y3)

= x3 (xy2 + 2y3) - x 2y (xy2 + 2y3)

= x4y2 + 2x3y3- [x3y3 + 2x2y4]

= x4y2 + 2x3y3- x 3y3-2x2y4 = x4y2 + x3y3-2x2y4

Thus, the answer isxiy2 + 2x3y3- x 3y3-2x2yi = x 4y 2 + x3y3- 2 x2yi

Q24. (x 3- 2 x 2 + 5 a :-7) ( 2 x -3 )

SOLUTION:


(a;3- 2a;2 + 5a;- 7) (2a:- 3)

= 2a: (a:3-2 a :2 + 5 a :-7) - 3 (x 3- 2 x 2 + 5 a ;-7)

= 2a:4-4 a :3 + 10a;2- 14a:-3a:3 + 6a:2- 15a; + 21 = 2x4- 4 x 3- 3 x 3 + 10a;2 + 6 x 2- 14a:- 15x + 21

= 2a:4-7 a :3 + 16x2- 2 9 x + 21

Thus, the answer is 2a:4- 7x3 + 16x2- 29x + 21.

Q25.2a:4-7 a :3 + 16x2- 29x + 21

SOLUTION:


(5a ;+ 3) ( x - 1 ) (3 a :-2)

= [(5a ;+ 3) ( x - 1 ) ] (3 x -2 )

= [5a: ( x - 1) + 3 ( x - 1)] (3a;- 2)

= [5x2- 5x + 3x- 3] ( 3 x - 2)

= 3x (5x2 + 2x- 3) - 2 (5a:2 + 2 x -3 )

= 15a:3-6 a :2- 9 a :- [ l0 x 2- 4 x - 6 ]

= 15x 3- 6 x 2- 9 x - 10x2 + 4 x + 6 = 15a:3-1 6 a :2-5 a : + 6

Thus, the answer is 15x 3- 6 x 2- 9 x - 10a:2 +4x + 6 = 15x 3- 1 6 x 2- 5 x + 6

Q26. ( 5 - x ) ( 6 - 5a:) (2 -x )

SOLUTION:


( 5 - x) (6 - 5a;) ( 2 - x)= [ ( 5 - x ) ( 6 - 5a;)] ( 2 - x )

= [5 (6 - 5x) - x (6 - 5x)] ( 2 - x )

= (3 0 -2 5 x -6 x + 5 x2) ( 2 - x )

= (3 0 -3 1 x + 5 x2) ( 2 - x )

= 2 (3 0 -3 1 x + 5 x 2) - x (3 0 -3 1 x + 5 x2)

= 6 0 -6 2 x + 10x 2- 3 0 x + 31x2- 5 x 3

= 6 0 -9 2 x + 41x 2- 5 x 3

Thus, the answer is 6 0 - 92x + 41x2- 5x3.

Q27. (2 x 2 + 3 x - 5) (3 x 2- 5x + 4)

SOLUTION:


(2a;2 + 3a;- 5) (3a;2- 5a; + 4)

= 2x2 (3a:2- 5s + 4) + 3x (3 s 2- 5s + 4) - 5 (3a;2- 5x + 4)

= 6a;4- 10s3 + 8 s 2 + 9 s 3- 15s2 + 1 2 s - 15s2 + 2 5 s - 20 = 6 s4- 10s3 + 9 s 3 + 8 s 2- 15s2- 15s2 + 25s + 1 2 s - 20

= 6 x4- x 3- 22s2 + 3 6 s - 20

Thus, the answer is 6 s4- x 3- 22s2 + 3 6 s - 20.

Q28. ( 3 s - 2) ( 2 s - 3) + ( 5 s - 3) (s + 1)

SOLUTION:


( 3 s - 2) ( 2 s - 3) + ( 5 s - 3) (s + 1)

= [ ( 3 s - 2) ( 2 s - 3)] + [ ( 5 s - 3) (s + 1)]

= [3s ( 2 s - 3) - 2 ( 2 s - 3)] + [5s (s + 1) - 3 (s + 1)]

= 6 s2- 9 s - 4s + 6 + 5 s 2 + 5 s - 3 s - 3= 6 s2 + 5 s 2- 9 s - 4s + 5 s - 3 s - 3 + 6= 11s2- 1 1 s + 3

Thus, the answer is 11s2- 11s + 3.

Q29. ( 5 s - 3) (s + 2) - (2s + 5) ( 4 s - 3)

SOLUTION:


( 5 s - 3) (s + 2) - (2s + 5) ( 4 s - 3)

= [ ( 5 s - 3) (s + 2)] - [(2s + 5) ( 4 s - 3)]

= [5s (s + 2) - 3 (s + 2)] - [2s ( 4 s - 3) + 5 ( 4 s - 3)]

= 5s2 + 1 0 s - 3 s - 6 + 8 s 2 + 6 s -2 0 s + 15

= 5s2- 8 s 2 + 1 0 s -3 s + 6s — 2 0 s - 6 + 15 = —3 s2- 7 s + 9

Thus, the answer is - 3 s 2- 7s + 9.

Q30. (3s + 2y) (4s + 3y) - ( 2 s - y) ( 7 s - 3y)

SOLUTION:


(3 s + 2y) (4 s + 3y) - ( 2 s - y) ( 7 s - 3y)= [(3s + 2y) (4 s + 3y)] - [ (2 s - y) (7 s -3 y ) ]

= [3s (4 s + 3y) + 2y (4s + 3y)] - [2s ( 7 s - 3y) - y ( 7 s - 3j/)[

= 12s2 + 9 xy + 8 xy + 6 y2- 14s2 + 6 xy + 7xy- 3 y2 = 12s2- 14s2 + 9 xy + 8 xy + 6 xy + 7 xy + 6 y2- 3 y2 = - 2 s2 + 30 xy + 3 y2

Thus, the answer is - 2 s 2 + 30xy + 3y2.

Q31. (s 2- 3s + 2) ( 5 s - 2) - (3 s 2 + 4 s - 5) ( 2 s - 1)

SOLUTION:


(a;2- 3a; + 2) (5a ;- 2) - (3a;2 + 4a;- 5) (2 a ;-1)

= [(a;2-3 a ; + 2) (5 a ;-2)] - [(3a;2 + 4 a ;-5) (2 a ;-1)]

= [5x ( x 2- 3 x + 2) - 2 (a;2-3 a ; + 2 )] - [2x (3a;2 + 4 a ;-5) - 1 (3a;2 + 4 x -

= [5a;3-1 5 a ;2 + 10a;-2a;2 + 6a;— 4] - [6a;3 + 8a;2-1 0 a ;-3 a ;2-4 a ; + 5]

= 5a;3- 15a:2 + 10a;-2a:2 + 6 x - 4 - 6 x 3- 8 x 2 + lO x + 3 x2 + 4 a ;-5= - x3- 22x2 + 30a;- 9

Thus, the answer is - a;3- 22x2 + 3 0 * - 9.

Q32. ( x 3- 2 x 2 + 3 * - 4 ) ( * - l ) - ( 2 * - 3 ) ( x 2- x + l )

SOLUTION:


( * 3- 2 * 2 + 3 * - 4 ) ( * - l ) - ( 2 * - 3 ) (x2- x + l )

= [ ( * 3- 2 * 2 + 3 * - 4 ) ( * - l ) ] - [ ( 2 * - 3 ) (x2- x + l ) ]

= [x ( * 3- 2 * 2 + 3 * - 4 ) - 1 (x 3- 2 * 2 + 3 * - 4 ) ] - [2a; ( * 2- * + l ) - 3 ( x 2-

= * 4- 2 * 3 + 3 * 2- 4 * - * 3 + 2a:2-3 a : + 4 - 2 x 3 + 2a;2-2 a ; + 3 * 2- 3 * + 3

= * 4- 2x3- 2x3- x 3 + 3 x 2 + 2 x2 + 2 x 2 + 3 x 2- 4 x - 3 x - 2 x - 3x + 4 + 3 = x 4- 5x3 + 10x2- 12x + 7

Thus, the answer is x 4- 5 x 3 + 10x2- 12x + 7..



Q1 Write the following squares o f binomials as trinomials

We know that, (o + b)2 = a2 + 2ab + b2 and

(a-b )2 = a2-2ab + b2

(i) (x + 2)2

Sol:

(a: + 2 )2 is in the form of (a + 6) 2 = a2 + 2ab + b2

here, a = x , b = 2

=> x2 + 2 x x x 2 + 62

=> a:2 + 4a: + b2

(ii) (8a + 36)2

Sol:

(8a + 36)2 is in the form of (a + 6)2 = a2 + 2a6 + b2

here, a = 8a, b = 3b

=> (8a )2 + 2 x (8a) x (36) + (36)2

=> 642 + 48a6 + 3662

(iii) (2m + 1)2

Sol:

(2 m + l ) 2 is in the form of (a + 6)2 = a2 + 2a6 + 62

here, a =2m, b = 1

=> (2 m )2 + 2 x (2m ) x (1) + ( l ) 2

=> 4 m 2 + 4 m + 1

( iv ) ( 9 a + { ) 2

Sol:

(9a + | - )2 is in the form of (a + 6)2 = a2 + 2a6 + 62

here, a = 9a, b =

= > (9 a )2 + 2 x ( 9 a ) x ( ± ) + ( ± ) 2

=>81a2 + 3 a + ^

(v) ( * + f ) 2

Sol:

(x + y ) 2 is in the form of (a + 6)2 = a2 + 2a6 + 62

here, a = x, b = y

=> a;2 + 2 x a; x ( y ) + ( - y ) 2

=> a:2 + x3 + y

(vi) ( f - f ) 2

Sol:

(■|— - | ) 2 is in the form of ( a -6 )2 = a2-2ab + b?

here, a = y , b = %

= > ( f ) 2- 2 x ( f ) x ( f ) + ( f ) 2

=> ?L-1X V + «116 9

(Vii) (3a;- ^ ) 2

Sol:

(3a:- ^ - ) 2 is in the form of ( a -6 )2 = a2-2ab + b2

here, a = 3x, b = ^

= > (3 * )a- 2 x 3 * x ( £ ) + ( £ ) »

=>9a;2- 2 + ^

( v i i i ) ( f - | ) 2

Sol:

(■|— ¥ ) 2 's 'n the form of ( a -6 )2 = a2-2ab + li2

here, a = ■£, b = —v x

=>( f ) 2- 2 x ( f ) x ( f ) + ( f ) 2

=> 4 - 2 + 4yi x1

( ix ) ( f - f ) 2

Sol:

(■y— y ) 2 is in the form of (a-b ) 2 = a2-2ab + 62

here, a = y , b = “

= > ( f ) 2- 2 x ( f ) x ( f ) + ( f ) 2

_ 9a2 15 a b , 2 5 t^" 4 4 + 16

(x) (a26 -6 c 2)2

Sol:

(a26 -6 c2)2 is in the form of (a -b )2 = a2- 2ab + b2

here, a = a2b, b = be2

=> (a2b)2-2 x (a2b) x (6c2) + (6c2) 2

=> a462- 2a262c2 + 62c4

Q2 Find the product o f the following binomials

(i) (2x + y)(2x + y)

sol:

(2x + y) (2x + y) can be written as (2x + y)2

=> (2x + y f

we know that, (a + 6)2 — a2 + 2ab + 62

=> ( 2 * )2 + 2 x (2 * ) x (y) + y2

=> 4x2 + 4xy + y2

(ii) (a + 2b) (a - 2b)

sol:

we know that (a + b) (a - b) = a2- b2

here, a = a, b = 2b

=> a2- (2b)2

=> a2- Ab2

(iii) (a2 + be)- (a2- be)

sol:


here, a = a2, b = be

=>(a2)2- ( 6 c ) 2

=> a4- 62c2

(lv) ( £ - ? ) ( £ + ? )

sol:

we know that (a + b) (a - b) = a2- 62

here, a = -y , b = y

= > ( f ) 2- ( f ) 2

1 6 x 2 9 y2~ 2 5 16

( v ) (2 * + f ) (2 ® -f )

sol:


here, a = 2x, b = |-

= > (2 * )2- ( f ) 2

=> 4x2- r

(vi) (2a3 + 63)(2 a 3- 6 3)

sol:


here, a = 2a3, b = b3

=> (2a3)2- (ft3)2

=> 4a6- b6

sol:


here, a= xA, b = \X £

= > (* 4) 2- ( J r ) 2

= > *8- ^

( v i i iH ^ + ^ X z 3- ^ )

sol:


here, a = a:3, b =X3

= > (* * )2- ( £ ) 2

= > *6- - VX8

Q3 Using the formula for squaring a binomial, evaluate the following

(I) (102)2

sol:

(102)2 can be written as (100 + 2 )2

we know that, (a + b)2 — a2 + 2ab + b2

here, a = 100, b = 2

=> (100 + 2 )2

=> (100)2 + 2 x (100) x 2 + 22

=>10000 + 400 + 4

=>10404

(II) (99 )2

sol:

(99 )2 can be written as (1 0 0 -1)2

we know that, ( a - b)2 = a2- 2ab + b2

here, a = 100, b = 1

=> (1 0 0 -1)2

=>(100)2- 2 x (100) x l + l 2

= >10000-200 + 1

=> 9801

(ill) (1001)2

sol:

(1001)2 can be written as (1000 + l ) 2

we know that, (a + b)2 = a2 + 2ab + b2

here, a = 1000, b = 1

=> (1000 + 1)2

=> (1000)2 + 2 x (1000) x 1 + l 2

=>1000000 + 2000 + 1

=>1002001

(iv) (999)2

sol:

(999)2 can be written as (1 0 0 0 -1)2

we know that, (a - b)2 = a2- 2ah + b2

here, a = 1000, b = 1

=> (1 0 0 0 -1)2

=> (1000)2- 2 x (1000) x 1 + l 2

=>1000000-2000 + 1

=> 998001

(v) (703)2

sol:

(703)2 can be written as (700 + 3 )2


here, a = 700, b = 3

=> (700 + 3 )2

=> (700)2 + 2 x (700) x 3 + 32

=> 490000 + 4200 + 9

=> 494209

Q4 Simplify the following using the formula: (a + b) (a - b) = a2- b2

(i) (82 )2- (18 )2

sol:

(82 )2- ( 1 8 ) 2

here, a = 82, b = 18

=>(82+ 18) (8 2 - 18)

=> 100 x 64

=> 6400

(II) (467)2- (33 )2

sol:

(467)2- (33 )2

here, a = 467, b = 33

=> (467 + 33) (467 - 33)

=> 500 x 434

=> 217000

(ill) (79 )2- (69 )2

sol:

(79 )2- (69 )2

here, a = 79, b = 69

=> (79 + 69) (79 - 69)

=> 148 x 10

=>1480

(iv) 197 x 203

sol:

Since, 197+2032

197 x 203 can be written as (200 + 3) (200 - 3)

=> (200 + 3) (200 - 3)

=> (200)2- (3 )2

=> 40000 - 9

=> 39991

(v) 113 x 87

sol:

113 x 87 can be written as (100 + 13) (1 0 0 - 13)

=> (100 + 13) (1 0 0 - 13)

=> (100)2- ( 1 3 ) 2

=>10000- 169

=> 9831

(vi) 95 x 105

sol:

95 x 105 can be written as (100 + 5) (100 - 5)

=> (100+ 5) (1 0 0 - 5)

=> (100)2- (5 )2

= >1 000 0-2 5

=> 9975

(vii) 1.8 x 2.2

sol:

1.8 x 2.2 can be written as (2 + 0.2) (2 - 0.2)

=> (2 + 0.2) (2 - 0.2)

=> (2 )2- (0 .2 )2

=> 4 - 0.04

=> 3.96

(viii) 9.8 x 10.2

sol:

9.8 x 10.2 can be written as (10 + 0.2) (10 - 0.2)

=> (10 + 0 .2 )(1 0 -0 .2 )

=> (10 )2- (0 .2 )2

= > 1 00 -0 .04

=> 99.96

Q5 Simplify the following using identities

, n ( 5 8 ) 2—( 4 2 ) 2

W 16

sol:

The numerator is in the form of (a + b) (a - b) = a2- 6 2

( 5 8 ) 2- ( 4 2 ) 2 _ ( 5 8 + 4 2 ) ( 5 8 - 4 2 )

16 16

^ =100

(58)2-(42)2 100x1616 16

= > ( ^ ± ^ = 1 0 016

(ii) (178 x 1 7 8 )- (22 x 22)

sol:

we know that, (a + b) (a - b) = a2- b2

=> (178 x 1 7 8 )-(2 2 x 22) = (178)z- ( 2 2 ) 2

=> (178 x 1 7 8 )- (22 x 22) = (178 + 22) (178 - 22)

=> (178 x 1 7 8 )-(2 2 x 22) = 200 x 156

=> (178 x 1 7 8 )-(2 2 x 22) =31200

(1 9 8 x 1 9 8 ) - ( 1 0 2 x 1 0 2 )

96(ill)

sol:

we know that, (a + b) (a - b) = a2- b2_ ( 1 9 8 x 1 9 8 ) - ( 1 0 2 x 1 0 2 ) _ ( 1 9 8 ) 2- ( 1 0 2 ) 2

- > 9 6 9 6

_ _ ( 1 9 8 x l 9 8 ) - ( 1 0 2 x l 0 2 ) _ ( 1 9 8 + 1 0 2 ) ( 1 9 8 - 1 0 2 )

- > 9 6 96

_ _ ( 1 9 8 x 1 9 8 ) - ( 1 0 2 x 1 0 2 ) _ 3 0 0 x 9 6

- > 9 6

_ _ ( 1 9 8 x 1 9 8 ) - ( 1 0 2 x 1 0 2 )

- > 9 6

96

= 300

(iv) (1.73 x 1 .7 3 )-(0 .2 7 x 0.27)

sol:


=> (1.73 x 1 .7 3 )-(0 .2 7 x 0.27) = (1 .73 )2- (0 .2 7 )2

=> (1.73 X 1 .7 3 )- (0.27 X 0.27) = (1.73 + 0.27) (1.73 - 0.27)

=> (1.73 x 1 .7 3 )-(0 .2 7 x 0.27) = 2 x 1.46

=> (1.73 x 1 .7 3 )-(0 .2 7 x 0.27) =2.92

( 8 . 6 3 x 8 . 6 3 ) - ( l . 3 7 x 1 .3 7 )(V)

sol

0 .7 2 6

we know that, (a + b) (a - b) = a2- b2( 8 .6 3 x 8 .6 3 ) - ( 1 . 3 7 x 1 .3 7 ) ( 8 .6 3 ) 2- ( 1 . 3 7 ) 2

0 .7 2 6

( 8 .6 3 x 8 .6 3 ) - ( 1 . 3 7 x 1 .3 7 )

0 7 2 6

( 8 .6 3 x 8 .6 3 ) - ( 1 . 3 7 x 1 .3 7 )

0 .7 2 6

( 8 .6 3 x 8 .6 3 ) - ( 1 . 3 7 x 1 .3 7 )

0 7 2 6

( 8 .6 3 x 8 .6 3 ) - ( 1 . 3 7 x 1 .3 7 )

0 .7 2 6

( 8 .6 3 + 1 .3 7 ) ( 8 .6 3 - 1 .3 7 )

0 .7 2 6

_ 1 0 x 7 .2 6 " 0 .7 2 6

10 X 10

1000 .7 2 6

Q6 Find the value of x, if:

(I) 4a; = (52 )2- (48 )2

sol:

we know that, (a + b) (a - b) = a2- ti2

=> 4x = (52 )2- (48 )2

=> 4x = (52 + 48) (52 - 48)

=> 4x = 100 x 4

=> 4x = 400

=> x = —

=> x = 100

(II) 1 4 * = (47 )2- (33)

sol:


=> 14a: = (47 )2- (33 )2

=> 14x = (47 + 33) (47 - 33)

=> 14x = 80 x 14

=> 14x = 1120

1120=> x= -u => x = 80

(iii) 5a; = (50 )2- (40 )2

sol:


=> 5x = (50 )2- (40 )2

=> 5x = (50 + 40) (50 - 40)

=> 5x = 90 X 10

=> 5x = 900

=> x = —X 5

=> x = 180

Q7 If x + j = 20 , find the value o f a:2 +

sol:

Given that,

a; + | = 20

squaring on both sides

= > ( a ; + i ) 2 = (20 )2

=> ( * + i ) 2 = 400

we know that, (a + ft)2 = a2 + 2ab + b2

=> x2 + 2 x x x ^ + ( ^ ) 2 = 400

=> a:2 + 2 + (-1)2 = 400

=>x2 + \ = 4 0 0 - 2X2

=> a;2 + j - = 398X 2

hence, a;2 + = 398X1Q8 If a;- = 3, find the values o f x2 + a:4 +

sol:

Given that,

x - ± = 3X


= > ( a ; - i ) 2 = (3 )2

= > ( x - i ) 2 = 9

we know that, ( a - 6)2 = a2- 2ab + b2

=>x2-2 x x x - + ( - ) 2 = 9

=>a:2- 2 + ( ± ) 2 = 9

=> x2 + = 9 + 2X2

=>a;2 + = 11X2

Again, squaring on both sides

= > ( * 2 + ^ ) 2 = ( H ) 2

=>(x2 + ± ) 2 = 121

=> a:2 + 2 x (x2) x (p - ) + ( p - ) 2 = 121

=> x* + 2 + ± = 121X4

=>x* + ± = 121-2X4

=>x* + ± = 119

hence, a:4 + = 119

Q9 If a:2 + = 18, find the values of x + x -

sol:

Given that,

x2 + 4 r = 18, find the values o f x + —, x - —x2 X Xconsider, x + —

X

squaring the above equation

( x + i ) 2 =a;2 + 2 x I X i + ( I ) 2

= a:2 + 2 + -V

= > ( * + ± ) 2 =a!2 + 2 + £

=> (a: + ■■)2 = 18 + 2

=> (x + \ ? = 20

=>a; + ^ = ± \ /2 0

consider, a:- —X

squaring the above equation

(x - ± ) 2 = x 2- 2 x x x ± + ( ± ) 2

= x 2- 2 +

= > {x -\ )2- x 2-2 + ^

= > ( a i - | ) 2 = 1 8 - 2

= > ( * - i ) 2 =16

=> x - ^ = ± \ /2 0

=> x - - = ± 4X

Q10 If x + y = 4 and xy = 2,find the value of a:2 + y2

sol:

Given that,

x + y = 4 and xy = 2


=> x2 + y2 = (x + y)2- 2xy

=> x2 + y2 = 42- (2 x 2)

=> x2 + y2 = 1 6 -4

=> x2 + y2 = 12

Q11 If x - y = 7 and xy = 9,find the value o f x2 + y2

sol:

Given that,

x - y = 7 and xy = 9

we know that, ( a - b)2 = a2- 2a6 + 62

=> x2- y 2 — (x -y )2 + 2 xy

=> x2- y 2 = 72 + (2 x 9)

=> x2- y 2 = 4 9 + 18

=>x2- y 2 = 67

Q12 If 3x + 5y = 11 and xy = 2,find the value o f 9a;2 + 25y2

sol:

Given that,

3x + 5y = 11 and xy = 2

we know that, (a + b)2 = a2 + 2ab + b2

(Zx + 5y)2 = (3a:)2 + 2 x (3a:) x (5y) + {by)2

=> (3a: + by)2 = 9x2 + ZOxy + 2by2

=> 9a:2 + 25y2 = (3a: + 5 y )2- 10xy

=> 9a:2 + 25y2 = ( l l ) 2- (30 x 2)

=>9x2 + 2by2 = 1 2 1 -6 0

=> 9a:2 + 2by2 = 61

Q13 Find the values o f the following expressions

(i) 16a;2 + 24a: + 9 when x = ^

Sol:

Given, 16a:2 + 24a: + 9 and x = j

we know that, (o + b)2 = a2 + 2ab + b2

16a:2 + 24a: + 9 = (4a: + 3 )2

=> 16a:2 + 24a: + 9 = ( 4 ( ^ ) + 3 )2

=> 16a:2 + 24a: + 9 = (7 + 3 )2

=> 16a:2 + 24a: + 9 = (10 )2

=> 16a:2 + 24a:+ 9 = 100

(ii) 64a:2 + 81 y2 + 144xy when x = 11 and y = f

sol:

Given, 64a;2 + 81 y2 + 144a; y and x = 11, y = |


64a:2 + 81y2 + 144xy = (8x + 9y)2

=> 64x2 + 81 y2 + \44xy = (8(11) + 9 ( | ) ) 2

=> 64a:2 + 81 j/2 + 144xy = (88 + 12)2

=> 64a:2 + 81y2 + 144xy = (100)2

=> 64a:2 + 81y2 + 144a;y = 10000

(Hi) 81a:2 + 16y2- 72xy when x = |

and y = |

sol:

Given that, 81a:2 + 16y2- 72xy and x = -|,

y = f

we know that, ( a - 6)2 = a2- 2ab + b2

81a:2 + 16y2-72xy = (9x-4y)2

=> 81a:2 + 16y2-72xy = ( 9 ( f ) - 4 ( f ) ) 2

=> 81a:2 + lby2-72xy = ( 6 - 3 ) 2

=> 81a:2 + 16y2-72xy = 32

=> 81a:2 + lby2-72xy = 9

Q14 If x + -7 = 9, find the value o f a;4 +X x4

sol:

Given,

* + - =9X


( * + ! ) 2 = 92

= > (a s + ± )2 = 81

= > i2 + 2 x i x i + ( i ) 2 = 81

=> x2 + 2 + -V = 81X*=>x2 + \ = 8 1 - 2X2

=>*2 + 4 = 79X2

Again, squaring on both sides

( * 2 + 4 ) 2 = (79 )2

=> (z 2 + ^ ) 2 = 6241

=> ( * 2) 2 + 2 x (a:2) x (^+ ) + ( + - ) 2 = 6241

=> x4 + 2 + 4 = 6241X4

=>x4 + ± = 6 2 4 1 - 2X4

=> * 4 + 4 = 6239X4

Q15 If x + + = 12, find the value o f as- ^

sol:

Given that,

* + 7 = 1 2


( * + £ ) 2 = (12 )2

= > ( a s + i ) 2 = 144

=>x2 + 2 x x x ^ + ( i ) 2 = 144

=> a:2 + 2 + 4 = 144X2

=>*2 + 4 = 144_2X2

=> x2 + 4 = 142X2

Here,

( , 4 ) 2 = , 2- 2 x J : x i + ( l ) 2

= a:2- 2 + 4XZ

= > ( * - i ) 2 = * 2 - 2 + 4v X f x 2

=> ( z - + ) 2 = 1 4 2 -2

= > ( z - + ) 2 = 140

=> * _ 7 = + V 4 0

Q16 If 2x + 3y = 14 and 2x - 3y = 2, find the value of xy

Sol:

we know that, (a + b)(a-b) = a2-b 2

Given, 2x + 3y = 14 and 2x - 3y = 2

squaring (2x + 3y) and (2x - 3y) and then subtracting them, we get

(2x + 3y)2- {2x -3y )2 = [{2x + 3 y) + (2 z -3 y ) ] [ (2 z + 3y)-(2x-3y)}

=> (2 z + 3y)2- (2x- 3y)2 = 4a; x 6y

=> (2 z + 3y)2- ( 2 z - 3y)2 = 24xy

=> (14 )2- ( 2 ) 2 = 2Axy

=> (14 + 2) (14 - 2) = 24xy

=> 16 x 12 = 24xy

=> 24xy = 192

1 9 2=> xy = - | -

=> xy = 8

hence, xy = 8

Q17 If x2 + y2 = 29 and xy = 2, Find the value of

( i) x + y

sol

Given,

x2 + y2 = 29 and xy = 2

squaring the (x + y)

( i + i/)2 = i 2+ 2 x j x i / + !/2 => (x + y)2 = 29 + (2 x 2)

=> (x + y)2 = 29 + 4

=> (x + y)2 = 33

=>x + y = ± \ /3 3

(ii) x - y

sol:

Given,

x2 + y2 = 29 and xy = 2

squaring the (x - y)

(x -y )2 = x 2- 2 x x x y + y2

=> {x -y f = 2 9 - ( 2 x 2 )

=> (x -y )2 = 2 9 -4

=> (x -y )2 = 25

=> x -y = ± v ^ 5

=>x + y = ± 5

(iii) + y4

Sol:

Given,

x2 + y2 = 29 and xy = 2

(x2 + 112)2 = x4 + 2 x2y2 + 114

=>x4 + y4 = (x2+ y 2)2-2x2y2

=> x4 + y4 = (x2 + y2)2-2 (xy )2

=> x4 + y4 — (29 )2- 2 ( 2 ) 2

=> x4 + y4 = 8 4 1 - 8

=> x4 + y4 = 833



Q1) Find the following products:

(i) (x + 4) (x + 7)

(II) (x - 11) (x + 4)

(ill) (x + 7) (x - 5)

(iv) (x - 3) (x - 2)

(v) (y2 - 4) (y2 - 3)

(vi) ( * + ! ) ( * + f )

(vii) (3x + 5) (3x + 11)

(viii) (2x2 - 3) (2x2 + 5)

(ix) (z2 + 2) (z2 - 3)

(x) (3x - 4y) (2x - 4y)

(xi) (3x2 - 4xy) (3x2 - 3xy)

(xii) ( * + i ) (a: + 5)

(xiii) ( z + £ ) ( z + f )

(xiv) (x2 + 4) (x2 + 9)

(xv) (y2 + 12) (y2 + 6)

(xvi) (y2 + | ) {y2~ f )

(xvii) (p2 + 16) (p2- ! )

Solution:

(i) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.

(x + 4) (x + 7)

= a:2 + (4 + 7)x + 4 x 7

= a:2 + 11a; + 28

(ii) Here, we will use the identity (x - a) (x + b) = x2 + (b - a)x - ab.

(x - 11) (x + 4)

= a:2 + (4 — l l ) a : - 1 1 x 4

= a;2- 7a;- 44

(iii) Here, we will use the identity (x + a) (x - b) = x2 + (a - b)x - ab.

(x + 7 ) ( x - 5 )

= a:2 + (7 — 5)a :- 7 x 5

= a:2 + 2a:- 35

(iv) Here, we will use the identity (x - a) (x - b) = x2 - (a + b)x + ab.

(x - 3) (x - 2)

= a:2- (3 + 2)a; + 3 x 2

= a;2-5 a ; + 6

(v) Here, we will use the identity (x - a) (x - b) = x2 - (a + b)x - ab.

(V2 - 4) (y2 - 3)

= (y2)2~ (4 + 3 ) (y 2) + 4 x 3

- y4~7y2 + 1 2

(vi) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.

(a: + f ) (a: + f )

= a:2 + ( f + £ ) a : + f x £ )

= o:2 + f a : + l

(vii) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.

(3x + 5) (3x + 11)

= (3 x )2 + (5 + 11) (3a;) + 5 x 11

= 9a:2 + 4 8 * + 55

(viii) Here, we will use the identity (x - a) (x + b) = x2 + (b - a)x - ab.

(2x2 - 3) (2x2 + 5)

= (2a:2) 2 + (5 -3 ) (2a:2) - 3 x 5

= 4x4 + Ax2- 15

(ix) Here, we will use the identity (x + a) (x - b) = x2 + (a - b)x - ab.

(z2 + 2) (z2 - 3)

= (z2)2 + (2-3)(z2)-2x3 = z 4- z2- 6

(x) Here, we will use the identity (x - a) (x - b) = x2 - (a + b)x - ab.

(3x - 4y) (2x - 4y)

= (4y - 3x) (4y - 2x) (Taking common -1 from both parentheses)

= (4y)2- (3a; + 2x) (4y) + 3x x 2x

= 16y2- (12xy + 8xy) + 6x2

= 16 y2- 20 xy + 6x2

(xi) Here, we will use the identity (x - a) (x - b) = x2 - (a + b)x - ab.

(3x2 - 4xy) (3x2 - 3xy)

= (3a;2) 2- (4xy + 3xy) (3a;2) + 4xy x 3xy

= 9a;4- (12 x3y + 9 x3y) + 12 x2y2

= 9xi-21x3y+12x2y2

(xii) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.

( * + |) (* + 5)

= a;2 + ( | + 5)a; + | x 5

= a;2 + f x + 1

(xiii) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.

( * + ! ) ( * + { )

= ,2 + (f + i)a; + f x {

= z2 + l l * + 1

(xiv) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.

(x2 + 4) (x2 + 9)

= (a;2)2 + (4 + 9)(a:2) + 4 x 9

= a;4 + 13a;2 + 36

(xv) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.

( ^ + 12) (y2 + 6)

= (y2)2 + (12 + 6)(j/2) + 12 x 6 = yi + 18xy2 + 72

(xvi) Here, we will use the identity (x + a) (x - b) = x2 + (a - b)x -ab.

(y2 + f ) (y2- f )

= (y2)2 + ( j - f ) ( y 2) - j x f

=y4- 7i y 2- 2(xvii) Here, we will use the identity (x + a) (x - b) = x2 + (a - b)x -ab.

(p? + 16)(p?-I)= (p2)2 + (16-i)(p»)- 16xi = P4 + f p 2- 4

Q2. Evaluate the following:

( i) 102x106

(ii) 109x107

(iii) 35 x37

(iv) 53 x 55

(v) 103x96

(vi) 34 x 36

(vii) 994x1006

Solution:

(i) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab

102x106

= (100+ 2) (100+ 6)

= 1002 + (2 + 6)100 + 2 x 6

=10000 + 800 +12 = 10812

(ii) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab

109x107

= (100 + 9) (100 + 7)

= 1002 + (9 + 7)100+ 9 x 7

=10000 + 1600 + 63 = 11663

(iii) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab

35x37

= (30 + 5) (30 + 7)

= 302 + (5 + 7)30 + 5 x 7

= 900 + 360 + 35 = 1295

(iv) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab

53x55

= (50 + 3) (50 + 5)

= 502 + (3 + 5)50 + 3 x 5

= 2500 + 400 + 15 = 2915

(v) Here, we will use the identity (x + a) (x - b) = x2 + (a - b)x - ab

103x96

= (100 + 3) (10 0 - 4 )

= 1002 + (3 - 4)100 - 3 x 4

= 10000- 1 0 0 - 12 = 9888

(vi) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab

34x36

= (30+ 4) (30+ 6)

= 302 + (4 + 6)30 + 4 x 6

= 900 + 300 + 24 = 1224

(vii) Here, we will use the identity (x - a) (x + b) = x2 + (b - a)x - ab

994x1006

= (1000 - 6) x (1000 + 6)

= 10002 + (6 - 6) x 1000 - 6 x 6

= 1000000-36 = 999964

class 8 algebraic expressions and identities …...class 8 algebraic expressions and identities...

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