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Q1. Identify the terms, their coefficients for each of the following expressions. (i) 5xyz2 – 3zy (ii) 1 + x + x2 (iii) 4x2 y2 – 4x2 y2 z2 + z2 (iv) 3 – pq + qr – p (v) (x/2) + (y/2) – xy (vi) 0.3a – 0.6ab + 0.5b
Solution :
| Sl. No. | Expression | Term | Coefficient |
| i) | 5xyz2 – 3zy |
Term: 5xyz
2
Term: -3zy |
5 -3 |
| ii) | 1 + x + x2 |
Term: 1
Term: x Term: x2 |
1 1 1 |
| iii) | 4x2 y2 – 4x2 y2 z2 + z2 |
Term: 4
x2
y
2
Term: -4 x2 y 2 z 2 Term : z 2 |
4 -4 1 |
| iv) | 3 – pq + qr – p | Term : 3 -pq qr -p | 3 -1 1 -1 |
| v) | (x/2) + (y/2) – xy | Term : x/2 Y/2 -xy | ½ 1/2 -1 |
| vi) | 0.3a – 0.6ab + 0.5b | Term : 0.3a -0.6ab 0.5b | 0.3 -0.6 0.5 |
2. Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories? x + y, 1000, x + x2 + x3 + x4 , 7 + y + 5x, 2y – 3y2 , 2y – 3y2 + 4y3 , 5x – 4y + 3xy, 4z – 15z2 , ab + bc + cd + da, pqr, p2 q + pq2 , 2p + 2q
Solution:
Let us first define the classifications of these 3 polynomials:
Monomials contain only one term.
Binomials contain only two terms.
Trinomials contain only three terms.
| x + y | two terms | Binomial |
| 1000 | one term | Monomial |
| x + x2 + x3 + x4 | four terms | Polynomial, and it does not fit in the listed three categories |
| 2y – 3y2 | two terms | Binomial |
| 2y – 3y2 + 4y 3 | three terms | Trinomial |
| 5x – 4y + 3xy | three terms | Trinomial |
| 4z – 15z 2 | two terms | Binomial |
| ab + bc + cd + da | four terms | Polynomial, and it does not fit in the listed three categories |
| pqr | one term | Monomial |
| p2 q + pq2 | two terms | Binomial |
| 2p + 2q | two terms | Binomial |
| 7 + y + 5x | three terms | Trinomial |
3. Add the following.
(i) ab – bc, bc – ca, ca – ab
(ii) a – b + ab, b – c + bc, c – a + ac
(iii) 2p2 q2 – 3pq + 4, 5 + 7pq – 3p2 q2
(iv) l2 + m2 , m2 + n2 , n2 + l2 , 2lm + 2mn + 2nl
Solution:
i) (ab – bc) + (bc – ca) + (ca-ab)
= ab – bc + bc – ca + ca – ab
= ab – ab – bc + bc – ca + ca
= 0
ii) (a – b + ab) + (b – c + bc) + (c – a + ac)
= a – b + ab + b – c + bc + c – a + ac
= a – a +b – b +c – c + ab + bc + ca
= 0 + 0 + 0 + ab + bc + ca
= ab + bc + ca
iii) 2p2 q 2 – 3pq + 4, 5 + 7pq – 3p2 q2
= (2p2 q2 – 3pq + 4) + (5 + 7pq – 3p2 q2 )
= 2p2 q2 – 3p2 q2 – 3pq + 7pq + 4 + 5
= – p2 q2 + 4pq + 9
iv) (l2 + m2 ) + (m2 + n2 ) + (n2 + l2 ) + (2lm + 2mn + 2nl)
= l2 + l2 + m2 + m2 + n2 + n2 + 2lm + 2mn + 2nl
= 2l2 + 2m2 + 2n2 + 2lm + 2mn + 2nl
4. (a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3
(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
(c) Subtract 4p2 q – 3pq + 5pq2 – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq2 + 5p2 q
Solution:
(a) (12a – 9ab + 5b – 3) – (4a – 7ab + 3b + 12)
= 12a – 9ab + 5b – 3 – 4a + 7ab – 3b – 12
= 12a – 4a -9ab + 7ab +5b – 3b -3 -12
= 8a – 2ab + 2b – 15
b) (5xy – 2yz – 2zx + 10xyz) – (3xy + 5yz – 7zx)
= 5xy – 2yz – 2zx + 10xyz – 3xy – 5yz + 7zx
=5xy – 3xy – 2yz – 5yz – 2zx + 7zx + 10xyz
= 2xy – 7yz + 5zx + 10xyz
c) (18 – 3p – 11q + 5pq – 2pq2 + 5p2 q) – (4p2 q – 3pq + 5pq2 – 8p + 7q – 10)
= 18 – 3p – 11q + 5pq – 2pq2 + 5p2 q – 4p2 q + 3pq – 5pq2 + 8p – 7q + 10
=18+10 -3p+8p -11q – 7q + 5 pq+ 3pq- 2pq^2 – 5pq^2 + 5 p^2 q – 4p^2 q
= 28 + 5p – 18q + 8pq – 7pq2 + p2 q
1. Find the product of the following pairs of monomials.
(i) 4, 7p
(ii) – 4p, 7p
(iii) – 4p, 7pq
(iv) 4p3 , – 3p
(v) 4p, 0
Solution:
(i) 4 , 7 p = 4 × 7 × p = 28p
(ii) – 4p × 7p = (-4 × 7 ) × (p × p )= -28p2
(iii) – 4p × 7pq =(-4 × 7 ) (p × pq) = -28p2 q
(iv) 4p3 × – 3p = (4 × -3 ) (p3 × p ) = -12p4
(v) 4p × 0 = 0
2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths, respectively.
(p, q) ; (10m, 5n) ; (20x2 , 5y2 ) ; (4x, 3x2 ) ; (3mn, 4np)
Solution:
Area of rectangle = Length x breadth. So, it is multiplication of two monomials.
The results can be written in square units.
(i) p ×q = pq
(ii)10m × 5n = 50mn
(iii) 20x2 × 5y2 = 100x2 y2
(iv) 4x × 3x2 = 12x3
(v) 3mn × 4np = 12mn2 p
3. Complete the following table of products:
Solution:
4. Obtain the volume of rectangular boxes with the following length, breadth and height, respectively.
(i) 5a, 3a2 , 7a 4
(ii)2p, 4q, 8r
(iii) xy, 2x2 y, 2xy 2
(iv)a, 2b, 3c
Solution:
Volume of rectangle = length x breadth x height. To evaluate volume of rectangular boxes, multiply all the monomials.
(i) 5a x 3a 2 x 7a 4 = (5 × 3 × 7) (a × a 2 × a 4 ) = 105a7
(ii) 2p x 4q x 8r = (2 × 4 × 8 ) (p × q × r ) = 64pqr
(iii) y × 2x2 y × 2xy 2 =(1 × 2 × 2 )( x × x2 × x × y × y × y 2 ) = 4x4 y4
(iv) a x2b x3c = (1 × 2 × 3 ) (a × b × c) = 6abc
5. Obtain the product of
(i) xy, yz, zx
(ii) a, – a2 , a3
(iii) 2, 4y, 8y2 , 16y3
(iv) a, 2b, 3c, 6abc
(v) m, – mn, mnp
Solution:
(i) xy × yz × zx = x2 y 2 z 2
(ii) a × – a2 × a3 = – a6
(iii) 2 × 4y × 8y2 × 16y3 = 1024 y6
(iv) a × 2b × 3c × 6abc = 36a2 b2 c2
(v) m × – mn × mnp = –m3 n2 p
1. Carry out the multiplication of the expressions in each of the following pairs.
(i) 4p, q + r
(ii) ab, a – b
(iii) a + b, 7a²b²
(iv) a2 – 9, 4a
(v) pq + qr + rp, 0
Solution :
(i)4p(q + r) = 4pq + 4pr
(ii)ab(a – b) = a2 b – a b2
(iii)(a + b) (7a2 b2 ) = 7a3 b2 + 7a2 b3
(iv) (a2 – 9)(4a) = 4a3 – 36a
(v) (pq + qr + rp) × 0 = 0 ( Anything multiplied by zero is zero )
2. Complete the table.
Solution:
| First expression | Second expression | Product | |
| (i) | a | b + c + d |
a(b+c+d)
= a×b + a×c + a×d = ab + ac + ad |
| (ii) | x + y – 5 | 5xy |
5 xy (x + y – 5)
= 5 xy x x + 5 xy x y – 5 xy x 5 = 5 x2 y + 5 xy2 – 25xy |
| (iii) | p | 6p2 – 7p + 5 |
p (6 p 2
-7 p +5)
= p× 6 p2 – p× 7 p + p×5 = 6 p3 – 7 p2 + 5 p |
| (iv) | 4 p2 q2 | P2 – q2 |
4p2
q2
* (p2
– q2
)
=4 p4 q2 – 4p2 q4 |
| (v) | a + b + c | abc |
abc(a + b + c)
= abc × a + abc × b + abc × c = a2 bc + ab2 c + abc2 |
3. Find the product.
i) a2 x (2a22 ) x (4a26 )
ii) (2/3 xy) ×(-9/10 x2 y2 )
(iii) (-10/3 pq3 /) × (6/5 p3 q)
(iv) (x) × (x2 ) × (x3 ) × (x4 )
Solution:
i) a2 x (2a22 ) x (4a26 )
= (2 × 4) ( a2 × a22 × a26 )
= 8 × a2 + 22 + 26
= 8a50
ii) (2xy/3) ×(-9x2 y2 /10)
=(2/3 × -9/10 ) ( x × x2 × y × y2 )
= (-3/5 x3 y3 )
iii) (-10pq3 /3) ×(6p3 q/5)
= ( -10/3 × 6/5 ) (p × p3 × q3 × q)
= (-4p4 q4 )
iv) ( x) x (x2 ) x (x3 ) x (x4 )
= x1 + 2 + 3 + 4
= x10
4. (a) Simplify 3x (4x – 5) + 3 and find its values for (i) x = 3 (ii) x =1/2
(b) Simplify a (a2 + a + 1) + 5 and find its value for (i) a = 0 , (ii) a = 1 (iii) a = – 1 .
Solution:
a) 3x (4x – 5) + 3
=3x ( 4x) – 3x( 5) +3
=12x2 – 15x + 3
(i) Putting x=3 in the equation we gets 12x2 – 15x + 3 =12(32 ) – 15 (3) +3
= 108 – 45 + 3
= 66
(ii) Putting x=1/2 in the equation we get
12x2 – 15x + 3 = 12 (1/2)2 – 15 (1/2) + 3
= 12 (1/4) – 15/2 +3
= 3 – 15/2 + 3
= 6- 15/2
= (12- 15 ) /2
= -3/2
b) a(a2 +a +1)+5
= a x a2 + a x a + a x 1 + 5 =a3 +a2 +a+ 5
(i) putting a=0 in the equation we get 03 +02 +0+5=5
(ii) putting a=1 in the equation we get 13 + 12 + 1+5 = 1 + 1 + 1+5 = 8
(iii) Putting a = -1 in the equation we get (-1)3 +(-1)2 + (-1)+5 = -1 + 1 – 1+5 = 4
5. (a) Add: p ( p – q), q ( q – r) and r ( r – p)
(b) Add: 2x (z – x – y) and 2y (z – y – x)
(c) Subtract: 3l (l – 4 m + 5 n) from 4l ( 10 n – 3 m + 2 l )
(d) Subtract: 3a (a + b + c ) – 2 b (a – b + c) from 4c ( – a + b + c )
Solution:
a) p ( p – q) + q ( q – r) + r ( r – p)
= (p2 – pq) + (q2 – qr) + (r2 – pr)
= p2 + q2 + r2 – pq – qr – pr
b) 2x (z – x – y) + 2y (z – y – x)
= (2xz – 2x2 – 2xy) + (2yz – 2y2 – 2xy)
= 2xz – 4xy + 2yz – 2x2 – 2y2
c) 4l ( 10 n – 3 m + 2 l ) – 3l (l – 4 m + 5 n)
= (40ln – 12lm + 8l2 ) – (3l2 – 12lm + 15ln)
= 40ln – 12lm + 8l2 – 3l2 +12lm -15 ln
= 25 ln + 5l2
d) 4c ( – a + b + c ) – (3a (a + b + c ) – 2 b (a – b + c))
= (-4ac + 4bc + 4c2 ) – (3a2 + 3ab + 3ac – ( 2ab – 2b2 + 2bc ))
=-4ac + 4bc + 4c2 – (3a2 + 3ab + 3ac – 2ab + 2b2 – 2bc)
= -4ac + 4bc + 4c2 – 3a2 – 3ab – 3ac +2ab – 2b2 + 2bc
= -7ac + 6bc + 4c2 – 3a2 – ab – 2b2
1. Multiply the binomials.
(i) (2x + 5) and (4x – 3)
(ii) (y – 8) and (3y – 4)
(iii) (2.5l – 0.5m) and (2.5l + 0.5m)
(iv) (a + 3b) and (x + 5)
(v) (2pq + 3q2 ) and (3pq – 2q2 )
(vi) (3/4 a2 + 3b2 ) and 4( a2 – 2/3 b2 )
Solution :
(i) (2x + 5)(4x – 3)
= 2x x 4x – 2x x 3 + 5 x 4x – 5 x 3
= 8x² – 6x + 20x -15
= 8x² + 14x -15
ii) ( y – 8)(3y – 4)
= y x 3y – 4y – 8 x 3y + 32
= 3y2 – 4y – 24y + 32
= 3y2 – 28y + 32
(iii) ( 2.5l– 0.5m)(2.5l + 0.5m)
= 2.5l x 2.5 l + 2.5l x 0.5m – 0.5m x 2.5l – 0.5m x 0.5m
= 6.25l2 + 1.25 lm – 1.25 lm – 0.25 m 2
= 6.25l2 – 0.25 m 2
iv) (a + 3b) (x + 5)
= ax + 5a + 3bx + 15b
v) (2pq + 3q2 ) (3pq – 2q2 )
= 2pq x 3pq – 2pq x 2q2 + 3q2 x 3pq – 3q2 x 2q2
= 6p2 q2 – 4pq3 + 9pq3 – 6q4
= 6p2 q2 + 5pq3 – 6q4
(vi) (3/4 a² + 3b² ) and 4( a² – 2/3 b² )
=(3/4 a² + 3b² ) x 4( a² – 2/3 b² )
=(3/4 a² + 3b² ) x (4a² – 8/3 b² )
=3/4 a² x (4a² – 8/3 b² ) + 3b² x (4a² – 8/3 b² )
=3/4 a² x 4a² -3/4 a² x 8/3 b² + 3b² x 4a² – 3b² x 8/3 b²
=3a4 – 2a² b² + 12 a² b² – 8b 4
= 3a4 + 10a² b² – 8b4
2. Find the product.
(i) (5 – 2x) (3 + x)
(ii) (x + 7y) (7x – y)
(iii) (a2 + b) (a + b2 )
(iv) (p2 – q2 ) (2p + q)
Solution:
(i) (5 – 2x) (3 + x)
= 5 (3 + x) – 2x (3 + x)
=15 + 5x – 6x – 2x2
= 15 – x -2 x 2
(ii) (x + 7y) (7x – y)
= x(7x-y) + 7y ( 7x-y)
=7x2 – xy + 49xy – 7y2
= 7x2 – 7y2 + 48xy
iii) (a2 + b) (a + b2 )
= a2 (a + b2 ) + b(a + b2 )
= a3 + a2 b2 + ab + b3
= a3 + b3 + a2 b2 + ab
iv) (p2 – q2 ) (2p + q)
= p2 (2p + q) – q2 (2p + q)
=2p3 + p2 q – 2pq2 – q3
= 2p3 – q3 + p2 q – 2pq2
3. Simplify.
(i) (x2 – 5) (x + 5) + 25
(ii) (a2 + 5) (b3 + 3) + 5
(iii)(t + s2 )(t2 – s)
(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
(v) (x + y)(2x + y) + (x + 2y)(x – y)
(vi) (x + y)(x2 – xy + y2 )
(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y
(viii) (a + b + c)(a + b – c)
Solution:
i) (x2 – 5) (x + 5) + 25
= x3 + 5x2 – 5x – 25 + 25
= x3 + 5x2 – 5x
ii) (a2 + 5) (b3 + 3) + 5
= a2 b3 + 3a2 + 5b3 + 15 + 5
= a2 b3 + 5b3 + 3a2 + 20
iii) (t + s2 )(t2 – s)
= t (t2 – s)+ s2 (t2 – s)
= t3 – st + s2 t2 – s3
= t3 – s3 – st + s2 t2
iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
= (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
=(ac – ad + bc – bd) + (ac + ad – bc – bd) + (2ac + 2bd)
= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
= 4ac
v) (x + y)(2x + y) + (x + 2y)(x – y)
= 2x2 + xy + 2xy + y2 + x2 – xy + 2xy – 2y2
= 3x2 + 4xy – y2
vi) (x + y)(x2 – xy + y2 )
= x3 – x2 y + xy2 + x2 y – xy2 + y3
= x3 + y3
vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y
= 2.25x2 + 6xy + 4.5x – 6xy – 16y2 – 12y – 4.5x + 12y = 2.25x2 – 16y2
viii) (a + b + c)(a + b – c)
= a2 + ab – ac + ab + b2 – bc + ac + bc – c2
= a2 + b2 – c2 + 2ab
1. Use a suitable identity to get each of the following products.
(i) (x + 3) (x + 3)
(ii) (2y + 5) (2y + 5)
(iii) (2a – 7) (2a – 7)
(iv) (3a – 1/2)(3a – 1/2)
(v) (1.1m – 0.4) (1.1m + 0.4)
(vi) (a2 + b2 ) (- a2 + b2 )
(vii) (6x – 7) (6x + 7)
(viii) (- a + c) (- a + c)
(ix) (1/2x + 3/4y) (1/2x + 3/4y)
(x) (7a – 9b) (7a – 9b)
Solution:
(i) (x + 3) (x + 3) = (x + 3)2
= x2 + 6x + 9
Using (a+b)2 = a2 + b2 + 2ab
ii) (2y + 5) (2y + 5) = (2y + 5)2
= 4y2 + 20y + 25
Using (a+b)2 = a2 + b2 + 2ab
iii) (2a – 7) (2a – 7) = (2a – 7)2
= 4a2 – 28a + 49
Using (a-b)2 = a2 + b2 – 2ab
iv) (3a – 1/2)(3a – 1/2) = (3a – 1/2)2
= 9a2 -3a+(1/4)
Using (a-b)2 = a2 + b2 – 2ab
v) (1.1m – 0.4) (1.1m + 0.4)
= 1.21m2 – 0.16
Using (a – b)(a + b) = a2 – b2
vi) (a2 + b2 ) (– a2 + b2 )
= (b2 + a2 ) (b2 – a2 )
= -a4 + b4
Using (a – b)(a + b) = a2 – b2
vii) (6x – 7) (6x + 7)
=36x2 – 49
Using (a – b)(a + b) = a2 – b2
viii) (– a + c) (– a + c) = (– a + c)2
= c2 + a2 – 2ac
Using (a-b)2 = a2 + b2 – 2ab
= (x2 /4) + (9y2 /16) + (3xy/4)
Using (a+b)2 = a2 + b2 + 2ab
x) (7a – 9b) (7a – 9b) = (7a – 9b)2
= 49a2 – 126ab + 81b2
Using (a-b)2 = a2 + b2 – 2ab
2. Use the identity (x + a) (x + b) = x2 + (a + b) x + ab to find the following products.
(i) (x + 3) (x + 7)
(ii) (4x + 5) (4x + 1)
(iii) (4x – 5) (4x – 1)
(iv) (4x + 5) (4x – 1)
(v) (2x + 5y) (2x + 3y)
(vi) (2a2 + 9) (2a2 + 5)
(vii) (xyz – 4) (xyz – 2)
Solution:
(i)(x + 3) (x + 7)
= x2 + (3+7)x + 21
= x2 + 10x + 21
ii) (4x + 5) (4x + 1)
= 16x2 + 4x + 20x + 5
= 16x2 + 24x + 5
iii) (4x – 5) (4x – 1)
= 16x2 – 4x – 20x + 5
= 16x2 – 24x + 5
iv) (4x + 5) (4x – 1)
= 16x2 + (5-1)4x – 5
= 16x2 +16x – 5
v) (2x + 5y) (2x + 3y)
= 4x2 + (5y + 3y)2x + 15y2
= 4x2 + 16xy + 15y2
vi) (2a2 + 9) (2a2 + 5)
= 4a4 + (9+5)2a2 + 45
= 4a4 + 28a2 + 45
vii) (xyz – 4) (xyz – 2)
= x2 y2 z2 + (-4 -2)xyz + 8
= x2 y2 z2 – 6xyz + 8
3. Find the following squares by using the identities.
(i) (b – 7)2
(ii) (xy + 3z)2
(iii) (6x2 – 5y)2
(iv) [(2m/3) + (3n/2)]2
(v) (0.4p – 0.5q)2
(vi) (2xy + 5y)2
Solution:
Using identities:
(a – b)2 = a2 + b2 – 2ab (a + b)2 = a2 + b2 + 2ab
(i) (b – 7)2 = b2 – 14b + 49
(ii) (xy + 3z)2 = x2 y2 + 6xyz + 9z2
(iii) (6x2 – 5y)2 = 36x4 – 60x2 y + 25y2
(iv) [(2m/3}) + (3n/2)]2 = (4m2 /9) +(9n2 /4) + 2mn
(v) (0.4p – 0.5q)2 = 0.16p2 – 0.4pq + 0.25q2
(vi) (2xy + 5y)2 = 4x2 y2 + 20xy2 + 25y2
4. Simplify.
(i) (a2 – b2 )2
(ii) (2x + 5)2 – (2x – 5)2
(iii) (7m – 8n)2 + (7m + 8n)2
(iv) (4m + 5n)2 + (5m + 4n)2
(v) (2.5p – 1.5q)2 – (1.5p – 2.5q)2
(vi) (ab + bc)2 – 2ab²c
(vii) (m2 – n2 m)2 + 2m3 n2
Solution:
i) (a2 – b2 )2 = a4 + b4 – 2a2 b2
ii) (2x + 5)2
– (2x – 5)2
= 4x2
+ 20x + 25 – (4x2
– 20x + 25) = 4x2
+ 20x + 25 – 4x2
+ 20x – 25 = 40x
iii) (7m – 8n)2
+ (7m + 8n)2
= 49m2
– 112mn + 64n2
+ 49m2
+ 112mn + 64n2
= 98m2
+ 128n2
iv) (4m + 5n)2
+ (5m + 4n)2
= 16m2
+ 40mn + 25n2
+ 25m2
+ 40mn + 16n2
= 41m2
+ 80mn + 41n2
v) (2.5p – 1.5q)2
– (1.5p – 2.5q)2
= 6.25p2
– 7.5pq + 2.25q2
– 2.25p2
+ 7.5pq – 6.25q2
= 4p2
– 4q2
vi) (ab + bc)2 – 2ab²c = a2 b2 + 2ab2 c + b2 c2 – 2ab2 c = a2 b2 + b2 c2
vii) (m2
– n2
m)2
+ 2m3
n2
= m4
– 2m3
n2
+ m2
n4
+ 2m3
n2
= m4
+ m2
n4
5. Show that.
(i) (3x + 7)2 – 84x = (3x – 7)2
(ii) (9p – 5q)2 + 180pq = (9p + 5q)2
(iii) (4/3m – 3/4n)2 + 2mn = 16/9 m2 + 9/16 n2
(iv) (4pq + 3q)2 – (4pq – 3q)2 = 48pq2
(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Solution:
i) LHS = (3x + 7)2 – 84x
= 9x2
+ 42x + 49 – 84x
= 9x2
– 42x + 49
= RHS
LHS = RHS
ii) LHS = (9p – 5q)2
+ 180pq
= 81p2
– 90pq + 25q2
+ 180pq
= 81p2
+ 90pq + 25q2
RHS = (9p + 5q)2
= 81p2
+ 90pq + 25q2
LHS = RHS
LHS = RHS
iv) LHS = (4pq + 3q)2 – (4pq – 3q)2
= 16p2 q2 + 24pq2 + 9q2 – 16p2 q2 + 24pq2 – 9q2
= 48pq2
= RHS
LHS = RHS
v) LHS = (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)
= a2 – b2 + b2 – c2 + c2 – a2
= 0
= RHS
6. Using identities, evaluate.
(i) 71²
(ii) 99²
(iii) 1022
(iv) 998²
(v) 5.2²
(vi) 297 x 303
(vii) 78 x 82
(viii) 8.92
(ix) 10.5 x 9.5
Solution:
i) 712
= (70+1)2
= 702 + 140 + 12
= 4900 + 140 +1
= 5041
ii) 99²
= (100 -1)2
= 1002 – 200 + 12
= 10000 – 200 + 1
= 9801
iii) 1022
= (100 + 2)2
= 1002 + 400 + 22
= 10000 + 400 + 4 = 10404
iv) 9982
= (1000 – 2)2
= 10002 – 4000 + 22
= 1000000 – 4000 + 4
= 996004
v) 5.22
= (5 + 0.2)2
= 52 + 2 + 0.22
= 25 + 2 + 0.04 = 27.04
vi) 297 x 303
= (300 – 3 )(300 + 3)
= 3002 – 32
= 90000 – 9
= 89991
vii) 78 x 82
= (80 – 2)(80 + 2)
= 802 – 22
= 6400 – 4
= 6396
viii) 8.92
= (9 – 0.1)2
= 92 – 1.8 + 0.12
= 81 – 1.8 + 0.01
= 79.21
ix) 10.5 x 9.5
= (10 + 0.5)(10 – 0.5)
= 102 – 0.52
= 100 – 0.25
= 99.75
7. Using a2 – b2 = (a + b) (a – b), find
(i) 512 – 492
(ii) (1.02)2 – (0.98)2
(iii) 1532 – 1472
(iv) 12.12 – 7.92
Solution:
i) 512 – 492
= (51 + 49)(51 – 49) = 100 x 2 = 200
ii) (1.02)2 – (0.98)2
= (1.02 + 0.98)(1.02 – 0.98) = 2 x 0.04 = 0.08
iii) 1532 – 1472
= (153 + 147)(153 – 147) = 300 x 6 = 1800
iv) 12.12 – 7.92
= (12.1 + 7.9)(12.1 – 7.9) = 20 x 4.2= 84
8. Using (x + a) (x + b) = x2 + (a + b) x + ab, find
(i) 103 x 104
(ii) 5.1 x 5.2
(iii) 103 x 98
(iv) 9.7 x 9.8
Solution:
i) 103 x 104
= (100 + 3)(100 + 4)
= 1002 + (3 + 4)100 + 12
= 10000 + 700 + 12
= 10712
ii) 5.1 x 5.2
= (5 + 0.1)(5 + 0.2)
= 52 + (0.1 + 0.2)5 + 0.1 x 0.2
= 25 + 1.5 + 0.02
= 26.52
iii) 103 x 98
= (100 + 3)(100 – 2)
= 1002 + (3-2)100 – 6
= 10000 + 100 – 6
= 10094
iv) 9.7 x 9.8
= (9 + 0.7 )(9 + 0.8)
= 92 + (0.7 + 0.8)9 + 0.56
= 81 + 13.5 + 0.56
= 95.06
NCERT Solutions for Class 8 Maths Chapter 9 is mainly about the study of solving polynomial-related problems. The chapter builds a strong foundation for the students to deal with higher grade Maths problems.
Students can utilise the NCERT Solutions for Class 8 Maths Chapter 9 for any quick references to comprehend and solve complex problems.