EXERCISE 9.1
1. Identify the terms, their coefficients for each of the following expressions. (i) 5xyz2 – 3zy (ii) 1 + x + x2(iii) 4x2y2 – 4x2y2z2 + z2 (iv) 3 – pq + qr – p (v) (x/2) + (y/2) – xy (vi) 0.3a – 0.6ab + 0.5b
EXERCISE 9.1
1. Identify the terms, their coefficients for each of the following expressions. (i) 5xyz2 – 3zy (ii) 1 + x + x2(iii) 4x2y2 – 4x2y2z2 + z2 (iv) 3 – pq + qr – p (v) (x/2) + (y/2) – xy (vi) 0.3a – 0.6ab + 0.5b
Explanation:
i) 5xyz2 – 3zy
The terms are 5xyz2 and -3zy and their coefficients are 5 and -3
ii) 1 + x + x2
The terms are 1, x and x2 and their coefficients are 1, 1 and 1
iii) 4x2y2 – 4x2y2z2 + z2
The terms are 4x2y2 , – 4x2y2z2 and z2 and their coefficients are 4, -4 and 1.
iv) 3 – pq + qr – p
The terms are 3, – pq, qr and p and their coefficients are 3, -1, 1 and -1
v) (x/2) + (y/2) – xy
The terms are x2 , y2 and – xy and their coefficients are ½, ½ and -1.
vi) 0.3a – 0.6ab + 0.5b
The terms are 0.3a, – 0.6ab and 0.5b and their coefficients are 0.3, -0.6 and 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, p2q + pq2 , 2p + 2q
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, p2q + pq2 , 2p + 2q
Explanation:
The types of polynomials given in the questions are explained as,
Monomial is an expression with only one term which is non zero.
Binomials is an expression with two terms
Trinomials is an expression with three terms
x + y- contains 2 terms and thus is binomial
1000- contains 1 term and thus is monomial
x + x2 + x3 + x4 – contains 4 terms and does not fit in any of the 3 categories
7 + y + 5x- contains 3 terms and is trinomial
2y – 3y2- contains 2 terms and is binomial
2y – 3y2 + 4y3- contains 3 terms and is trinomial
5x – 4y + 3xy- contains 3 terms and is trinomial
4z – 15z2- contains 2 terms and is binomial
ab + bc + cd + da- contains 4 terms and does not fit in any of the 3 categories
pqr- contains 1 term and thus is monomial
p2q + pq2- contains 2 terms and is binomial
2p + 2q- contains 2 terms and is binomial
7 + y + 5x-- contains 3 terms and is trinomial
3. Add the following.
(i) ab – bc, bc – ca, ca – ab
(ii) a – b + ab, b – c + bc, c – a + ac
(iii) 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2
(iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl
3. Add the following.
(i) ab – bc, bc – ca, ca – ab
(ii) a – b + ab, b – c + bc, c – a + ac
(iii) 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2
(iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl
Explanation:
i) (ab – bc) + (bc – ca) + (ca-ab)
First we open the brackets,
= ab – bc + bc – ca + ca – ab
= ab – ab – bc + bc – ca + ca
= 0
ii) (a – b + ab) + (b – c + bc) + (c – a + ac)
On opening the brackets we get,
= 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) (2p2q2 – 3pq + 4) + (5 + 7pq – 3p2q2)
On opening the brackets we get,
= 2p2q2 – 3pq + 4+ 5+ 7pq – 3p2q2
= 2p2q2– 3p2q2 – 3pq+ 7pq+ 4+ 5
= – p2q2 + 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 4p2q – 3pq + 5pq2 – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq2 + 5p2q
4.(a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3
(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
(c) Subtract 4p2q – 3pq + 5pq2 – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq2 + 5p2q
Explanation:
(a) (12a – 9ab + 5b – 3) – (4a – 7ab + 3b + 12)
On opening the brackets we get,
= 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)
On opening the brackets we get,
= 5xy – 2yz – 2zx + 10xyz – 3xy – 5yz + 7zx
=5xy – 3xy – 2yz – 5yz – 2zx + 7zx + 10xyz
= 2xy – 7yz + 5zx + 10xyz
c) (18 – 3p – 11q + 5pq – 2pq2 + 5p2q) – (4p2q – 3pq + 5pq2 – 8p + 7q – 10)
On opening the brackets we get,
= 18 – 3p – 11q + 5pq – 2pq2 + 5p2q – 4p2q + 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 + p2q
EXERCISE- 9.2
5.Find the product of the following pairs of monomials.
(i) 4, 7p
(ii) – 4p, 7p
(iii) – 4p, 7pq
(iv) 4p3, – 3p
(v) 4p, 0
EXERCISE- 9.2
5.Find the product of the following pairs of monomials.
(i) 4, 7p
(ii) – 4p, 7p
(iii) – 4p, 7pq
(iv) 4p3, – 3p
(v) 4p, 0
Explanation:
i) 4 x 7p
= 28p
ii) -4p x 7p
= -28p2
iii) -4p x 7pq
= -28p2q
iv) 4p3 x -3p
= -12p4
v) 4p x 0
= 0 (Any multiple when multiplied by 0 its product is also 0)
6. 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)
6. 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)
Explanation:
We know, Area of a rectangle = Length of rectangle x breadth of rectangle.
So, it is multiplication of two monomials.
(i) p × q = pq
(ii)10m × 5n
= 10 × 5 mn
= 50mn
(iii) 20x2 × 5y2
= 20× 5 x2y2
= 100x2y2
(iv) 4x × 3x2
= 12x3
(v) 3mn × 4np
= 12mn2p
7. Complete the following table of products:
7. Complete the following table of products:
Explanation:
8. Obtain the volume of rectangular boxes with the following length, breadth and height, respectively.
(i) 5a, 3a2, 7a4
(ii) 2p, 4q, 8r
(iii) xy, 2x2y, 2xy2
(iv) a, 2b, 3c
8. Obtain the volume of rectangular boxes with the following length, breadth and height, respectively.
(i) 5a, 3a2, 7a4
(ii) 2p, 4q, 8r
(iii) xy, 2x2y, 2xy2
(iv) a, 2b, 3c
Explanation:
Volume of a rectangle is calculated by multiplying its length, breadth and height. So, V= length x breadth x height.
(i) 5a x 3a2 x 7a4
= (5 × 3 × 7) × (a × a2 × a4 )
= 105 a1+2+4
= 105a7
(ii) 2p x 4q x 8r
= (2 × 4 × 8 ) × (p × q × r )
= 64pqr
(iii) y × 2x2y × 2xy2
=(1 × 2 × 2 ) × ( x × x2 × x × y × y × y2 )
= 4x4y4
(iv) a x 2b x 3c
= (1 × 2 × 3 ) × (a × b × c)
= 6abc
9. 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
9. 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
Explanation:
(i) xy × yz × zx
= x2 y2 z2
(ii) a × – a2 × a3
= – a1+2+3
= -a6
(iii) 2 × 4y × 8y2 × 16y3
= 2 × 4 × 8 × 16 y1+2+3
= 1024y6
(iv) a × 2b × 3c × 6abc
= (2× 3× 6) a2 b2 c2
= 36a2b2c2
(v) m × – mn × mnp
= –m3 n2 p
Exercise 9.3
10.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
Exercise 9.3
10.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
Explanation:
(i)4p×(q + r)
= 4pq + 4pr
(ii)ab× (a – b)
= a2 b – a b2
(iii)(a + b) (7a2b2)
= a(7a2b2) + b( 7a2b2)
= 7a3b2 + 7a2b3
(iv) (a2 – 9)(4a)
= 4a3 – 36a
(v) (pq + qr + rp) × 0
= 0
(Any multiple when multiplied by 0 its product is also 0)
11. Complete the table.
11. Complete the table.
Explanation:
i) Given, first expression= a
second expression= b+c+d
Therefore, product= a(b+c+d)
= ab+ac+ad
ii) Given, first expression= x+y-5
second expression= 5xy
Therefore, product= 5xy(x+y-5)
= 5x2y+5xy2-25xy
iii) Given, first expression= p
second expression= 6p2-7p+5
Therefore, product= p (6p2-7p+5)
= 6p3-7p2+5p
iv) Given, first expression= 4p2q2
second expression= p2-q2
Therefore, Product= 4p2 q2 x (p2 – q2 )
= 4 p4 q2– 4p2 q4
v) Given, first expression=a + b + c
second expression= abc
Therefore, product= abc( a + b+ c)
=a2bc + ab2c + abc2
12. Find the product.
i) a2 x (2a22) x (4a26)
ii) (2/3 xy) ×(-9/10 x2y2)
(iii) (-10/3 pq3/) × (6/5 p3q)
(iv) (x) × (x2) × (x3) × (x4)
12. Find the product.
i) a2 x (2a22) x (4a26)
ii) (2/3 xy) ×(-9/10 x2y2)
(iii) (-10/3 pq3/) × (6/5 p3q)
(iv) (x) × (x2) × (x3) × (x4)
Explanation:
i) a2 x (2a22) x (4a26)
= (2 × 4) × ( a2 × a22 × a26 )
= 8 × a2 + 22 + 26
= 8a50
ii) (2xy/3) ×(-9x2y2/10)
=(2/3 × -9/10 ) ( x × x2 × y × y2 )
=(2/3 × -9/10 ) ( x2+1 × y2+1 )
= (-3/5 x3y3)
iii) (-10pq3/3) ×(6p3q/5)
= ( -10/3 × 6/5 ) (p × p3× q3 × q)
= (-4p4q4)
iv) ( x) x (x2) x (x3) x (x4)
= x 1 + 2 + 3 + 4
= x10
13. (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.
13. (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.
Explanation:
a) 3x (4x – 5) + 3
= 12x2 – 15x + 3
(i) Putting x=3 in 12x2 – 15x + 3
12(3)2 – 15(3) + 3
108 – 45 + 3
66
(ii) Putting x=1/2 in 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
= a3+ a2+ a+ 5
(i) putting a=0 in a3+ a2+ a+ 5
=03+02+0+5
=5
(ii) putting a=1 in a3+ a2+ a+ 5
13 + 12 + 1+5
= 1 + 1 + 1+5
= 8
(iii) Putting a = -1 in a3+ a2+ a+ 5
(-1)3+(-1)2 + (-1)+5
= -1 + 1 – 1+5
= 4
14. (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 )
14. (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 )
Explanation:
a) p ( p – q) + q ( q – r) + r ( r – p)
= (p2 – pq) + (q2 – qr) + (r2 – pr)
= 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 – 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
EXERCISE- 9.4
15. 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)
EXERCISE- 9.4
15. 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)
Explanation:
(i) (2x + 5)(4x – 3)
= 2x ( 4x – 3) +5 ( 4x – 3)
= 8x² – 6x + 20x -15
= 8x² + 14x -15
ii) (y – 8) (3y – 4)
= y(3y – 4) -8 (3y – 4)
= 3y2 – 4y – 24y + 32
= 3y2 – 28y +32
iii) (2.5l – 0.5m) (2.5l + 0.5m)
= 2.5l (2.5 l + 0.5m) – 0.5m (2.5l + 0.5m)
= 6.25l2 + 1.25 lm – 1.25 lm – 0.25 m2
= 6.25l2– 0.25 m2
iv) (a + 3b) (x + 5)
= a (x + 5) +3b (x + 5)
= ax + 5a + 3bx + 15b
v) (2pq + 3q2) (3pq – 2q2)
= 2pq ( 3pq – 2q2) + 3q2 ( 3pq - 2q2)
= 6p2q2 – 4pq3 + 9pq3 – 6q4
= 6p2q2 + 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 – 8b4
= 3a4 + 10a² b² – 8b4
16. Find the product.
(i) (5 – 2x) (3 + x)
(ii) (x + 7y) (7x – y)
(iii) (a2+ b) (a + b2)
(iv) (p2 – q2) (2p + q)
16. Find the product.
(i) (5 – 2x) (3 + x)
(ii) (x + 7y) (7x – y)
(iii) (a2+ b) (a + b2)
(iv) (p2 – q2) (2p + q)
Explanation:
(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 + a2b2 + ab + b3
= a3 + b3 + a2b2 + ab
iv) (p2– q2) (2p + q)
= p2 (2p + q) – q2 (2p + q)
=2p3 + p2q – 2pq2 – q3
= 2p3 – q3 + p2q – 2pq2
17. 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)
17. 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)
Explanation:
i) (x2– 5) (x + 5) + 25
= x3 + 5x2 – 5x – 25 + 25
= x3 + 5x2 – 5x
ii) (a2+ 5) (b3+ 3) + 5
= a2b3 + 3a2 + 5b3 + 15 + 5
= a2b3 + 5b3 + 3a2 + 20
iii) (t + s2)(t2 – s)
= t (t2 – s) + s2(t2 – s)
= t3 – st + s2t2 – s3
= t3 – s3 – st + s2t2
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 – x2y + xy2 + x2y – 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
EXERCISE 9.5
18. 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)
EXERCISE 9.5
18. 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)
Explanation:
i) ( x+3)(x+3)
= x(x+3)+3(x+3)
= x2+3x+3x+9
= x2+6x+9
Alternatively, we can use the (a+b)2=a2+2ab+b2 formula to easily derive the answer since, ( x+3)(x+3)= (x+3)2
ii) (2y + 5) (2y + 5)
= 2y (2y + 5)+5 (2y + 5)
= 4y2+10y+10y+25
= 4y2+20y+25
Alternatively, we can use the (a+b)2=a2+2ab+b2 formula to easily derive the answer since, (2y + 5) (2y + 5)= (2y + 5)2
iii) (2a – 7) (2a – 7)
= 2a (2a – 7) -7(2a – 7)
= 4a2- 14a – 14a+49
= 4a2- 28a+49
Alternatively, we can use the (a-b)2=a2-2ab+b2 formula to easily derive the answer since (2a – 7) (2a – 7)= (2a – 7)2
iv) (3a – 1/2)(3a – 1/2)
= 3a (3a – 1/2) -1/2(3a – 1/2)
= 9a2- 3a/2-3a/2+1/4
= 9a2- 3a + ¼
v) (1.1m – 0.4) (1.1m + 0.4)
= (1.1m)2 – (0.4)2 [Because, (a – b)(a + b) = a2 – b2]
= 1.21m2 – 0.16
vi) (a2+ b2) (- a2+ b2)
We first bring it into (a+b)(a-b) form
= ( b2+a2) ( b2- a2)
= b2-a2 [Because, (a – b)(a + b) = a2 – b2]
vii) (6x – 7) (6x + 7)
Using the identity, (a-b)(a+b) = a2 – b2
= (6x)2 – 72
= 36x2 – 49
viii) (- a + c) (- a + c)
We can write this expression as
( c – a ) (c – a )
= ( c – a ) 2
Using the identity, (a-b) 2 = a2 + b2 – 2ab
= c2 + a2 – 2ac
ix) (1/2x + 3/4y) (1/2x + 3/4y)
= (1/2x + 3/4y)2
Using the identity, (a+b) 2 = a2 + b2 + 2ab
= (x/2)2 + (3y/4)2 + 2 (1/2x)(3/4y)
= (x2/4) + (9y2/16) + (3xy/4)
x) (7a – 9b) (7a – 9b)
= (7a – 9b)2
Using the identity, (a-b) 2 = a2 + b2 – 2ab
= (7a)2 + (9b)2- 2 (7a)(9b)
= 49a2 – 126ab + 81b2
19. 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)
19. 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)
Explanation:
(i)(x + 3) (x + 7)
Here, a=3 and b=7. So, the expression can be framed as
= x2 + (3+7)x+7x3
= x2+10x+21
ii) (4x + 5) (4x + 1)
Here, a=5 and b=1 and x=4x. So, the expression can be framed as
= (4x)2 + (5+1)4x + 5
= 16x2 + 20x+4x + 5
= 16x2 + 24x + 5
iii) (4x – 5) (4x – 1)
Here, a= -5 and b= -1. So, the expression can be framed as
= (4x)2 + (-5 + (-1))4x + 5
= 16x2 – 24x + 5
iv) (4x + 5) (4x – 1)
Here, a=5 and b=-1 and x=4x. So, the expression can be framed as
= (4x)2 + (5-1)4x – 5
= 16x2 +(4)4x – 5
= 16x2 +16x – 5
v) (2x + 5y) (2x + 3y)
Here, a=5y and b=3y and x=2x. So, the expression can be framed as
= (2x)2 + (5y + 3y)2x + 5y x 3y
= 4x2 + 16xy + 15y2
vi) (2a2+ 9) (2a2+ 5)
Here, a=9 and b=5 and x= 2a2. So, the expression can be framed as
= (2a2)2 + (9+5)2a2 + 9 x 5
= 4a4 + 28a2 + 45
vii) (xyz – 4) (xyz – 2)
Here, a=-4 and b=-2 and x= xyz. So, the expression can be framed as
= x2y2z2 + (-4 -2)xyz + 8
= x2y2z2 – 6xyz + 8
20. 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
20. 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
Explanation:
Using identities:
(a – b) 2 = a2 + b2 – 2ab (a + b) 2 = a2 + b2 + 2ab
(i) (b – 7)2
Using identity (a – b) 2 = a2 + b2 – 2ab,
= (b)2 – 2(7)(b) + (7)2
= b2- 24b + 49
(ii) (xy + 3z)2
Using identity (a + b) 2 = a2 + b2 + 2ab
= (xy)2 + 2(xy)(3z) + (3z)2
= x2y2 + 6xyz + 9z2
(iii) (6x2 – 5y)2
Using identity (a – b) 2 = a2 + b2 – 2ab
= (6x)2 + (5y)2 – 2 (6x)(5y)
= 36x4 – 60x2y + 25y2
(iv) [(2m/3) + (3n/2)]2
Using identity (a + b) 2 = a2 + b2 + 2ab
= (4m2/9) +(9n2/4) + 2mn
(v) (0.4p – 0.5q)2
Using identity (a – b) 2 = a2 + b2 – 2ab,
= 0.16p2 – 0.4pq + 0.25q2
(vi) (2xy + 5y)2
Using identity (a + b) 2 = a2 + b2 + 2ab
=4x2y2 + 20xy2 + 25y2
21. 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 – n2m)2 + 2m3n2
21. 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 – n2m)2 + 2m3n2
Explanation:
i) (a2– b2)2
Using identity (a – b) 2 = a2 + b2 – 2ab,
= (a2+b2-2ab)2
= a4 + b4 – 2a2b2
ii) (2x + 5)2 – (2x – 5)2
= [(2x)2 + 2(5) (2x) + (5)2 ] – [(2x)2 – 2(2x)(5) + (5)2]
= 4x2 + 20x + 25 – 4x2 + 20x – 25
= 20x+20x
= 40x
iii) (7m – 8n)2 + (7m + 8n)2
= 49m2 – 112mn + 64n2 + 49m2 + 112mn + 64n2
= 98m2 + 128n2
iv) (4m + 5n)2 + (5m + 4n)2
= [(4m)2 + 2(4m)(5n) + (5n)2] + [(5m)2 + 2(5m)(4n) + (4n)2]
= 16m2 + 40mn + 25n2 + 25m2 + 40mn + 16n2
= 41m2 + 80mn + 41n2
v) (2.5p – 1.5q)2 – (1.5p – 2.5q)2
= [(2.5p)2 – 2 (2.5p)(1.5q) + (1.5q)2] – [(1.5p)2-2(1.5p)(2.5q)+ (2.5q)2]
= 6.25p2 – 7.5pq + 2.25q2 – 2.25p2 + 7.5pq – 6.25q2
= 4p2 – 4q2
vi) (ab + bc)2– 2ab²c
= [(ab)2 + 2(ab)(bc) + (bc)2] – 2ab²c
= a2b2 + 2ab2c + b2c2 – 2ab2c
= a2b2 + b2c2
vii) (m2 – n2m)2 + 2m3n2
= m4 – 2m3n2 + m2n4 + 2m3n2
= m4 + m2n4
22. 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
22. 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
Explanation:
i) LHS = (3x + 7)2 – 84x
Using identity (a + b) 2 = a2 + b2 + 2ab
= (3x)2 + 72 + 2(3x)(7) – 84x
= 9x2 + 49 + 42x – 84x
= 9x – 42x + 49
= RHS
LHS = RHS
ii) LHS = (9p – 5q)2+ 180pq
Using identity (a - b) 2 = a2 + b2 - 2ab
= 81p2 – 90pq + 25q2 + 180pq
= 81p2 + 90pq + 25q2
RHS = (9p + 5q)2
Using identity (a + b) 2 = a2 + b2 + 2ab
= 81p2 + 90pq + 25q2
LHS = RHS
LHS = RHS
iv) LHS = (4pq + 3q)2– (4pq – 3q)2
= 16p2q2 + 24pq2 + 9q2 – 16p2q2 + 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
23. 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
23. 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
Explanation:
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
24. 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
24. 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
Explanation:
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
25. 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
25. 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
Explanation:
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
Sample