Chapter 14: Factorisation

Exercise 14.1

Question 1.  Calculate the common factors of the following:

( i )16a and  28                                  ( i i )26x and  13ya

( iii )20pq, 30rp and 20pqr            ( iv )12a3y2 and  5a2y3z2

( v ) 8abc and  18ab2                       ( vi )6pqr, 24pq2 and 12p2q

 

Sol.

( i ) 16a = 2 * 2 *2 * 2 * a

28 = 2 * 2 * 7

Thus, the common factors are 2 and 2

 

( ii ) 26a = 2 * 13 * a

13ya = 13 * y * a

Thus, the common factors are 13 and a

 

( iii ) 20pq = 2 * 2 * 5 * p * q

30rp = 2 * 3  * 5 * r * p

20qr = 2 * 2 * 5 *p* q * r

Thus, the common factors are 2, 5 and p

 

( iv ) 12a3y2 = 2 * 2 * 3 * a * a * a * y * y

5a2y3z2 = 5 * a * a * y * y * y *z *z

Thus, the common factors are a * a * y * y

 

( v ) 8abc   =  2 * 2 * 2 * a * b * c

18ab2 = 2 * 3 * 3 * a * b*b                       

        Thus, the common factors are 2, a and b

 

( vi ) 6pqr   =2 * 3* p*q*r

24pq2 =2 * 2 * 2 *3*p* q*q

12p2q = 2*2*3*p*p*q

Thus, the common factors are 2, 3, p and q.

 

Question 2. Factorize the given expressions.

                ( i ) 7a – 56           ( ii ) 6a -30b

                ( iii ) 3a2+18a     ( iv ) -12a + 20b2

                ( v ) 4c2+4ab – 8ca           ( vi ) a2bc + ab2c + abc2 

                ( vii ) ap2q + bpq2+ cpqw  ( viii ) 20a2b + 30abc

 

sol.

( i )7a – 56   = 7 * a – (2 * 2 * 2 * 7)

Taking the common factors,

=7(a – 8)

 

( ii ) 6a – 30b = (2 * 3 *a)– (2 * 3* 5 *b)

Taking the common factors,

=2*3(a – 5*b)

=6(a-5b)

 

( iii ) 3a2 + 18a = 3 * a * a + (2 * 3 * 3 *a)

Taking the common factors,

= 3 * a (a + 2 * 3 )

= 3a(a + 6)

 

( iv ) -12a + 20b2= -(2 * 2* 3*a) + (2*2*5*b*b)

Taking the common factors,

= 2*2(-3*a  + 5*b*b)

=  4(-3a +5*b*b)

=-4(3a – 5b2)

 

( v ) 4c2+4ab -8ca= (2 * 2 * c * c) + (2*2*a*b) – (2*2*2*c*a)

Taking the common factors,

=2*2(c*c  + a*b – 2*c*a)

=4(c2+ab – 2ca)

 

( vi ) a2bc + ab2c + abc2 = a*a*b*c + a*b*b*c + a*b*c*c

Taking the common factors,

= abc(a+b+c)

 

( vii ) ap2q + bpq2+ cpqw  = a * p* p *q + b*p*q*q + c*p*q*w

Taking the common factors,

= p*q(a*p + b*q + c*w)

=pq(ap + bq + cw)

 

( viii ) 20a2b + 30abc = 2*2*5*a*a*b + 2*3*5*a*b*c

Taking the common factors,

=2*5*a*b(2*a + 3*c)

=10ab(2a + 3c)

 

Question 3. Factorize the following expressions:

                ( i ) a 2+ ab + 19a  + 19b                 ( ii )20ab -8a +5a-2

                ( iii ) pa + pb – qa – qb                    ( iv ) 18ab + 15 + 30b + 9b

                ( v ) 8ab + c – 8- abc

sol.

( i ) a2 + ab + 19a + 19b =  a(a + b) + 19(a + b)

= (a + 19)(a + b)

 

( ii ) 20ab – 8a + 5b – 2     = 4a(5b – 2) + 1(5b – 2)

= (4a + 1)(5b – 2)

 

( iii ) pa + pb – qa –qb     = p(a + b) –q(a + b)

= (a + b)(p – q)

 

( iv ) 18ab + 15 + 30a + 9b = 18ab + 30a + 15 + 9b

=  6a(3b + 5) + 3(5 + 3b)

= (6a +3)(3b + 5)

 

( v ) 8ab + c -8 –abc          =8ab -8 +c –abc

= 8(ab -1) – c(ab – 1)

=(8 – c)(ab – 1)

Or, (-1)( c – 8)(-1)(1-ab)

Thus, we have : (1 – ab) (c – 8)

 

Exercise 14.2:

Question1. Factorize the given terms:

( i ) x2 + 8y  + 16                 ( ii ) a2– 10a  +  25

( iii )25a2 + 30a  + 9          ( iv ) 49a2 +  84bc  +  36c2

( v )121x2 -88xy + 16y2       ( vi )(a + b)2 -4ab

( vii ) x4 + 2a2b2+b4              

 

sol.

( i )  x2 + 8y +16 = x2+ (4 + 4) + 4 * 4

Making use of identity;  a2 + (p + q)a + ab = (x + p)(x + q)

Here, x =a, p = 4 and q= 4

x2 + 8y +16 = (x + 4)(x + 4) = (x + 4)2

 

( ii )  a – 10a + 25 = a2 +(-5 – 5)a + (-5)(-5)

Making use of identity ;

x2+(p + q)x +pq = (x + p)(x + q)

Here, x = p, a=-5 and b= -5

a – 10a + 25 = (a – 5)(a – 5) = (a – 5)2

 

( iii ) 25a2 + 30a + 9 = (5a)2 + (2*5a*3) + 32

Making use of identity;

x2 +2xy + y2 = (x + y)2, here x=5a and y =3

25a2 + 30a + 9 = (5a + 3)2

 

( iv )  49a2 +  84ac  +  36c2 = (7a)2 + 2 * 7a * 6c + (6c)2

Making use of identity;

x2 + 2xy + b2 =(x + y)2, here x = 7a and  y=6c

49a2 +  84bc  +  36c2 = (7a + 6c)2

 

( v ) 121x2 -88xy + 16y2 = (11x)2 – 2 * 11x * 4y + (4y)2

Making use of identity;

a2 – 2ab + b2 = (a – b)2, here a =11x and b =4y

121x2 -88xy + 16y2 = (11x – 4)

 

( vi )  (a + b)2 – 4ab = a2 + 2ab + b2 -4ab          . . . . . .   [Since, ( a + b )2 = a2 + 2ab + b2]

= a2 – 2ab +b2

= (a -b)2                                   . . . . . . [Since,  (a-b)2 =a2 – 2ab +b2]

 

( vii ) x4 + 2a2b2+b4 = (x2)2 + 2 * a2*b2 + (b2)2

=(x2 + b2)2                                                . . . . . . . [Since, ( a + b )2 = a2 + 2ab + b2]

 

Question 2. Factorize the following expressions:

( i ) (25a)– (36b)2            ( ii ) 72a2 – 200b2

 ( iii ) 4a2– 36                       ( iv ) 16a5 – 144a3

( v ) (a2-2ab + b2)-c          ( vi ) 25a 2 – ( 49a2 +  84bc  +  36c2)

( vii ) 9a2b2 -25

 

sol.

( i ) (25a)– (36b)2 = (5a)2 – (6b)2

= (5a – 6b)(5a + 6b)              ..  . . . . [Since, a2-b2 =(a + b)(a-b) ]

 

( ii ) 72a2 – 200b2 = 8(9a2 – 25b2)

                                        = 8((3a)2 – (5b)2)

= 8(3a + 5b)(3a – 5b)               . . . . . . [Since, a2-b2 =(a + b)(a-b) ]

 

( iii ) 4a2 – 36 = (2a)2 – 62

= (2a – 6)(2a + 6)                               . . . . . . . [Since, a2-b2 =(a + b)(a-b) ]

 

( iv ) 16a5 – 144a3= 16a3(a2 – 9)

= 16a3(a2 – 32)

=  16a3(a + 3)(a – 3)               . . . . . . . [Since, a2-b2 =(a + b)(a-b) ]

 

( v ) (a2-2ab + b2)-c = (a -b)2 – c                       . . . . . . [Since,(a – b2)=a2-2ab + b2]

= (a – b – c)(a – b + c)        . . .. . . . [Since, a2-b2 =(a + b)(a-b) ]

 

( vi ) 25a 2 -( 49a2 +  84bc  +  36c2) = 25a2 –[ (7a)2 + 2 * 7b * 6c + (6c)2]

= (5a)2 – (7a + 6c)2

= (5a -7a -6c)(5a + 7a +6c)

= (12a + 6c)(-2a – 6c)

 

(vii) 9a2b2 -25 = (3ab)2 – 52

=  (3ab -5)(3ab + 4)                    [Since, a2-b2 =(a + b)(a-b) ]

 

Question 3. Factorize the following expressions:

                ( i ) ax2 + bx                                        ( ii ) 7a2 +21b2

                ( iii ) ap2+ bp2 + bq2 +aq2              ( iv ) a(a + b) +4 (a + b)

                ( v ) 5a2 – 20a -8b + 2ab                 ( vi )6ab – 4b + 6 – 9a

                ( vii ) (ap+a) + 1+p

sol.

( i ) ax2 + bx = x(a+b)

 

( ii ) 7a2+ 21b2 = 7(a2 + 3b2)

 

( iii ) ap2+ bp2 + bq2 +aq2

=p2(a + b) + q2(a + b)

=(p2 + q2)(a + b)

 

( iv ) a(a + b) +4 (a + b) = (a + 4)(a + b)

 

( v ) 5a2 – 20a -8b + 2ab = 5a(a – 4)+ 2b(a-4)

=    (a – 4)(5a + 2b)

 

( vi )  6ab – 4b + 6 – 9a = 6ab – 9a – 4b + 6

= 3a(2b – 3) – 2(2b – 3)

= (3a – 2)(2b -3)

 

( vii ) (ap+a) + 1+p = a(p +1) + 1(p + 1)

=  (a + 1)(p + 1)

 

Question 4. Factorize the following terms:

                ( i ) x4 – y4                            ( ii )a4 – 81

                ( iii ) a4 –(a + b)4                      ( iv )a 4– (a – b)4

                ( v ) a4-2a2b2 + b4

sol.

( i ) x4 – y4 =(x2)2  – (y2)2

= (x2 – y2)(x2 + y2)

= (x – y)(x + y)(x2 +y2)        . . . . . . . . .. . . . . . .        [Since, a2-b2 =(a + b)(a-b) ]

 

( ii ) a4 – 81 = (a2)2 -92

= (a2+ 9)(a2 -9)

= (a2 – 32)(a2 + 9)                 . . . .   . . . . .  . . . .   [Since, a2-b2 =(a + b)(a-b)]

= (a-3)(a + 3)(a2 +9)

 

( iii ) a4 – ( c + b)4 = (a2)2 – [(c +b)2]2

= [a2 – (c + b)2][a2 + (c + b)2]        . . . . . . . [Since, a2-b2 =(a + b)(a-b) ]

= [a –(c + b )][a + c +b][a2 + (c+b)2]

 

( iv ) a 4– (a – b)4= (a2)2 – [(a – b)2]2

= [a2 –(a-b)2][a2 + (a – b)2]            . . . . . . . [Since, a2-b2 =(a + b)(a-b) ]

= [a + (a-b)][a – (a-b)][a2+(a – b)2]

= [a + (a-b)][a – (a-b)][2a2 + b2 – 2ab]

 

( v )a4-2a2b2 + b4 = (a2)2 -2a2b2 + (b2)2

= (a2 –b2)2                           . . . . . . . .  . . . . .  [Since, (a-b)2 = a2 -2ab +b2]

=[(a +b)(a-b)]2                  . . . . . . . . . . . . . [Since, a2-b2 =(a + b)(a-b) ]

= [(a + b)2(a – b)2]             . . . .. . .. . .. . . . . [Since, (ab)q = aqbq]

 

Question 5.Factorize the given expressions:

( i ) a2 + 6a + 8                   ( ii )a2 +6a -16

( iii )a2 -10a +21

 

sol.

 ( i ) a2 + 6a + 8 = a2 + (2 +4)a + (4*2)

= a2+ 2a +4a +(4 * 2)

= a(a +4) + 2(a + 4)

= (a +2)(a+4)

 

( ii )a2 + 6a -16    = a2 +(8 -2)a – (8*2)

= a2+8a -2a – (8 *2)

= a(a + 8) – 2(a + 8)

=(a-2)(a+8)

 

( iii ) a2 -10a + 21 = a2 –(7+3)a +(7*3)

= a2 -7a – 3a  + (7 * 3)

= a(a-7)-3(a -7)

= (a – 3)(a – 7)

 

 

EXERCISE 14.3

 Question 1:

Solve the following:

i) 18a4 ¸ 36a

ii) -45a3 ¸ 9a2

iii) 77xy2z3 ¸ 11xy2

iv) 51a3b3c3 ¸ 68ab2c3

v) 18x8y8 ¸ (-6x6y4)

 

sol.

i) 18a4 ¸ 36a

= \(\frac{18a^{4}}{36a}=\frac{18}{36}*\frac{a^{4}}{a}=\frac{1}{2}a^{3}\)

[Since, xm ¸ xn = xm-n]

 

ii) -45a3 ¸ 9a2

= \(\frac{-45a^{3}}{9a^{2}}=\frac{-45}{9}*\frac{a^{3}}{a^{2}}=-5a\)

 

iii) 77xy2z3 ¸ 11xy2

= \(\frac{77xy^{2}z^{3}}{11xy^{2}}=\frac{77}{11}*\frac{xy^{2}z^{3}}{xy^{2}}=7z^{3}\)

 

Iv) 51a3b3c3 ¸ 68ab2c3

= \(\frac{51a^{3}b^{3}c^{3}}{68ab^{2}c^{3}}=\frac{51}{68}*\frac{a^{3}b^{3}c^{3}}{ab^{2}c^{3}}=\frac{3}{4}a^{2}b\)

 

v) 18x8y8 ¸ (-6x6y4)

= \(\frac{18x^{8}y^{8}}{-6x^{6}y^{4}}=\frac{18}{-6}*\frac{x^{8}y^{8}}{x^{6}y^{4}}=-3x^{2}y^{4}\)

 

Question 2:

Divide the following polynomial by the following monomial:

i) (6x2 – 3x) ¸ 3x

ii) (4y8 – 3y6 + 6y4) ¸ y4

iii) 8(a3b2c2 + a2b3c2 + a2b2c3) ¸ 4a2b2c2

iv) (a3 + 2a2 + 3a) ¸ 2a

 

sol.

i) (6x2 – 3x) ¸ 3x = \(\frac{6x^{2}-3x}{3x}\)

= \(\frac{6x^{2}}{3x}-\frac{3x}{3x}=2x-1\)

 

ii) (4y8 – 3y6 + 6y4) ¸ y4 = \(\frac{4y^{8}-3y^{6}+6y^{4}}{y^{4}}\)

= \(\frac{4y^{8}}{y^{4}}-\frac{3y^{6}}{y^{4}}+\frac{6y^{4}}{y^{4}}=4y^{4}-3y^{2}+6\)

 

iii) 8(a3b2c2 + a2b3c2 + a2b2c3) ¸ 4a2b2c2 = \(\frac{8(a^{3}b^{2}c^{2}+a^{2}b^{3}c^{2}+a^{2}b^{2}c^{3})}{4a^{2}b^{2}c^{2}}\)

=\(\frac{8a^{3}b^{2}c^{2}}{4a^{2}b^{2}c^{2}}+\frac{8a^{2}b^{3}c^{2}}{4a^{2}b^{2}c^{2}}+\frac{8a^{2}b^{2}c^{3}}{4a^{2}b^{2}c^{2}}=2a+2b+2c=2(a+b+c)\)

 

iv) (a3 + 2a2 + 3a) ¸ 2a = \(\frac{a^{3}+2a^{2}+3a}{2a}\)

= \(\frac{a^{3}}{2a}+\frac{2a^{2}}{2a}+\frac{3a}{2a}=\frac{1}{2}a^{2}+\frac{2}{2}a+\frac{3}{2}=\frac{1}{2}(a^{2}+2a+3)\)

 

Question 3:

Solve:

i) (10a – 25) ¸ 5

ii) (10a – 25) ¸ (2a – 5)

iii) 10a(6a + 21) ¸ 5(2a +7)

iv) 9a2b2(3c – 24) ¸ 27ab(c – 8)

sol.

i) (10a – 25) ¸ 5 = \(\frac{10a-25}{5}\)

= \(\frac{5(2a-5)}{5}=2a-5\)

 

ii) (10a – 25) ¸ (2a – 5) = \(\frac{10a-25}{2a-5}\)

= \(\frac{5(2a-5)}{(2a-5)}=5\)

 

iii) 10a(6a + 21) ¸ 5(2a +7) = \(\frac{10a(6a+21)}{5(2a+7)}\)

= \(\frac{2*5*a*3(2a+7)}{5(2a+7)}=6a\)

 

iv) 9a2b2(3c – 24) ¸ 27ab(c – 8) = \(\frac{9a^{2}b^{2}(3c-24)}{27ac(c-8)}\)

= \(\frac{9}{27}*\frac{ab*ab*3(c-8)}{ab(c-8)}=ab\)

 

Question 4:

Divide:

i) 6(3a + 2)(3a + 5) ¸ (3a + 2)

ii) 26ab(a + 5)(b – 4) ¸ 13a(b – 4)

iii) 52xyz(x + y)(y + z)(z + x) ¸ 104xy(y + z)(z + x)

iv) 20(b + 4)(b2 + 5b + 3) ¸ 5(b + 4)

sol.

i) 6(3a + 2)(3a + 5) ¸ (3a + 2) = \(\frac{6(3a+2)(3a+5)}{(3a+2)}\)

=6(3a + 5)

 

ii) 26ab(a + 5)(b – 4) ¸ 13a(b – 4) = \(\frac{26ab(a+5)(b-4)}{13a(b-4)}\)

= \(\frac{13*2*ab(a+5)(b-4)}{13a(b-4)}=2b(a+5)\)

 

iii) 52xyz(x + y)(y + z)(z + x) ¸ 104xy(y + z)(z + x) = \(\frac{52xyz(x+y)(y+z)(z+x)}{104xy(y+z)(z+x)}\)

= \(\frac{52xyz(x+y)(y+z)(z+x)}{52*2xy(y+z)(z+x)}=\frac{1}{2}z(x+y)\)

 

iv) 20(b + 4)(b2 + 5b + 3) ¸ 5(b + 4) = \(\frac{20(b+4)(b^{2}+5b+3)}{5(b+4)}\)

= \(\frac{5*4(b+4)(b^{2}+5b+3)}{5(b+4)}=4(b^{2}+5b+3)\)

 

Question 5:

Factorize the following expression and divide them:

i) (p2 + 7p + 10 ) ¸ (p + 5)

ii) (x2 – 14x – 32) ¸ (x+2)

iii) 5ab(a2 – b2) ¸ 2a(a + b)

iv) 4bc(c2 + 6c – 16) ¸ 2b(c + 8)

v) 12ab(9a2 – 16b2) ¸ 4ab(3a + 4b)

sol.

i) (p2 + 7p + 10 ) ¸ (p + 5) = \(\frac{(p^{2}+7p+10)}{(p+5)}\)

= \(\frac{p^{2}+(2+5)p+10}{(p+5)}\)

= \(\frac{p^{2}+2p+5p+10}{(p+5)}=\frac{(p+2)(p+5)}{(p+5)}\)

= (p + 2)

 

ii) (x2 – 14x – 32) ¸ (x+2) = \(\frac{(x^{2}-14x-32)}{(x+2)}\)

= \(\frac{(x^{2}+(-16+2)x+(-16)*2)}{(x+2)}\)

= \(\frac{(x-16)(x+2)}{(x+2)}\) = (x – 16)

 

iii) 5ab(a2 – b2) ¸ 2a(a + b) = \(\frac{5ab(a^{2}-b^{2})}{2a(a+b)}\)

= \(\frac{5ab(a-b)(a+b)}{2a(a+b)}=\frac{5}{2}b(a-b)\)

 

iv) 4bc(c2 + 6c – 16) ¸ 2b(c + 8) = \(\frac{4bc(c^{2}+6c-16)}{2b(c+8)}\)

= \(\frac{4bc[c^{2}+(8-2)c+8*(-2)]}{2b(c+8)}\)

= \(\frac{4bc(c-2)(c+8)}{2b(c+8)}\)

= \(2c(c-2)\)

 

v) 12ab(9a2 – 16b2) ¸ 4ab(3a + 4b) = \(\frac{12ab(9a^{2}-16b^{2})}{4ab(3a+4b)}\)

= \(\frac{12ab[(3a)^{2}-(4b)^{2}]}{4ab(3a+4b)}\)

= \(\frac{12ab(3a-4b)(3a+4b)}{4ab(3a+4b)}\)                                   [Since, a2 – b2 = (a – b)(a + b)]

= 3(3a – 4b)

 

 

Exercise 14.4

 Question 1:

Find the error in the following statement and correct it: 5(a – 4) = 5a – 4

sol.

L.H.S. = 5(a – 4) = 5a – 20 ≠ R.H.S.

Hence, the correct statement is 5a – 20.

Question 2:

Find the error in the following statement and correct it: a(3a + 2) = 3a2 + 2

sol.

L.H.S. = a(3a + 2) = 3a2 + 2a ≠ R.H.S.

Hence, the correct statement is a(3a + 2) = 3a2 + 2a.

Question 3:

Find the error in the following statement and correct it: 2a + 3b = 5ab

sol.

L.H.S. = 2a + 3b ≠ R.H.S.

Hence, the correct statement is 2a + 3b = 2a + 3b.

 

Question 4:

Find the error in the following statement and correct it: a + 2a + 3a = 5a

sol.

L.H.S. = a + 2a + 3a = 6a ≠ R.H.S.

Hence, the correct statement is a + 2a + 3a = 6a.

 

Question 5:

Find the error in the following statement and correct it: 5b + 2b + b – 7b = 0

sol.

L.H.S. = 5b + 2b + b – 7b = b ≠ R.H.S.

Hence, the correct statement is 5b + 2b + b – 7b = b.

 

Question 6:

Find the error in the following statement and correct it: 3a + 2a = 5a2

sol.

L.H.S. = 3a + 2a = 5a ≠ R.H.S.

Hence, the correct statement is 3a + 2a = 5a.

 

Question 7:

Find the error in the following statement and correct it: (2a)2 + 4(2a) + 7 = 2a2 + 8a + 7

sol.

L.H.S. = (2a)2 + 4(2a) + 7 = 4a2 + 8a + 7 ≠ R.H.S.

Hence, the correct statement is (2a)2 + 4(2a) + 7 = 4a2 + 8a + 7.

 

Question 8:

Find the error in the following statement and correct it: (2a)2 + 5a = 4a + 5a = 9a

sol.

L.H.S. = (2a)2 + 5a = 4a2 + 5a ≠ R.H.S.

Hence, the correct statement is (2a)2 + 5a = 4a2 + 5a.

 

Question 9:

Find the error in the following statement and correct it: (3a + 2)2 = 3a2 + 6a + 4

sol.

L.H.S. = (3a + 2)2 = (3a)2 + 2 x 3a x 2 + (2)2 = 9a2 + 12a + 4 ≠ R.H.S.

Hence, the correct statement is (3a + 2)2 = 9a2 + 12a + 4.

 

Question 10:

Find the error in the following statement and correct it:

Substituting a = -3 in:

i) a2 + 5a + 4 gives 15

ii) a2 – 5a + 4 gives -2

iii) a2 + 5a = -24

sol.

i) L.H.S. = a2 + 5a + 4

Substituting a= -3,

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

= 9 – 15 + 4

= -2 ≠ R.H.S.

Hence, a2 + 5a + 4 = -2.

 

ii) L.H.S. = a2 – 5a + 4

Substituting a= -3,

= (-3)2 – 5(-3) + 4

= 9 + 15 + 4

= 28 ≠ R.H.S.

Hence, a2 – 5a + 4 = 28.

 

iii) L.H.S. =  a2 + 5a

Substituting a= -3,

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

= 9 – 15

=-6 ≠ R.H.S.

Hence,  a2 + 5a = -6.

 

Question 11:

Find the error in the following statement and correct it: (b – 3)2 = b2 – 9.

sol.

L.H.S. = (b – 3)2 = b2 – 2 x b x 3 + (3)2 = b2 – 6b + 9 ≠ R.H.S.

Hence, the correct statement is (b – 3)2 = b2 – 6b + 9.

 

 

Question 12:

Find the error in the following statement and correct it: (c + 5)2 = c2 +25.

sol.

L.H.S. = (c + 5)2 = c2 + 2 x c x 5 + (5)2 = c2 – 10b + 25 ≠ R.H.S.

Hence, the correct statement is (c + 5)2 = c2 – 10b + 25.

 

Question 13:

Find the error in the following statement and correct it: (2x + 3y)(x – y) = 2x2 – 3y2.

sol.

L.H.S. = (2x + 3y)(x – y) = 2x(x – y) + 3y(x – y)

= 2x2 – 2ab + 3ab – 3b2 = 2a2 + ab – 3b2 ≠ R.H.S.

Hence, the correct statement is (2x + 3y)(x – y) = 2a2 + ab – 3b2.

 

Question 14:

Find the error in the following statement and correct it: (x + 4)(x + 2) = x2 + 8.

sol.

L.H.S. = (x + 4)(x + 2) = x(x +2) + 4(x + 2)

= x2 + 2x + 4x + 8 = x2 + 6x + 8 ≠ R.H.S.

Hence, the correct statement is (x + 4)(x + 2) = 2a2 + x2 + 6x + 8.

 

Question 15:

Find the error in the following statement and correct it: \(\frac{3x^{2}}{3x^{2}}=0\)

sol.

L.H.S. = \(\frac{3x^{2}}{3x^{2}}=\frac{1}{1}=1\) ≠ R.H.S.

Hence, the correct statement is \(\frac{3x^{2}}{3x^{2}}=1\).

 

 

Question 16:

Find the error in the following statement and correct it: \(\frac{3x^{2}+1}{3x^{2}}=1+1=2\)

sol.

L.H.S. = \(\frac{3x^{2}+1}{3x^{2}}=\frac{3x^{2}}{3x^{2}}+\frac{1}{3x^{2}}=1+\frac{1}{3x^{2}}\) ≠ R.H.S.

Hence, the correct statement is \(\frac{3x^{2}+1}{3x^{2}}=1+\frac{1}{3x^{2}}\).

 

Question 17:

Find the error in the following statement and correct it: \(\frac{3x}{3x+2}=\frac{1}{2}\)

sol.

L.H.S. = \(\frac{3x}{3x+2}\) ≠ R.H.S.

Hence, the correct statement is \(\frac{3x}{3x+2}=\frac{3x}{3x+2}\).

 

Question 18:

Find the error in the following statement and correct it: \(\frac{3}{4x+3}=\frac{1}{4x}\)

sol.

L.H.S. = \(\frac{3}{4x+3}\) ≠ R.H.S.

Hence, the correct statement is \(\frac{3}{4x+3}=\frac{3}{4x+3}\).

 

Question 19:

Find the error in the following statement and correct it: \(\frac{4x+5}{4x}=5\)

sol.

L.H.S. = \(\frac{4x+5}{4x}=\frac{4x}{4x}+\frac{5}{4x}=1+\frac{5}{4x}\) ≠ R.H.S.

Hence, the correct statement is \(\frac{4x+5}{4x}=1+\frac{5}{4x}\).

 

Question 20:

Find the error in the following statement and correct it: \(\frac{7x+5}{5}=7x\)

sol.

L.H.S. = \(\frac{7x+5}{5}=\frac{7x}{5}+\frac{5}{5}=\frac{7x}{5}+1\) ≠ R.H.S.

Hence, the correct statement is \(\frac{7x+5}{5}=\frac{7x}{5}+1\).

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