NCERT Solutions For Class 8 Maths Chapter 14

NCERT Solutions Class 8 Maths Factorisation

Ncert Solutions For Class 8 Maths Chapter 14 PDF Download

Class 8 Maths Chapter 14 Factorisation NCERT Solutions provides the ready solution for all the questions present in the NCERT textbook. When an algebraic expression is factored it is denoted as the product of its factors. These factors can be algebraic variables, numbers or algebraic expressions. Equations like 21xy, 5xy, 12x(y + 2), 54(y + 14) (12x + 32) are represented in the factor form. Whereas, for expressions like 13x + 23y, 21x + 25x, 12x + 5y + 6, etc it is not clear what their factors are. Let us consider the equation 2yx + 3x + 2y + 3. For the given equation it is difficult to predict the factorisation. However, on rearranging the equation as 3x + 3 + 2xy + 2y, enables us to form groups i.e. (3x + 3) and (2xy + 2y) leading to factorisation. This is termed as the regrouping. NCERT Solutions for Class 8 Maths chapter 14 are prepared by our team of subject experts and are explained in detail in a logical manner to help students to understand the concepts easily.

An irreducible factor cannot be expressed more as a product of factors. A precise way of factorising an equation is common factor method. It comprises of 3 levels: (i) Write term of the equation as the product of irreducible factors (ii) Order the common factors (iii) Merge the left factors according to the distributive law. Any regrouping of terms in the given expression may or may not lead to factorisation. It’s always better to observe the expression with the desired regrouping by trial and error method. In equations having factors of type (m + p) (m + q), the numerical term gives ‘pq’. Its factors, p and q, must be so determined that their aggregate, with signs taken care of, is the coefficient of m. Also, the division of a polynomial by a monomial can be carried out either by

dividing every term of a polynomial by a monomial or by using the common factor method.

Chapter 14 Class 8 Maths NCERT Solutions is the best reference materials for students to learn more in detail about the Factorisation. Students can avail this material from the link mentioned in the below article. They are available in PDF files for free and students can either download Class 8 Maths chapter 14 NCERT solutions pdf files or by practising online.

NCERT Solutions Class 8 Maths Chapter 14 Exercises

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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