NCERT Solutions for Class 8 Maths Chapter 14 Factorization

NCERT solutions for class 8 maths chapter 14 factorization topic 14.2.1 method of common factor

Question:(i) Factorise:

Answer:

We have
$12x = 2 \times 2 \times 3 \times x$
$36 = 2 \times 2 \times 3 \times 3$

So, we have $2 \times 2 \times 3$ common in both
Therefore,

$2 \times 2 \times 3 (x + 3)$

Question:(ii) Factorise : 22y-32z

Answer:

We have,
$2 \times 11 \times y$
$3 \times 11 \times z$
So, we have 11 common in both
Therefore,

Question:(iii) Factorise :

$( iii) \: \: 14 pq + 35 pqr$

Answer:

We have
$2 \times 7 \times p \times q$
$5 \times 7 \times p \times q \times r$
So, we have

$7 \times p \times q$ common in both
Therefore,

NCERT solutions for class 8 maths chapter 14 factorization-Exercise: 14.1

Question:1(i) Find the common factors of the given terms.

Answer:

We have
$12x ={\color{Red} 2 \times 2 \times 3}\times x$
$36 = {\color{Red} 2 \times 2 \times 3}\times 3$
So, the common factors between the two are

$2\times2\times3=12$

Question:1(ii) Find the common factors of the given terms

Answer:

We have,
$2y = {\color{Red} 2 \times y}$
$22xy = {\color{Red} 2} \times 11 \times x {\color{Red} \times y}$
Therefore, the common factor between these two is 2y

Question:1(iii) Find the common factors of the given terms

Answer:

We have,
$14pq = {\color{Red} 2 \times 7 \times p \times q}$
$28p^2q^2 = 2 \times {\color{Red} 2 \times 7 \times p} \times p{\color{Red} \times q} \times q$
Therefore, the common factor is

$2\times7\times p\times q=14pq$

Question:1(iv) Find the common factors of the given terms.

Answer:

We have,
$2x = 2 \times x$
$3x^2 = 3 \times x \times x$
$4 = 2 \times 2$
Therefore, the common factor between these three is 1

Question:1(v) Find the common factors of the given terms

Answer:

We have,
$6abc ={\color{Red} 2 \times 3 \times a \times b }\times c$
$24ab^2 = 2 \times 2\times {\color{Red} 2\times 3 \times a \times b} \times b$
$12a^2b = 2 \times {\color{Red} 2\times 3 \times a} \times a{\color{Red} \times b}$
Therefore, the common factors is

$2 \times 3 \times a \times b = 6ab$

Question:1(vi) Find the common factors of the given terms

Answer:

We have,
$16x^3 = 2 \times 2 \times {\color{Red} 2 \times 2 \times x} \times x \times x$
$4x^2 = {\color{Red} 2 \times 2 \times x} \times x$
$32x = 2 \times 2 \times 2 \times{\color{Red} 2 \times 2 \times x}$
Therefore, the common factors is

$2 \times 2 \times x = 4x$

Question:1(vii) Find the common factors of the given terms

Answer:

We have,
$10pq ={\color{DarkRed} 2 \times 5} \times p \times q$
<img alt="20qr = 2\times{\color{DarkRed} 2 \times 5 }\times q \times r" height="17"

src="https://lh4.googleusercontent.com/Ss8ADbZ4hBQ70-uoJgKlTa_FN-nCGSJnPuZy52TF0xY9yYMom99vCtS000Dk52edALKSK5d1g7Ii9OBkAvP679IyFyXIecTCxYAJWA2UjOgjt5Ft30fG5FiUH-dM0vZZS7UeSGQ" style="margin-left: 0px; margin-top: 0px;" width="192" />
$30rp ={\color{DarkRed} 2}\times 3{\color{DarkRed} \times 5} \times r \times p$
Therefore, the common factors between these three is

$2 \times 5 =10$

Question:1(viii) Find the common factors of the given terms

Answer:

We have,
$3x^{2}y^{2}$ $= 3 \times {\color{Red} x \times x \times y \times y}$
$10x^{3}y^{2}$ $=2 \times 5 \times x \times {\color{Red} x\times x \times y \times y}$
$6x^{2}y^{2}z$ $=2 \times 3 \times{\color{Red} x \times x \times y \times y} \times z$
Therefore, the common factors between these three are $x\times x\times y \times y =$ $x^{2}y^{2}$

Question:2(i) Factorise the following expressions

Answer:

We have,
$7x = 7 \times x \\ 42=7\times 2 \times 3=7\times 6\\ 7x-42=7x-7\times 6=7(x-6)$

Therefore, 7 is a common factor

Question:2(ii) Factorise the following expressions

Answer:

We have,
$6p = 2 \times 3 \times p$
$12q = 2 \times 2 \times 3 \times q$
$\therefore$ on factorization

$6p -12q = (2\times 3 \times p) - (2\times 2 \times 3 \times q) = (2\times 3)(p-2q) = 6(p-2q)$

Question:2(iii) Factorise the following expressions

Answer:

We have,
$7a^2 = 7 \times a \times a$
$14a = 2 \times 7 \times a$
$\therefore$ $7a^2+14a = (7\times a \times a)+(2 \times 7 \times a) = (7 \times a)(a+2)$

Question:2(iv) Factorise the following expressions

Answer:

We have,
$-16z = -1 \times 2 \times 2 \times 2 \times 2 \times z$
$20z^3 = 2 \times 2 \times 5 \times z \times z \times z$
$\therefore$ on factorization we get,
$-16z+20z^3 = (-1 \times 2 \times 2 \times 2 \times 2 \times z)+(2 \times 2 \times 5 \times z \times z \times z )$
$= (2\times 2 \times z)(-1 \times 2 \times 2+ 5 \times z \times z )$

Question:2(v) Factorise the following expressions

Answer:

We have,
$20l^2m = 2 \times 2 \times 5 \times l \times l \times m$
$30alm = 2 \times 3 \times 5 \times a \times l \times m$
$\therefore$ on factorization we get,
$20l^2m+30alm =(2\times 2 \times 5 \times l \times l \times m) + (2 \times 3 \times 5 \times a \times l \times m)$
$=(2\times 5 \times l \times m)(2\times l + 3 \times a )$

Question:2(vi) Factorise the following expressions

Answer:

We have,
$5x^2y = 5 \times x\times x \times y$
$15xy^2 =3\times 5 \times x\times y \times y$
$\therefore$ on factorization we get,
$5x^2y - 15xy^2 = (5 \times x \times x \times y ) - (3\times 5 \times x \times y \times y )$
$=(5\times x \times y) ( x - 3\times y )$

Question:2(vii) Factorise the following expressions

Answer:

We have,
$10a^2 = 2 \times 5 \times a \times a$
$15b^2 = 3 \times 5 \times b \times b$
$20c^2 = 2\times 2 \times 5 \times c \times c$
$\therefore$ on factorization we get,
$10a^2-15b^2+20c^2 = (2\times 5 \times a \times a)-(3\times 5 \times b \times b)+(2\times 2 \times 5 \times c \times c)$ $=5 (2 \times a \times a-3 \times b \times b+2\times 2 \times c \times c)$

Question:2(viii) Factorise the following expressions

Answer:

We have,
$-4a^2 = -1\times 2 \times 2 \times a\times a$
$4ab = 2 \times 2 \times a\times b$
$4ca = 2 \times 2 \times c\times a$
$\therefore$ on factorization we get,
$-4a^2+4ab-4ca = (-1 \times 2 \times 2 \times a\times a )+( 2 \times 2 \times a\times b )- (2 \times 2 \times c\times a)$

src="https://lh3.googleusercontent.com/DjxprQzpQEmOgRkrVWxMQkssFhWtJsvauh543sKzwcsl2dJ_DFmaqC77QGOLprtXa_fcPnz8i_T5CobT71IXB0c4mSkSLAKJtD02fc5GcsDjBj9cLwGyw_ky-W3IEb-yUFnDqrI" style="margin-left: 0px; margin-top: 0px;" width="565" />

$=(2 \times 2 \times a) (-1 \times a + b - c)$

Question:2(ix) Factorise the following expressions

Answer:

We have,
$x^2yz =x \times x \times y \times z$
$xy^2z =x \times y \times y \times z$
$xyz^2 =x \times y \times z \times z$
Therefore, on factorization we get,
$x^2yz+xy^2z+xyz^2 =(x \times x \times y \times z)+(x \times y \times y \times z)+(x \times y \times z \times z)$

$=( x \times y \times z)(x + y + z)$

Question:2(x) Factorise the following expressions

Answer:

We have,
$ax^2y = a \times x \times x \times y$
$bxy^2 = b \times x \times y \times y$
$cxyz = c \times x \times y \times z$
Therefore, on factorization we get,
$ax^2y+bxy^2+cxyz = ( a \times x \times x \times y)+( b \times x \times y \times y)+(c \times x \times y \times z)$ $= (x\times y)( a \times x+ b \times y+c \times z)$

Question:3(i) Factorise

Answer:

We have,
$x^2 = x \times x$
$xy = x \times y$
$8x = 8 \times x$
$8y = 8 \times y$
Therefore, on factorization we get,
$x^2+xy+8x+8y = (x \times x)+(x\times y )+(8 \times x)+(8 \times y)$

Question:3(ii) Factorise

Answer:

We have,
$15xy = 3 \times 5 \times x \times y$
$6x = 2 \times 3 \times x$
$5y = 5 \times y$

Therefore, on factorization we get,
$15xy - 6x +5y-2 = (3\times 5 \times x \times y)-(2 \times 3 \times x)+(5\times y)-2$
$=(5 \times y)(3\times x + 1)-2(3\times x + 1)$

Question:3(iii) Factorise

Answer:

We have,

Therefore, on factorization we get,

Question:3(iv) Factorise

Answer:

We have,

Therefore, on factorization we get,

Question:3(v) Factorise

Answer:

We have,

Therefore, on factorization we get,

NCERT solutions for class 8 maths chapter 14 factorization-Exercise: 14.2

Question:1(i) Factorise the following expressions

Answer:

We have,

$(a+4)^{2}$
Therefore,

Question:1(ii) Factorise the following expressions

Answer:

We have,

$(p-5)^{2}$
Therefore,

Question:1(iii) Factorise the following expressions

Answer:

We have,

$(5m+3)^{2}$
Therefore,

Question:1(iv) Factorise the following expressions

Answer:

We have,

$(7y+ 6z)^{2}$
Therefore,

Question:1(v) Factorise the following expressions

Answer:

We have,

$= 4(x-1)(x-1) \\\ \ \ = 4(x-1)^{2}$

Question:1(vi) Factorise the following expressions

Answer:

We have,

$(11b -4c)^{2}$
Therefore,
$(11b -4c)^{2}$

Question:1(vii) Factorise the following expressions

Answer:

We have,
= $l^{2} + 2ml + m^{2} - 4lm$ $(using \ (a+b)^{2} = a^{2} + 2ab + b^{2})$
= $l^{2} - 2lm + m^{2}$
= $(l-m)^{2}$ $(using \ (a-b)^{2} = a^{_2} -2ab + b^{2})$

Question:1(viii) Factorise the following expressions

Answer:

We have,
= $a^{4}$ + $a^{2}b^{2}$ + $a^{2}b^{2}$ + $b^{4}$
= $a^{2}(a^{2 }+ b^{2}) + b^{2}(a^{2}+b^{2})$ = $(a^{2}+b^{2})(a^{2}+b^{2})$ = $(a^{2}+b^{2})^{2}$

Question:2(i) Factorise :

Answer:

This can be factorized as follows
= $(2p)^{2} - (3q)^{2}$ $(using \ (a)^{2} - (b)^{2} = (a-b)(a+b))$

Question:2(ii) Factorise the following expressions

Answer:

We have,
$(9a^{2} - 16b^{2})$ $((3a)^{2} - (4b)^{2})$
$(using \ (a)^{2} - (b)^{2} = (a-b)(a+b))$

Question:2(iii) Factorise

Answer:

This can be factorised as follows
= $(7x)^{2} - (6)^{2}$ $(using \ (a)^{2} - (b)^{2} = (a-b)(a+b) )$

Question:2(iv) Factorise

Answer:

The given question can be factorised as follows
$= 16x^3(x^{2}- 9)$
$= 16x^3((x)^{2}- (3)^{2})$ $(using \ (a)^{2}- (b)^{2} = (a-b)(a+b))$

Question:2(v) Factorise

Answer:

We have,
(using $a^{2} - b^{2} = (a-b)(a+b)$ )
<img alt="= (l + m - l + m)(l + m + l - m)" height="18"

src="https://lh6.googleusercontent.com/7n9Osk0x9mzggIf5EhzuRIVjoamvGsAg7oAvqBzW_vEutnQcNMHfAdWJIIohDCWF9uxPvzMQlxC0urWVe6ZCEt6T7FrQicXsyfSsz1T8fGexhBVmKHM83VtwCm-b78X5ttNx7CE" style="margin-left: 0px; margin-top: 0px;" width="265" />

Question:2(vi) Factorise

Answer:

We have,
= $(3xy)^{2} -(4)^{2}$ (using $(a)^{2} -(b)^{2} = (a-b) (a+b)$ )

Question:2(vii) Factorise

Answer:

We have,
= $(x-y)^{2} - z^{2}$ $(using \ (a-b)^{2} = a^{2} -2ab + b^{2})$
$(using \ (a)^{2} - (b)^{2} = (a -b ) (a+b))$

Question:2(viii) Factorise

Answer:

We have,
= $25a^{2} - (2b-7c)^{2}$ $(using \ (a-b)^{2} = a^{2} -2ab + b^{2})$
= $(5a)^{2} - (2b-7c)^{2}$ $(using \ (a)^{2} - (b)^{2} = (a -b ) (a+b))$
)

Question:3(i) Factorise the following expressions

Answer:

We have,
$ax^2 = a \times x \times x$
$bx = b \times x$
Therefore,
$= (a \times x \times x) + (b \times x)$
$= x(a \times x + b)$

Question:3(ii) Factorise the following expressions

Answer:

We have,
$7p^2 = 7 \times p \times p$
$21q^3 = 3 \times 7 \times q \times q$
Therefore,
$= (7 \times p \times p) + (3 \times 7 \times q \times q)$
$(p^{2}+ 3q^{2})$

Question:3(iii) Factorise the following expressions

Answer:

We have,
$2x^3 = 2 \times x \times x \times x$
$2xy^2 = 2 \times x \times y \times y$
$2xz^2 = 2 \times x \times z \times z$
Therefore,
$= (2 \times x \times x \times x) + ( 2 \times x \times y \times y) + ( 2 \times x \times z \times z)$
$= (2 \times x) [(x \times x) + (y \times y ) + (z \times z)]$

Question:3(iv) Factorise the following expressions

Answer:

We have,

$(m^{2 }+n^{2})$

Question:3(v) Factorise the following expressions

Answer:

We have,

Question:3(vi) Factorise the following expressions

Answer:

We have,

Take ( y+z) common from this
Therefore,

Question:3(vii) Factorise the following expressions

Answer:

We have,

Therefore,

Question:3(viii) Factorise

Answer:

We have,

Therefore,

Question:3(ix) Factorise the following expressions

Answer:

We have,

Therefore,

Question:4(i) Factorise

Answer:

We have,
= $(a^{2})^{2} - (b^{2})^{2} = (a^{2} - b^{2})(a^{2} + b^{2}) = (a-b)(a+b)(a^{2} + b^{2})$
$using \ (x^{2} - y^{2}) = (x-y)(x+y)$

Question:4(ii) Factorise

Answer:

We have,
=
$using \ a^{2} - b^{2} = (a-b)(a+b)$

Question:4(iii) Factorise

Answer:

We have,
=

$(using \ a^{2} -b^{2} = (a-b)(a+b))$

Question:4(iv) Factorise

Answer:

We have,
= $(x^{2})^{2} - ((x-z)^{2})^{2}$ $using \ a^{2}-b^{2} = (a-b)(a+b)$
= $(x^{2} - (x-z)^{2})(x^{2}+(x-z)^{2})$
= $(x+(x-z))(x - (x-z))(x^{2}+(x-z)^{2})$
= $x^{2}+(x-z)^{2}$

Question:4(v) Factorise

Answer:

We have,
= $a^{4} - a^{2}b^{2} - a^{2}b^{2} + b^{4}$
= $a^{2}(a^{2} - b^{2}) - b^{2}(a^{2} - b^{2})$
= $(a^{2} - b^{2}) (a^{2}-b^{2})$ $using \ a^{2}-b^{2} = (a-b)(a+b)$
= $(a^{2} - b^{2})^{2}$
= $((a - b)(a+b))^{2}$
= $(a - b)^{2}(a+b)^{2}$

Question:5(i) Factorise the following expression

Answer:

We have,
= $p^{2} + 2p + 4p + 8$

Therefore,

Question:5(ii) Factorise the following expression

Answer:

We have,
= $q^{2} - 7q -3q + 21$

Therefore,

Question:5(iii) Factorise the following expression

Answer:

We have,
= $p^{2} + 8p - 2p - 16$

Therefore,

NCERT solutions for class 8 maths chapter 14 factorization topic 14.3.1 division of a monomial by another monomial

Question:(i) Divide $24 xy^2 z^3 \: \: by \: \: 6 yz^2$

Answer:

We have,

Question:(ii) Divide $63 a ^ 2 b^ 4 c ^6 \: \: by \: \: 7 a ^2 b^ 2 c ^3$

Answer:

We have,

NCERT solutions for class 8 maths chapter 14 factorization-Exercise: 14.3

Question:1(i) Carry out the following divisions

$28 x ^ 4 \div 56 x$

Answer:

This is done using factorization.

Question:1(ii) Carry out the following divisions

$-36 y^3 \div 9 y^2$

Answer:

We have,
$y^{3}$ $= -1 \times 2 \times 2 \times 3 \times 3 \times y \times y \times y$
$y^{2 }$ $= 3 \times 3 \times y \times y$
Therefore,

Question:1(iii) Carry out the following divisions

$66 pq^2 r ^ 3 \div 11 q r ^2$

Answer:

We have,
$66pq^2r^3 = 2 \times 3 \times 11 \times p \times q \times q \times r \times r \times r$
$11qr^2 = 11 \times q \times r \times r$
Therefore,

Question:1(iv) Carry out the following divisions

$34 x^ 3 y^3 z ^ 3 \div 51 x y^2 z ^ 3$

Answer:

We have,

Question:1(v) Carry out the following divisions

$12 a ^ 8 b^ 8 \div ( -6 a ^ 6 b ^ 4 )$

Answer:

We have,

Question:2(i) Divide the given polynomial by the given monomial

$( 5x ^2 -6x ) \div 3x$

Answer:

We have,

$\therefore \frac{5x^{2}-6}{3x} = \frac{x(5x-6)}{3x} = \frac{5x-6}{3}$

Question:2(ii) Divide the given polynomial by the given monomial

$( 3 y ^8 - 4 y^6 + 5 y ^4 )\div y ^ 4$

Answer:

We have,
$3y^{8} - 4y^{6} + 5y^{4} = y^{4}(3y^{4}-4y^{2} + 5)$
$\therefore \frac{y^{4}(3y^{4}-4y^{2}+5)}{y^{4}} = (3y^{4}-4y^{2}+5)$

Question:2(iii) Divide the given polynomial by the given monomial

$8 ( x ^3 y^2 z ^2 + x^2 y^3 z^2 + x ^2 y^2 z^3 ) \div 4 x ^2 y ^2 z ^2$

Answer:

We have,
$8(x^{3}y^{2}z^{2} + x^{2}y^{3}z^{2} + x^{2} y^{2}z^{3}) = 8x^{2}y^{2}z^{2}(x+y+z)$

Question:2(iv) Divide the given polynomial by the given monomial

$( x^3 +2 x ^2 + 3 x ) \div 2x$

Answer:

We have,
$x^{3} + 2x^{2} + 3x = x(x^{2} + 2x + 3)$

$\therefore \frac{x^{3} + 2x^{2} + 3x}{2x} = \frac{x(x^{2} + 2x + 3)}{2x} = \frac{x^{2} + 2x + 3}{2}$

Question:2(v) Divide the given polynomial by the given monomial

$( p ^ 3 q ^6 - p ^ 6 q ^ 3 ) \div p ^3 q ^3$

Answer:

We have,
$(p^{3}q^{6} - p^{6}q^{3}) = p^{3}q^{3}(q^{3} - p^{3})$

Question:3(i) workout the following divisions

$( 10 x - 25) \div 5$

Answer:

We have,

Therefore,
$\frac{10x-25}{5}= \frac{5(2x-5)}{5} = 2x - 5$

Question:3(ii) workout the following divisions

$( 10 x -25 ) \div ( 2x -5 )$

Answer:

We have,

Therefore,
$\frac{10x-25}{2x-5} = \frac{5(2x-5)}{2x-5} = 5$

Question:3(iii) workout the following divisions

$10 y ( 6y +21 ) \div 5 ( 2y + 7 )$

Answer:

We have,
$10y(6y + 21) = 2 \times y \times 5 \times 3(2y + 7)$
Therefore,
$\frac{10y(6y+21)}{5(2y+7)} = \frac{2 \times 5 \times y \times 3(2y+7)}{5(2y+7)} = 6y$

Question:3(iv) workout the following divisions

$9 x ^2 y^2 ( 3z -24 ) \div 27 xy ( z-8 )$

Answer:

We have,
$9x^{2}y^{2}(3z-24) = 9x^{2}y^{2} \times 3(z-8) = 27x^{2}y^{2}(z-8)$

$\therefore \frac{9x^{2}y^{2}(3z-24)}{27xy(z-8)} = \frac{27x^{2}y^{2}(z-8)}{27xy(z-8)} = xy$

Question:3(v) workout the following divisions

$96 abc ( 3a -12 ) ( 5 b -30 ) \div 144 (a-4 ) ( b- 6 )$

Answer:

We have,
$96abc(3a - 12)(5b - 30) = 2 \times 48abc \times 3(a - 4) \times 5(b - 6)$
$= 2 \times144abc (a - 4) \times 5(b - 6)$
Therefore,
$\frac{96abc(3a-12)(5b-30)}{144(a-4)(b-6)} = \frac{2 \times 144abc (a-4) \times 5 (b - 6)}{144(a-4)(b-6)} = 10abc$

Question:4(i) Divide as directed

$5 ( 2x +1 ) ( 3x +5 ) \div ( 2x +1)$

Answer:

We have,
$\frac{5(2x+1)(3x+5)}{2x+1} = 5(3x+5)$

Question:4(ii) Divide as directed

$26 xy ( x+5 ) ( y-4) \div 13 x ( y-4 )$

Answer:

We have,
$\frac{26xy(x+5)(y-4)}{13x(y-4)} = \frac{2 \times 13xy(x+5)(y-4)}{13x(y-4)} =2y(x+5)$

Question:4(iii) Divide as directed

$52 pqr ( p+ q ) ( q+ r ) ( r +p)\div 104 pq ( q+r ) ( r + p )$

Answer:

We have,
$\frac{52pqr(p+q)(q+r)(r+p)}{104pq(q+r)(r+p)} = \frac{r(p+q)}{2}$

Question:4(iv) Divide as directed

$20 ( y+4 ) ( y^2 + 5 y + 3 ) \div 5 (y +4 )$

Answer:

We have,
<img alt="\frac{20(y+4)(y^{2}+5y+3)}{5(y+4)} =\frac{4 \times 5(y+4)(y^{2}+5y+3)}{5(y+4)} = 4(y^{2}+5y+3)" height="45"

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Question:4(v) Divide as directed

$x ( x+1 ) ( x+2 ) ( x+3 ) \div x ( x+1 )$

Answer:

We have,
$\frac{x(x+1)(x+2)(x+3)}{x(x+1)} = (x+2)(x+3)$

Question:5(i) Factorise the expression and divide then as directed

$( y ^ 2 + 7 y + 1 0 ) \div ( y + 5 )$

Answer:

We have,

Question:5(ii) Factorise the expression and divide then as directed

$( m^2 - 14 m -32 ) \div ( m +2 )$

Answer:

We have,

Question:5(iii) Factorise the expression and divide then as directed

$( 5 p^2 -25p + 20 ) \div ( p-1 )$

Answer:

We have,

Question:5(iv) Factorise the expression and divide then as directed

$4 yz ( z^2 + 6z -16 ) \div 2y ( z+8 )$

Answer:

We first simplify our numerator
So,

Add and subtract 64 $\Rightarrow$

$using \ a^{2} -b^{2} = (a - b)(a + b)$

Now,
$\frac{4yz(z^{2}+6z-16)}{2y(z+8)} = \frac{4yz(z+8)(z-2)}{2y(z+8)}= 2z(z-2)$

Question:5(v) Factorise the expression and divide then as directed

$5 pq ( p^2 - q ^ 2 ) \div 2 p ( p + q )$

Answer:

We have,

Question:5(vi) Factorise the expression and divide then as directed

$12 xy ( 9 x^2 - 16 y^2 ) \div 4 xy ( 3 x + 4 y )$

Answer:

We first simplify our numerator,
( $9x^{2} -16y^{2}$ ) = $(3x)^{2} -(4y)^{2}$

using $(a)^{2} -(b)^{2} = (a-b)(a+b)$

Now,
$\frac{12xy(9x^{2} - 16y^{2})}{4xy(3x + 4y)} = \frac{12xy(3x+4y)(3x-4y)}{4xy(3x+4y)} = 3(3x-4y)$

Question:5(vii) Factorise the expression and divide then as directed

$39 y^2 ( 50 y^2 - 98 ) \div 26 y^2 ( 5y +7 )$

Answer:

We first simplify our numerator,
$39y^{2}(50y^{2} -98) = 39y^{2} \times 2(25y^{2} - 49)$ using $(a)^{2} -(b)^{2} = (a-b)(a+b)$
= $78y^{2} ((5y)^{2} - (7)^{2})$
= $78y^{2} (5y - 7)(5y+7)$
Now,
$\frac{39y^{2}(50y^{2}-98)}{26y^{2}(5y +7)} = \frac{78y^{2}(5y-7)(5y+7)}{26y^{2}(5y+7)} = 3(5y-7)$

NCERT solutions for class 8 maths chapter 14 factorization-Exercise: 14.4

Question:1 Find and correct the errors in the following mathematical statements

Answer:

Our L.H.S.

R.H.S.
It is clear from the above that L.H.S. is not equal to R.H.S.
So, correct statement is

Question:2 Find and correct the errors in the following mathematical statements

Answer:

Our L.H.S.

R.H.S.=
It is clear from the above that L.H.S. is not equal to R.H.S.
So, correct statement is

Question:3 Find and correct the errors in the following mathematical statements

Answer:

Our L.H.S.
R.H.S. =
It is clear from the above that L.H.S. is not equal to R.H.S.
SO, correct statement is

Question:4 Find and correct the errors in the following mathematical statements

Answer:

Our L.H.S.
R.H.S.
It is clear from the above that L.H.S. is not equal to R.H.S.
So, correct statement is

Question:5 Find and correct the errors in the following mathematical statements

$(Q5)\ 5 y + 2y + y - 7y = 0$

Answer:

Our L.H.S. is

R.H.S. = 0
IT is clear from the above that L.H.S. is not equal to R.H.S.
So, Correct statement is

Question:6 Find and correct the errors in the following mathematical statements

Answer:

Our L.H.S. is

R.H.S. =
It is clear from the above that L.H.S. is not equal to R.H.S.
So, correct statement is

Question:7 Find and correct the errors in the following mathematical statements

Answer:

Our L.H.S. is

R.H.S.
It is clear from the above that L.H.S. is not equal to R.H.S.
So, correct statement is

Question:8 Find and correct the errors in the following mathematical statements

Answer:

Our L.H.S. is
$\Rightarrow (2x)^{2}+5x = 4x^2+5x$
R.H.S. = 9x
It is clear from the above that L.H.S. is not equal to R.H.S.
So, the correct statement is
$(2x)^{2}+5x = 4x^2+5x$

Question:9 Find and correct the errors in the following mathematical statements

Answer:

LHS IS

$(3x + 2)^{2 } = (3x)^{2} + 2(3x)(2) +(2)^{2}$ using $(a + b)^{2 } = (a)^{2} + 2(a)(b) +(b)^{2}$

RHS IS

$\boldsymbol{LHS} \neq \boldsymbol{RHS}$

Correct statement is

$(3x + 2)^{2 } = (3x)^{2} + 2(3x)(2) +(2)^{2}$

Question:10(a) Find and correct the errors in the following mathematical statements

Substituting in $x ^ 2 + 5 x + 4 \: \: gives \: \: ( -3 ) ^ 2 + 5 ( -3 ) + 4 = 9 + 2 + 4 = 15$

Answer:

We need to substitute x = -3 in

$x^{2}+5x+4$
$=(-3)^{2}+5(-3)+4$

$= -2 \neq 15$

so the given statement is wrong
Correct statement is $(-3)^{2}+5(-3)+4= -2$

Question:10(b) find and correct the errors in the following mathematical statements

Substituiting x = -3 in $x ^2 -5 x + 4 \: \: gives \: \: ( -3)^2 - 5 ( -3 ) + 4 = 9 - 15 + 4 = -2$

Answer:

We need to substitute x = -3 in

so the given statement is wrong
Correct statement is

Question:10(c) find and correct the errors in the following mathematical statements

Substituting x = - 3 in $x ^2 + 5 x \: \: gives \: \: ( -3 ) ^ 2 + 5 ( -3 ) = -9-15 = -24$

Answer:

We need to Substitute x = - 3 in $x^{2} + 5x$
= $(-3)^{2} + 5(-3)$
= 9 - 15
= - 6 $\neq$ R.H.S
Correct statement is Substitute x = - 3 in $x^{2} + 5x$ gives -6

Question:11 Find and correct the errors in the following mathematical statements

Answer:

Our L.H.S. is $(y - 3 )^{2}$
= $(y )^{2} + 2(y)(-3) + (-3)^{2}$ using $(a-b)^{2} = (a )^{2} + 2(a)(-b) + (-b)^{2}$
= $y^{2}$ $\neq$ R.H.S.

Correct statement is

$(y - 3 )^{2}$ = $y^{2}$

Question:12 Find and correct the errors in the following mathematical statements

Answer:

Our L.H.S. is $(z+5)^{2}$
= $(z)^{2} + 2(z)(5) + (5)^{2}$ using $(a+b)^{2} = (a)^{2} + 2(a)(b) + (b)^{2}$
= $(z)^{2}$ $\neq$ R.H.S.
Correct statement is

$(z+5)^{2}$ = $(z)^{2}$

Question:13 Find and correct the errors in the following mathematical statements.

Answer:

Our L.H.S. is (2a + 3b)(a -b)
= $2a^{2} -2ab + 3ab - 3b^{2}$
= $2a^{2} +ab - 3b^{2}$ $\neq$ R.H.S.
Correct statement is (2a + 3b)(a -b) = $2a^{2} +ab - 3b^{2}$

Question:14 Find and correct the errors in the following mathematical statements.

Answer:

Oue L.H.S. is (a + 4)(a + 2)
= $a^{2} + 2a + 4a + 8$
= $a^{2} + 6a + 8$ $\neq$ R.H.S.
Correct statement is (a + 4)(a + 2) = $a^{2} + 6a + 8$

Question:15 Find and correct the errors in the following mathematical statements.

Answer:

Our L.H.S. is (a - 2) (a - 4)
= $a^{2} - 4a - 2a + 8$
= $a^{2} - 6a+ 8$ $\neq$ R.H.S.
Correct statement is (a - 2) (a - 4) = $a^{2} - 6a+ 8$

Question:16 Find and correct the errors in the following mathematical statements.

$\frac{3 x ^2}{3 x ^2 } = 0$

Answer:

Our L.H.S. is
$\Rightarrow \frac{3x^{2}}{3x^{2}}$
R.H.S. = 0
It is clear from the above that L.H.S. is not equal to R.H.S.
So, correct statement is
$\frac{3x^{2}}{3x^{2}} = 1$

Question:17 Find and correct the errors in the following mathematical statements.

$\frac{3 x ^2 + 1 }{3 x ^2 } = 1+1 = 2$

Answer:

Our L.H.S. is
$\Rightarrow \frac{3x^2+1}{3x^2}$
R.H.S. = 2
It is clear from the above stattement that L.H.S. is not equal to R.H.S.
So, correct statement is
$\frac{3x^{2}+1}{3x^{2}} = 1 + \frac{1}{3x^{2}} = \frac{3x^{2}+1}{3x^{2}}$

Question:18 find and correct the errors in the following mathematical statements.

$\frac{3 x }{3 x +2 } = 1/2$

Answer:

Our L.H.S.

$\Rightarrow \frac{3x}{3x+2}$

R.H.S. = 1/2

It can be clearly observed that L.H.S is not equal to R.H.S

So, the correct statement is,

$\frac{3x}{3x+2} = \frac{3x}{3x+2}$

Question:19 find and correct the errors in the following mathematical statements

$\frac{3}{4x +3}= \frac{1}{4x }$

Answer:

Our L.H.S. is $\Rightarrow \frac{3}{4x+3} = \frac{3}{4x+3} \neq$ R.H.S.

Correct statement is $\frac{3}{4x+3} = \frac{3}{4x+3}$

Question:20 find and correct the errors in the following mathematical statements

$\frac{4 x + 5 }{4x } = 5$

Answer:

Our L.H.S. is $\Rightarrow \frac{4x+5}{4x} = \frac{4x}{4x} + \frac{5}{4x} = 1 + \frac{5}{4x} \neq$ R.H.S.

Correct statement is $\frac{4x+5}{4x} = 1 + \frac{5}{4x} = \frac{4x+5}{4x}$

Question:21 find and correct the errors in the following mathematical statements

$\frac{7x +5}{5} = 7x$

Answer:

Our L.H.S. is $\Rightarrow \frac{7x+5}{5} = \frac{7x}{5} + \frac{5}{5} = \frac{7x}{5} + 1 \neq$ R.H.S.

Correct statement is $\frac{7x+5}{5} = \frac{7x}{5} + 1 = \frac{7x+5}{5}$