NCERT Solutions for Class 8 Maths Chapter 12 Factorization

Question:(i) Factorise:

$12 x +36$

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,

$12x + 36 =$ $2 \times 2 \times 3 (x + 3)$

$12x + 36 = 12(x + 3)$

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

Answer:

We have,
$22y=$ $2 \times 11 \times y$
$33z =$ $3 \times 11 \times z$
So, we have 11 commons in both
Therefore,

$22y - 33z = 11(2y - 3z)$

Question:(iii) Factorise :

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

Answer:

We have
$14pq =$ $2 \times 7 \times p \times q$
$35pqr =$ $5 \times 7 \times p \times q \times r$
So, we have

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

$14pq + 35pqr =7pq (2 + 5r)$

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

$(i) 12 x , 36$

Answer:

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

$2\times2\times3=12$

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

$(ii) 2y , 22xy$

Answer:

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

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

$(iii) 14 pq, 28 p^2 q^2$

Answer:

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

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

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

$(iv) 2x , 3x ^2 , 4$

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

$( v ) 6 abc , 24 ab^2 , 12 a^2 b$

Answer:

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

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

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

$(vi)16 x ^ 3 , -4 x ^ 2 , 32 x$

Answer:

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

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

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

$(vii) 10 pq , 20 qr , 30 rp$

Answer:

We have,
$10pq ={2 \times 5} \times p \times q$
$20qr = 2\times{2 \times 5 }\times q \times r$
$30rp ={ 2}\times 3{ \times 5} \times r \times p$
Therefore, the common factors between these three is

$2 \times 5 =10$

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

$(viii)3 x ^2 y^3 , 10 x ^3 y ^ 2 , 6 x^ 2 y^2 z$

Answer:

We have,
$3x^{2}y^{2}$ $= 3 \times { x \times x \times y \times y}$
$10x^{3}y^{2}$ $=2 \times 5 \times x \times { x\times x \times y \times y}$
$6x^{2}y^{2}z$ $=2 \times 3 \times{ x \times x \times y \times y} \times z$
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

$(i)7x -42$

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

$(ii)6 p - 12 q$

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

$(iii)7 a ^2 + 14 a$

Answer:

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

Question:2(iv) Factorise the following expressions

$(iv)-16 z + 20 z^3$

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,
$-16z+20z^3 = (-1 \times 2 \times 2 \times 2 \times 2 \times z)+(2 \times 2 \times 5 \times z \times z \times z )$
$= (2\times 2 \times z)(-1 \times 2 \times 2+ 5 \times z \times z )$
$= 4z(-4+5z^2 )$

Question:2(v) Factorise the following expressions

$20 l^2 m + 30 alm$

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,
$20l^2m+30alm =(2\times 2 \times 5 \times l \times l \times m) + (2 \times 3 \times 5 \times a \times l \times m)$
$=(2\times 5 \times l \times m)(2\times l + 3 \times a )$
$=10lm(2l+3a)$

Question:2(vi) Factorise the following expressions

$5 x^2 y - 15 xy^2$

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,
$5x^2y - 15xy^2 = (5 \times x \times x \times y ) - (3\times 5 \times x \times y \times y )$
$=(5\times x \times y) ( x - 3\times y )$
$=5xy (x-3y)$

Question:2(vii) Factorise the following expressions

$10 a ^2 - 15 b^2 +20 c^2$

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,
$10a^2-15b^2+20c^2 = (2\times 5 \times a \times a)-(3\times 5 \times b \times b)+(2\times 2 \times 5 \times c \times c)$ $=5 (2 \times a \times a-3 \times b \times b+2\times 2 \times c \times c)$
$=5(2a^2-3b^2+4c^2)$

Question:2(viii) Factorise the following expressions

$- 4 a ^2 + 4 ab - 4ca$

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,
$-4a^2+4ab-4ca = (-1 \times 2 \times 2 \times a\times a )+( 2 \times 2 \times a\times b )- (2 \times 2 \times c\times a)$

$=(2 \times 2 \times a) (-1 \times a + b - c)$
$= 4a(-a+b-c)$

Question:2(ix) Factorise the following expressions

$x^2 yz + xy^2 z + xyz^2$

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,
$x^2yz+xy^2z+xyz^2 =(x \times x \times y \times z)+(x \times y \times y \times z)+(x \times y \times z \times z)$

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

Question:2(x) Factorise the following expressions

$a x^2 y + bxy^2 + cxyz$

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,
$ax^2y+bxy^2+cxyz = ( a \times x \times x \times y)+( b \times x \times y \times y)+(c \times x \times y \times z)$ $= (x\times y)( a \times x+ b \times y+c \times z)$

$= xy(ax+by+cz)$

Question:3(i) Factorise $x ^ 2 + xy + 8 x + 8y$

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,
$x^2+xy+8x+8y = (x \times x)+(x\times y )+(8 \times x)+(8 \times y)$
$= x(x +y )+8(x+ y)$
$= (x+8)(x+y)$

Question:3(ii) Factorise

$15 xy -6 x +5 y -2$

Answer:

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

Question:3(iii) Factorise

$ax + bx - ay - by$

Answer:

We have,
$ax+bx-ay-by = a(x-y)-b(x-y)$
$=(a-b)(x-y)$
Therefore, on factorization we get,
$(a-b)(x-y)$

Question:3(iv) Factorise

$15 pq + 15 + 9q + 25p$

Answer:

We have,
$15pq + 15 + 9q + 25p = 5 p(3q + 5) + 3 (3q + 5)$
$= (3q + 5)(5p + 3)$
Therefore, on factorization we get,
$(3q + 5)(5p + 3)$

Question:3(v) Factorise

$z-7 + 7 xy - xyz$

Answer:

We have,
$z - 7 + 7xy - xyz = z(1 - xy) -7(1 - xy)$
$= (1 - xy)(z - 7)$
Therefore, on factorization we get,
$(1 - xy)(z - 7)$

Question:1(i) Factorise the following expressions

$a ^ 2 + 8a + 16$

Answer:

We have,
$a^2 + 8a + 16 = a^2+ 4a + 4a + 16$
$= a(a + 4) + 4 (a+4)$
$= (a+4)(a+4) =$ $(a+4)^{2}$
Therefore,
$a^2+8a+16 = (a+4)^2$

Question:1(ii) Factorise the following expressions

$p^2 -10 p + 25$

Answer:

We have,
$p^2 - 10p + 25 = p^2 - 5p - 5p + 25$
$= p(p - 5) -5 (p -5)$
$= (p - 5)(p - 5) =$ $(p-5)^{2}$
Therefore,
$p^2-10p+25 =(p-5)^2$

Question:1(iii) Factorise the following expressions

$25 m ^2 + 30 m + 9$

Answer:

We have,
$25m^2 + 30m + 9 = 25m^2 + 15m + 15m + 9$
$= 5m (5m + 3) +3(5m + 3)$
$= (5m + 3) (5m + 3) =$ $(5m+3)^{2}$
Therefore,
$25m^2+30m+9 = (5m+3)^2$

Question:1(iv) Factorise the following expressions

$49 y^2 + 84 yz + 36 z^2$

Answer:

We have,
$49 y^2 + 84 yz + 36 z^2$ $= 49y^2 + 42yz + 42yz + 36z^2$
$= 7y(7y + 6z) + 6z(7y + 6z)$
$= (7y + 6z)(7y + 6z) =$ $(7y+ 6z)^{2}$
Therefore,
$49y^2+84yz+36z^2=(7y+6z)^2$

Question:1(v) Factorise the following expressions

$4 x^2 - 8x + 4$

Answer:

We have,
$4 x^2 - 8x + 4$ $= 4x^2 - 4x - 4x + 4$
$= 4x(x - 1) -4(x - 1)$
$= 4(x-1)(x-1) \\\ \ \ = 4(x-1)^{2}$

Question:1(vi) Factorise the following expressions

$121 b^2 - 88 bc + 16 c^2$

Answer:

We have,
$121 b^2 - 88 bc + 16 c^2$ $= 121b^2 - 44bc - 44bc + 16c^2$
$= 11b(11b - 4c) - 4c(11b - 4c)$
$= (11b-4c)(11b-4c) =$ $(11b -4c)^{2}$
Therefore,
$121 b^2 - 88 bc + 16 c^2$ $=$ $(11b -4c)^{2}$

Question:1(vii) Factorise the following expressions

$( l+m ) ^2 - 4lm$

Answer:

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

Question:2(ii) Factorise the following expressions

$63 a ^2 - 112 b ^ 2$

Answer:

We have,
$63 a ^2 - 112 b ^ 2$ $= 7$ $(9a^{2} - 16b^{2})$ $= 7$ $((3a)^{2} - (4b)^{2})$ $=7 (3a - 4b)(3a + 4b)$
$(using \ (a)^{2} - (b)^{2} = (a-b)(a+b))$

Question:2(iii) Factorise

$49 x^2 - 36$

Answer:

This can be factorised as follows
$49 x^2 - 36$ = $(7x)^{2} - (6)^{2}$ $= (7x - 6)(7x + 6)$ $(using \ (a)^{2} - (b)^{2} = (a-b)(a+b) )$

Question:2(iv) Factorise

$16 x^5 - 144 x ^ 3$

Answer:

The given question can be factorized as follows
$16 x^5 - 144 x ^ 3$ $= 16x^3(x^{2}- 9)$
$= 16x^3((x)^{2}- (3)^{2})$ $= 16x^3(x-3)(x+3)$ $(using \ (a)^{2}- (b)^{2} = (a-b)(a+b))$

Question:2(v) Factorise

$(l+m) ^ 2 - ( l- m ) ^2$

Answer:

We have,
$(l+m) ^ 2 - ( l- m ) ^2$ $= [(l + m) - (l - m)][(l + m) + (l - m)]$ (using $a^{2} - b^{2} = (a-b)(a+b)$ )
$= (l + m - l + m)(l + m + l - m)$
$= (2m)(2l) = 4ml$

Question:2(vi) Factorise

$9 x ^2 y^2 - 16$

Answer:

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

Question:2(vii) Factorise

$( x ^2 -2xy + y^2 ) - z ^2$

Answer:

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

Question:2(viii) Factorise

$25 a ^2 -4 b ^2 + 28 bc - 49 c ^2$

Answer:

We have,
$25 a ^2 -4 b ^2 + 28 bc - 49 c ^2$ = $25a^{2} - (2b-7c)^{2}$ $(using \ (a-b)^{2} = a^{2} -2ab + b^{2})$
= $(5a)^{2} - (2b-7c)^{2}$ $(using \ (a)^{2} - (b)^{2} = (a -b ) (a+b))$
$=(5a - (2b - 7c))(5a + (2b - 7c)$ )
$= (5a - 2b + 7c)(5a + 2b - 7c )$

Question:3(i) Factorise the following expressions

$ax ^2 + bx$

Answer:

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

Question:3(ii) Factorise the following expressions

$7p^2 + 21 q ^2$

Answer:

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

Question:3(iii) Factorise the following expressions

$2 x^3 + 2xy^2 + 2 xz ^2$

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

Question:3(iv) Factorise the following expressions

$am^2 + bm ^2 + bn ^2 + an^2$

Answer:

We have,
$am^2 + bm ^2 + bn ^2 + an^2$ $= m^2(a + b) + n^2(a + b)$
$= (a + b)$ $(m^{2 }+n^{2})$

Question:3(v) Factorise the following expressions

$( lm + l ) + m + 1$

Answer:

We have,
$( lm + l ) + m + 1$ $= lm + l + m + 1$
$= l(m + 1) +1(m + 1)$
$= (m + 1)(l + 1)$

Question:3(vi) Factorise the following expressions

$y ( y + z ) + 9 ( y + z )$

Answer:

We have,
$y ( y + z ) + 9 ( y + z )$
Take ( y+z) common from this
Therefore,
$y ( y + z ) + 9 ( y + z )$ $= (y + z)(y + 9)$

Question:3(vii) Factorise the following expressions

$5 y ^ 2 - 20 y - 8z + 2yz$

Answer:

We have,
$5 y ^ 2 - 20 y - 8z + 2yz$ $= 5y(y - 4) + 2z(y - 4)$
$= (y - 4)(5y + 2z)$
Therefore,
$5 y ^ 2 - 20 y - 8z + 2yz$ $= (y - 4)(5y + 2z)$

Question:3(viii) Factorise

$10 ab + 4a + 5b + 2$

Answer:

We have,
$10 ab + 4a + 5b + 2$ $= 2a(5b + 2) + 1(5b + 2)$
$= (5b + 2)(2a + 1)$
Therefore,
$10 ab + 4a + 5b + 2$ $= (5b + 2)(2a + 1)$

Question:3(ix) Factorise the following expressions

$6 xy - 4 y + 6 - 9 x$

Answer:

We have,
$6 xy - 4 y + 6 - 9 x$ $= 2y(3x - 2) - 3 (3x - 2)$
$= (3x - 2)(2y - 3)$
Therefore,
$6 xy - 4 y + 6 - 9 x$ $= (3x - 2)(2y - 3)$

Question:4(i) Factorise $a ^ 4 - b ^ 4$

Answer:

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

Question:4(ii) Factorise $p ^ 4 - 81$

Answer:

We have,
$p ^ 4 - 81$ =
$(p^{2})^{2} - (9)^{2} = (p^{2} - 9)(p^{2}+9) = (p^{2}-(3)^{2})(p^{2}+9) = (p-3)(p+3)(p^{2}+9)$ [$using \ a^{2} - b^{2} = (a-b)(a+b)$]

Question:4(iii) Factorise $x ^4 - ( y + z )^4$

Answer:

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

Question:4(iv) Factorise $x ^ 4 - ( x-z ) ^ 4$

Answer:

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

Question:4(v) Factorise $a ^ 4 - 2 a^2 b^2 + b ^ 4$

Answer:

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

Question:5(i) Factorise the following expression

$p^ 2 + 6 p + 8$

Answer:

We have,
$p^ 2 + 6 p + 8$ = $p^{2} + 2p + 4p + 8$
$= p(p + 2) + 4(p + 2)$
$=(p + 2)(p + 4)$
Therefore,
$p^ 2 + 6 p + 8$ $=(p + 2)(p + 4)$

Question:5(ii) Factorise the following expression

$q ^ 2 - 10 q + 21$

Answer:

We have,
$q ^ 2 - 10 q + 21$ = $q^{2} - 7q -3q + 21$
$= q(q - 7) -3(q - 7)$
$=(q - 7)(q - 3)$
Therefore,
$q ^ 2 - 10 q + 21$ $=(q - 7)(q - 3)$

Question:5(iii) Factorise the following expression

$p^2 + 6 p - 16$

Answer:

We have,
$p^2 + 6 p - 16$ = $p^{2} + 8p - 2p - 16$
$= p(p + 8) -2(p + 8)$
$=(p - 2)(p + 8)$
Therefore,
$p^2 + 6 p - 16$ $=(p - 2)(p + 8)$

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

Answer:

We have,
$\frac{24xy^{2}z^{3}}{6yz^{2}} =\frac{2\times 2\times 2\times3\times y \times y \times z\times z\times z}{2\times 3 \times y \times z \times z}= 4xyz$

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

Answer:

We have,
$\frac{63a^{2}b^{4}c^{6}}{7a^{2}b^{2}c^{3}}=\frac{3\times 3 \times 7 \times a \times a \times b \times b\times b^2 \times c \times c \times c \times c^3}{7a^{2}b^{2}c^{3}} = 9b^{2}c^{3}$

Question:1(i) Carry out the following divisions

$28 x ^ 4 \div 56 x$

Answer:

$\frac{28x^{4}}{56x} = \frac{2 \times 2 \times 7 \times x \times x \times x \times x}{2 \times 2 \times 2 \times 7 \times x} = \frac{x^{3}}{2}$

This is done using factorization.

Question:1(ii) Carry out the following divisions

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

Answer:

We have,
$-36$ $y^{3}$ $= -1 \times 2 \times 2 \times 3 \times 3 \times y \times y \times y$
$9$ $y^{2 }$ $= 3 \times 3 \times y \times y$
Therefore,

$\frac{-36y^{3}}{9y^{2}} = \frac {-1 \times 2 \times 2 \times 3 \times 3 \times y \times y \times y}{3 \times 3 \times y \times y} = -4y$

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,
$\frac{66pq^{2}r^{3}}{11qr^{2}} = \frac{2 \times 3 \times 11 \times p \times q \times q \times r \times r \times r}{11 \times q \times r \times r} = 6pqr$

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,
$\therefore \frac{34x^{3}y^{3}z^{3}}{51xy^{2}z^{3}} = \frac{2 \times 17\times \ x \times x \times x \times y \times y \times y \times z\times z \times z}{3 \times 17 \times x \times y \times y \times z \times z\times z} = \frac{2x^{2}y}{3}$

Question:1(v) Carry out the following divisions

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

Answer:

We have,
$\frac{12a^8b^8}{-6a^4b^4}= \frac{2 \times 2 \times 3 \times a \times a \times a^{6} \times b \times b \times b \times b \times b^{4}}{-1 \times 2 \times 3 \times a^{6} \times b^{4}} = -2a^{2}b^{4}$

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

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

Answer:

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

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,
$8(x^{3}y^{2}z^{2} + x^{2}y^{3}z^{2} + x^{2} y^{2}z^{3}) = 8x^{2}y^{2}z^{2}(x+y+z)$
$\therefore \frac{8(x^{3}y^{2}z^{2} + x^{2}y^{3}z^{2} + x^{2} y^{2}z^{3})}{4x^{2}y^{2}z^{2}} =\frac{ 8x^{2}y^{2}z^{2}(x+y+z)}{4x^{2}y^{2}z^{2}} =2(x+y+z)$

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,
$(p^{3}q^{6} - p^{6}q^{3}) = p^{3}q^{3}(q^{3} - p^{3})$
$\therefore \frac{(p^{3}q^{6} - p^{6}q^{3})}{p^{3}q^{3}} = \frac{p^{3}q^{3}(q^{3} - p^{3})}{p^{3}q^{3}} = (q^{3} - p^{3})$

Question:3(i) workout the following divisions

$( 10 x - 25) \div 5$

Answer:

We have,
$10x -25 = 5(2x - 5)$
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,
$10x-25 = 5(2x - 5 )$
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,
$\frac{20(y+4)(y^{2}+5y+3)}{5(y+4)} =\frac{4 \times 5(y+4)(y^{2}+5y+3)}{5(y+4)} = 4(y^{2}+5y+3)$

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,
$\frac{y^{2}+7y+10}{y+5} = \frac{y^{2}+2y +5y +10}{y+5} =\frac{y(y+2)+5(y+2)}{y+5}\\ \\ \Rightarrow \frac{(y+5)(y+2)}{(y+5)} = (y+2)$

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

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

Answer:

We have,
$\frac{m^{2}-14m-32}{m+2} = \frac{m^{2}+2m-16m-32}{m+2} = \frac{m(m+2)-16(m+2)}{m+2}\\ \\\Rightarrow \frac{(m-16)(m+2)}{m+2} = m-16$

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

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

Answer:

We have,
$\frac{5p^{2}-25p+20}{p-1} = \frac{5p^{2} -5p -20p +20}{p-1} = \frac{5p(p-1)-20(p-1)}{p-1}\\ \\ \frac{(5p-20)(p-1)}{p-1} = 5p-20$

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,
$4yz( z^2+ 6z - 16)$
Add and subtract 64 $\Rightarrow$ $4yz( z^2- 64 + 6z - 16 + 64)$
$= 4yz(z^2-8^2 + 6z + 48)$
$= 4yz((z + 8)(z - 8) + 6(z + 8))$ $using \ a^{2} -b^{2} = (a - b)(a + b)$
$= 4yz (z + 8)(z - 8 + 6)$
$= 4yz(z + 8)(z - 2)$
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,
$\frac{5pq(p^{2} - q^{2})}{2p(p+q)} = \frac{5pq(p-q)(p+q)}{2p(p+q)} \ \ \ \ \ \ \ \ \ \ \ \ \ using \ a^{2}-b^{2}= (a-b)(a+b)$ $= \frac{5q(p-q)}{2}$

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

Using $(a)^{2} -(b)^{2} = (a-b)(a+b)$
$= 12xy((3x - 4y)(3x + 4y))$
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,
$\frac{39y^{2}(50y^{2}-98)}{26y^{2}(5y +7)} = \frac{78y^{2}(5y-7)(5y+7)}{26y^{2}(5y+7)} = 3(5y-7)$