Excercise 11.1
Solution 1
(i) $8 x y^{3}+12 x^{2} y^{2}$
H.C.F of $8 x y^{3}$ and $12 x^{2} y^{2}$ is $4 x y^{2}$
$\therefore$ Divide each expression by $4 x y^{2}$ and keep $4 x y^{2}$
outside the bracket.
$\Rightarrow \quad 4 x y^{2}(2 y+3 x)$
$\therefore 8 x y^{3}+12 x^{2} y^{2}=4 x y^{2}(2 y+3 x)$
(ii) $15 a x^{3}-9 a x^{2}$
H.C.F of exprassion $15 a x^{3}$ and $9 a x^{2}$ is $3 a x^{2}$
$\quad 15 a x^{3}-9 a x^{2}=3 a x^{2}(5 x-3)$
Solution 2
(i) $21 p y^{2}-56 p y$
$H \cdot C \cdot F$ of $21 p y^{2}$ and $56 p y$ is $7 p y$
$\therefore 21 p y^{2}-56 p y \Rightarrow 7 p y(3 y-8)$
(ii) $4 x^{3}-6 x^{2}$
H.c.F of $4 x^{3}$ and $6 x^{2}$ is $2 x^{2}$
$\therefore 4 x^{3}-6 x^{2} \Rightarrow 2 x^{2}(2 x-3)$
Solution 3
(i) $25 a b c^{2}-15 a^{2} b^{2} c$
H.c.F of $25 a b c^{2}$ and $15 a^{2} b^{2} c$ is $5 a b c$
$\therefore 25 a b c^{2}-15 a^{2} b^{2} c \Rightarrow 5 a b c(5 c-3 a b)$
(ii)
$x^{2} y z+x y^{2} z+x y z^{2}$
HC.F of $x^{2} y z$, $x y^{2} z$ and $x y z^{2}$ is $x y z$
$\therefore \quad x y z(x+y+z)$
Solution 4
(i) $8 x^{3}-6 x^{2}+10 x$
H.C.F of $8 x^{3}, 6 x^{2}$ and $10 x$ is $2 x$
$\Rightarrow \quad 2 x\left(4 x^{2}-3 x+5\right)$
(ii) $14 \mathrm{mn}+22 \mathrm{~m}-6 \mathrm{p}$
H.C.F of $14 \mathrm{mn}, 22 \mathrm{~m}$ and $62 \mathrm{P}$ is 2
$\Rightarrow 2(7 \mathrm{mn}+11 \mathrm{n}-31 \mathrm{P})$
Solution 5
(i) $18 p^{2} q^{2}-24 p q^{2}+30 p^{2} q$
$H \cdot c \cdot F$ of $18 p^{2} q^{2}, 24 p a^{2}$ and $30 p^{2} q$ is $6 p q$
$\Rightarrow \quad 6 p q(3 p q-4 q+5 p)$
(ii) $27 a^{3} b^{3}-18 a^{2} b^{3}+75 a^{3} b^{2}$
H. C .F of $27 a^{3} b^{3}, 18 a^{2} b^{2}$ and $7 s a^{3} b^{2}$ is $3 a^{2} b^{2}$
$\Rightarrow 3 a^{2} b^{2}(9 a b-6 b+25 a)
Solution 6
(i) $15 a(2 p-3 q)-10 b(2 p-3 q)$
$H \cdot c \cdot F \quad$ of $15 a(2 p-3 q)$ and $10 b(2 p-3 q)$ is
$5(2 p-3 q)$
$\Rightarrow 5(2 p-3 q)(3 a-2 b)$
(ii) $3 a\left(x^{2}+y^{2}\right)+6 b\left(x^{2}+y^{2}\right)$
$H \cdot C \cdot F$ of $3 a\left(x^{2}+y^{2}\right)$ and $6 b\left(x^{2}+y^{2}\right)$ is $3\left(x^{2}+y^{2}\right)$
$\therefore \Rightarrow 3\left(x^{2}+y^{2}\right)(a+2 b)$
Solution 7
(i) $6(x+2 y)^{3}+8(x+2 y)^{2}$
H.C.F of $6(x+2 y)^{3}$ and $8(x+2 y)^{2}$ is $2(x+2 y)^{2}$
$\therefore \quad 2(x+2 y)^{2} \quad(3(x+2 y)+4)$
$\Rightarrow \quad 2(x+2 y)^{2}(3 x+6 y+4)$
(ii)$14(a-3 b)^{3}-21 p(a-3 b)$
H.C.F of $14(a-3 b)^{3}$ and $21 p(a-3 b)$ is
$\quad F(a-3 b)$
$\rightarrow \quad 7(a-3 b)\left[2(a-3 b)^{2}-3 p\right]$
Solution 8
(i) $\quad 10 a(2 p+q)^{3}-15 b(2 p+q)^{2}+35(2 p+q)$
$H \cdot C \cdot F$ of $10 a(2 p+q)^{3}, 15 b(2 p+q)^{2}$ and
$35(2 p+q)$ is $5(2 p+q)$
$\Rightarrow 5(2 p+q)(2 a-3 b+7)$
Excercise 11.2
Solution 1
(i)
$\begin{aligned} & x^{2}+x y-x-y \\ \Rightarrow & x(x+y)-1(x+y) \\ \Rightarrow &(x+y)(x-1) \end{aligned}$
(ii)
$\begin{aligned} & y^{2}-y z-5 y+5 z \\ \Rightarrow & y(y-z)-5(y-z) \\ \Rightarrow &(y-z)(y-5) \end{aligned}$
Solution 2
(i)
$\begin{aligned} & 5 x y+7 y-5 y^{2}-7 x \\ \Rightarrow & 5 x y-5 y^{2}+7 y-7 x . \\ \Rightarrow & 5 x y-5 y^{2}-7 x+7 y \\ \Rightarrow & 5 y(x-y)-7(x-y) \\ \rightarrow &(x-y)(5 y-7) \end{aligned}$
(ii) $5 p^{2}-8 p q-10 p+16 q$
$5 p^{2}-10 p-8 p q+16 q$
$5 p(p-2)-8 q(p-2)$
$(p-2)(5 p-8 q)$
Solution 3
(i)
$\begin{aligned} & a^{2} b-a b^{2}+3 a-3 b \\ \Rightarrow & a b(a-b)+3(a-b) \\ \Rightarrow &(a-b)(a b+3) \end{aligned}$
(ii)
$\begin{aligned} & x^{3}-3 x^{2}+x-3 \\=& x^{2}(x-3)+1(x-3) \\ \Rightarrow &(x-3)\left(x^{2}+1\right) \end{aligned}$
Solution 4
(i)
$\begin{aligned} & 6 x y^{2}-3 x y-10 y+5 \\ \Rightarrow & 3 x y(2 y-1)-5(2 y-1) \\ \Rightarrow &(2 y-1)(3 x y-5) \end{aligned}$
(ii) $3 a x-6 a y-8 b y+4 b x$
$ \quad 3 a(x-2 y)+2 b(-2 y+x)$
$\Rightarrow 3 a(x-2 y)+2 b(x-2 y)$
$\Rightarrow \quad(x-2 y)(3 a+2 b)$
Solution 5
(i) $x^{2}+x y(1+y)+y^{3}$\
$\Rightarrow x^{2}+x y+x y^{2}+y^{3}$
$\Rightarrow \quad x(x+y)+y^{2}(x+y)$
$\Rightarrow \quad(x+y)\left(x+y^{2}\right)$
(ii) $y^{2}-x y(1-x)-x^{3}$
$\Rightarrow y^{2}-x y+x^{2} y-x^{3}$
$\Rightarrow y(y-x)+x^{2}(y-x)$
$\Rightarrow(y-x)+\left(y+x^{2}\right)$
Solution 6
(i) $a b^{2}+(a-1) \cdot b-1$
$a b^{2}+a b-b-1$
$a b(b+1)-(b+1)$
$(b+1)(a b-1)$
(ii) $2 a-4 b-x a+2 b x$
=$2(a-2 b)-x(a-2 b)$
=$(a-2 b)(2-x)$
Solution 7
(i) $\quad 5 p h-10qk+2 r p h-4 q r k$
$=5(p h-2 q k)+2 r(p h-2 q k)$
$=\left(p h-2 qk\right)(5+2 r)$
(ii) $x^{2}-x(a+2 b)+2 a b$
=$x^{2}-x a-2 b x+2 a b$
=$x(x-a)-2 b(x-a)$
=$(x-a)(x-2 b)$
Solution 8
(i) $a b\left(x^{2}+y^{2}\right)-x y\left(a^{2}+b^{2}\right)$
=$a b x^{2}+a b y^{2}-x y a^{2}-x y b^{2}$
=$a x(b x-a y)+b y(a y-b x)$
=$a x(b x-a y)-b y(b x-a y)$
=$(b x-a y)(a x-b y)$
(ii) $(a x+b y)^{2}+(b x-a y)^{2}$
=$a^{2} x^{2}+b^{2} y^{2}+2 a x \cdot b y+b^{2} x^{2}+a^{2} y^{2}-2 b x .a y$
=$x^{2}\left(a^{2}+b^{2}\right)+y^{2}\left(a^{2}+b^{2}\right)$
=$\left(a^{2}+b^{2}\right)\left(x^{2}+y^{2}\right)$
Solution 9
(i) (i) $a^{3}+a b(1-2 a)-2 b^{2}$
$= a^{3}+a b-2 a b-2 b^{2}$
$= a\left(a^{2}+b\right)-2 b\left(a^{2}+b\right)$
= $\left(a^{2}+b\right)(a-2 b)$
(ii) $3 x^{2} y-3 x y+12 x-12$
=$3 x y(x-1)+12(x-1)$
=$(x-1)(3 x y+12)$
=$(x-1) \cdot 3 \cdot(x y+4)$
=$\therefore \quad 3(x-1)(x y+4)$
Solution 10
(i) $a^{2} b+a b^{2}-a b c-b^{2} c+a x y+b x y$
=$\left(a^{2} b+a b^{2}\right)-\left(a b c+b^{2} c\right)+(a x y+b x y)$
=$a b(a+b)-b c(a+b)+x y(a+b)$
=$(a+b)(a b-b c+x y)$
(ii) $a x^{2}-b x^{2}+a y^{2}-b y^{2}+a z^{2}-b z^{2}$
=$x^{2}(a-b)+y^{2}(a-b)+x^{2}(a-b)$
=$(a-b)\left(x^{2}+y^{2}+z^{2}\right)$
Solution 11
(i) $x-1-(x-1)^{2}+a x-a$
=$1(x-1)-(x-1)^{2}+a(x-1)$
=$(x-1) \quad(1-(x-1)+a)$
=$(x-1)(1-x+1+a)$
=$(x-1)(2-x+a)$
(ii) $a x+a^{2} x+a b y+b y-(a x+b y)^{2}$
=$a x+b y+a^{2} x+a b y-(a x+b y)^{2}$
=$1(a x+b y)+a(a x+b y)-(a x+b y)^{2}$
=$(a+b y)(1+a-a x-b y)$
Excercise 11.3
Solution 1
(i) $\quad x^{2}-12 x+36$
$\Rightarrow(x)^{2}-2 \cdot 6 \cdot x+(6)^{2}$
by using $a^{2}-2 a b+b^{2}=(a-b)^{2}$
$\therefore \quad(x-6)^{2}$
(ii) $36 p^{2}-60 p q+25 q^{2}$
=$(6 p)^{2}-2 \cdot 6 p \cdot 5 q+(5 q)^{2}$
=$(6 p-5 q)^{2}$
(iii) $9 x^{2}+66 x y+121 y^{2}$
$(3 x)^{2}+23 x \cdot 11 y+(11 y)^{2}$
$=(3 x+11 y)^{2}$
(iv) $a^{4}+6 a^{2} b^{2}+9 b^{4}$
=$\left(a^{2}\right)^{2}+2 \cdot a^{2} \cdot 3 b^{2}+\left(3 b^{2}\right)^{2}$
=$\left(a^{2}+3 b^{2}\right)^{2}$
(v) $x^{2}+\frac{1}{x^{2}}+2$
=$(x)^{2}+2 \cdot x \cdot \frac{1}{x}+\left(\frac{1}{x}\right)^{2}$
$=\left(x+\frac{1}{x}\right)^{2}$
Solution 2
(i) $4 p^{2}-9$
=$(2 p)^{2}-3^{2}$
by using $a^{2}-b^{2}=(a+b)(a-b)$
=(2 p+3)(2 p-3)
(ii) $4 x^{2}-169 y^{2}$
=$(2 x)^{2}-(13 y)^{2}$
=$(2 x+13 y)(2 x-13 y)$
Solution 3
(i) $9 x^{2} y^{2}-25$
=$(3 x y)^{2}-5^{2}$
=$(3 x y+5)(3 x y-5)$
(ii) $16 x^{2}-\frac{1}{144}$
=$(4 x)^{2}-\left(\frac{1}{12}\right)^{2}$
=$\left(4 x^{2}+\frac{1}{12}\right)\left(4 x-\frac{1}{12}\right)$
Solution 4
(i) $20 x^{2}-45 y^{2}$
=$5\left(4 x^{2}-9 y^{2}\right)$
=$5\left((2 x)^{2}-(3 y)^{2}\right)$
=$5(2 x+3 y)(2 x-3 y)$
(ii) $\frac{9}{16}-25 a^{2} b^{2}$
=$\left(\frac{3}{4}\right)^{2}-(5 a b)^{2}$
=$\left(\frac{3}{4}+5 a b\right)\left(\frac{3}{4}-5 a b\right)$
Solution 5
(i) $(2 a+3 b)^{2}-16 c^{2}$
=$(2 a+3 b)^{2}-(4 c)^{2}$
=$(2 a+3 b-4 c)(2 a+3 b+4 c)$
(ii) $1-(b-c)^{2}$
=$1^{2}-(b-c)^{2}$
=$(1+b-c)(1-b-c)$
Solution 6
(i) $9(x+y)^{2}-x^{2}$
=$3^{2}(x+y)^{2}-x^{2}$
=$(3(x+y))^{2}-x^{2}$
=$(3(x+y)+x)(3(x+y)-x)$
=$(3 x+3 y+x)(3 x+3 y-x)$
=$(4 x+3 y)(2 x+3 y)$
(ii) $(2 m+3 n)^{2}-(3 m+2 n)^{2}$
$= \quad(2 m+3 n+3 m+2 n)(2 m+3 n-3 m-2 n)$
$=(5 m+5 n)(n-m)$
Soluyion 7
(i) $25(a+b)^{2}-16(a-b)^{2}$
$5^{2}(a+b)^{2}-4^{2}(a-b)^{2}$
$(5 a+5 b)^{2}-(4 a-4 b)^{2}$
$=(5 a+5 b+4 a-4 b)(5 a+5 b-4 a+4 b)$
$=(9 a-b)(a+a b)$
(ii) $9(3 x+2)^{2}-4(2 x-1)^{2}$
=$3^{2}(3 x+2)^{2}-2^{2}(2 x-1)^{2}$
=$(9 x+6)^{2}-(4 x-2)^{2}$
=$(9 x+6+4 x-2)(9 x+6-4 x+2)$
=$(13 x+4)(5 x+7)$
Solution 8
(i) $x^{3}-25 x$
=$x\left(x^{2}-25\right)$
=$x\left(x^{2}-5^{2}\right)$
=$x \quad((x+5)(x-5))$
(ii) $63 p^{2} q^{2}-7$
=$7\left(9 p^{2} q^{2}-1\right)$
=$7\left((3 p q)^{2}-1^{2}\right)$
=$7 \quad((3 p q+1)(3 p q-1))$
Solution 9
(i) $32 a^{2} b-72 b^{3}$
=$8 b\left(4 a^{2}-9 b^{2}\right)$
=$8 b\left((2 a)^{2}-(3 b)^{2}\right)$
=$8 b((2 a+3 b)(2 a-3 b))$
(ii) $9(a+b)^{3}-25(a+b)$
=$(a+b)\left(3^{2}\left(a+b^{2}\right)^{2}-5^{-2}\right)$
=$(a+b)\left((3 a+3 b)^{2}-5^{2}\right)$
=$(a+b)(3 a+3 b+5)(3 a+3 b-5)$
Solution 10
(i) $x^{2}-y^{2}-2 y-1$
=$x^{2}-\left(y^{2}+2 y+1\right)$ $a^{2}+b^{2}+2 a b=(a+b)^{2}$
$(a+b)(a-b)=a^{2}-b^{2}$
=$1 \quad x^{2}-(y+1)^{2}$
=$x \quad(x+y+1)(x-y-1)$
(ii) $p^{2}-4 p q+4 q^{2}-r^{2}$
=$\left(p^{2}-2 \cdot p \cdot 2 q+(2 q)^{2}\right)-r^{2}$
=$\Rightarrow \quad(p+2 q)^{2}-r^{2}$
=$\Rightarrow \quad(p-2 q+r)(p-2 q-r)$
Solution 11
(i) $9 x^{2}-y^{2}+4 y-4$
=$(3 x)^{2}-\left(y^{2}-2 \cdot y \cdot 2+2^{-2}\right)$
=$(3 x)^{2}-(y-2)^{2}$
=$(3 x+y-2)(3 x-y+2)$
(ii) $4 a^{2}-4 b^{2}+4 a+1$
=$(2 a)^{2}+2 \cdot 2 a \cdot 1+(1)^{2}-(2 b)^{2}$
=$(2 a+1)^{2}-(2 b)^{2}$
=$(2 a+1+2 b) \cdot(2 a+1-2 b)$
Solution 12
(i) $625-p^{4}$
=$(25)^{2}-\left(p^{2}\right)^{2}$
=$\left(25+p^{2}\right)\left(25-p^{2}\right)$
(ii) $5 y^{5}-405 y$
=$5 y\left(y^{4}-81\right)$
=$5 y \quad\left(\left(y^{2}\right)^{2}-9^{2}\right)$
=$5 y\left(y^{2}+9\right)\left(y^{2}-9\right)$
Solution 13
(i) $x^{4}-y^{4}+x^{2}-y^{2}$
=$\left(x^{4}-y^{4}\right)+\left(x^{2}-y^{2}\right)$
=$\left(\left(x^{2}\right)^{2}-\left(y^{2}\right)^{2}\right)+\left(x^{2}-y^{2}\right)$
=$\left(x^{2}+y^{2}\right)\left(x^{2}-y^{2}\right)+\left(x^{2}-y^{2}\right)$
=$\left(x^{2}-y^{2}\right)\left(x^{2}+y^{2}+1\right)$
(ii) $64 a^{2}-9 b^{2}+42 b c-49 c^{2}$
=$(8 a)^{2}-(3 b)^{2}+7 c(6 b-7 c)$
=$(8 a+3 b)(8 a-3 b)+7 c(6 b-7 c)$
Excercise 11.4
Solution 1
(i) $x^{2}+3 x+2$
$\Rightarrow \quad x^{2}+2 x+x+2$
$\Rightarrow \quad x(x+2)+1(x+2)$
$\Rightarrow(x+1)(x+2) .$
(ii) $z^{2}+10 z+24$
=$z^{2}+6 z+4 z+24$
=$z(z+6)+4(z+6)$
=$(z+6)(z+4)$
Solution 2
(i) $y^{2}-7 y+12$
=$y^{2}-4 y-3 y+12$
=$y(y-4)-3(y-4)$
=$(y-4)(y-3)$
(ii) $m^{2}-23 m+42$
=$m^{2}-21 m-2 m+42$
=$m(m-21)-2(m-21)$
=$m(m-21)(m-2)$
Solution 3
(i) $y^{2}-5 y-24$
=$y^{2}-8 y+3 y-24$
=$y(y-8)+3(y-8)$
=$(y-8)(y+3)$
(ii) $t^{2}+23 t-108$
=$t^{2}+27 t-4 t-108$
=$t(t+27)-4(t+27)$
=$(t+27)(t-4)$
Solution 4
(i) $3 x^{2}+14 x+8$
=$3 x^{2}+12 x+2 x+8$
=$3 x(x+4)+2(x+4)$
=$(x+4)(3 x+2)$
(ii) $3 y^{2}+10 y+8$
=$3 y^{2}+6 y+4 y+8$
=$3 y(y+2)+4(y+2)$
=$(y+2)(3 y+4)$
Solution 5
(i) $14 x^{2}-23 x+8$
=$14 x^{2}-16 x-7 x+8$
=$2 x(7 x-8)-1(7 x-8)$
=$(7 x-8)(2 x-1)$
(ii) $12 x^{2}-x-35$
=$12 x^{2}-21 x+20 x-35$
=$3 x(42 x-7 x)+5(4 x-7)$
=$(4 x-7) \quad(3 x+5)$
Solution 6
(i) $6 x^{2}+11 x-10$
=$6 x^{2}+15 x-4 x-10$
=$3 x(2 x+5)-2(2 x+5)$
=$(2 x+5)(3 x-2)$
(ii) $5-4 x-12 x^{2}$
=$5-10 x+6 x-12 x^{2}$
=$5(1-2 x)+6 x(1-2 x)$
=$(1-2 x)(5+6 x)$
(i) $1-18 y-63 y^{2}$
=$1-21 y+3 y-63 y^{2}$
=$1(1-21 y)+3 y(1-21 y)$
=$(1-21 y)(1+3 y)$
(ii) $3 x^{2}-5 x y-12 y^{2}$
=$3 x^{2}-9 x y+4 x y-12 y^{2}$
=$3 x(x-3 y)+4 y(x-3 y)$
=$(x-3 y)(3 x+4 y)$
Solution 8
(i) $x^{2}-3 x y-40 y^{2}$
=$x^{2}-8 x y+5 x y-40 y^{2}$
=$x(x-8 y)+5 y(x-8 y)$
=$(x-8 y)(x+5 y)$
(ii) $10 p^{2} q^{2}-21 p q+9$
=$10 p^{2} q^{2}-15 p q-6 p q+9$
=$5 p q(2 p q-3)-3(2 p q-3)$
=$(2 p q-3)(5 p q-3)$
Solution 9
(i) $2 a^{2} b^{2}+a b-45$
=$2 a^{2} b^{2}+10 a b-9 a b-45$
=$2 a b(a b+5)-9(a b+5)$
=$(a b+5)(2 a b-9)$
(ii) $x(12 x+7)-10$
=$12 x^{2}+7 x-10$
=$12 x^{2}+15 x-8 x-10$
=$3 x(4 x+5)-2(4 x+5)$
=$(4 x+5)(3 x-2)$
Solution 10
(i) $(a+b)^{2}-11(a+b)-42$
=$(a+b)^{2}-14(a+b)+3(a+b)-42$
=$(a+b)(a+b-14)+3(a+b-14)$
=$(a+b-14)(a+b+3)$
(ii) $8+6(p+q)-5(p+q)^{2} \quad$
=$8+10(p+q)-4(p+q)-5(p+a)^{2}$
=$2(4+5(p+q))-(p+q)(4+5(p+a))$
=
$(4+5(p+q))(2-(p+q))$
Solution 11
(i) $(x-2 y)^{2}-6(x-2 y)+5-1 \times 5$
=$(x-2 y)^{2}-5(x-2 y)-(x-2 y)+5$
=$(x-2 y)(x-2 y-5)-1(x-2 y-5)$
=$(x-2 y-5)(x-2 y-1)$
(ii) $7+10(2 x-3 y)-8(2 x-3 y)^{2} \quad 0 \times 8=-56$
$7+14(2 x-3 y)-4(2 x-3 y)-8(2 x-3 y)^{2}$
$7(1+2(2 x-3 y))-4(2 x-3 y)(1+2(2 x-3 y))$
$(1+2(2 x-3 y))(7-4(2 x-3 y))$
$(1+4 x-6 y)(7-8 x+12 y)$
Excercise 11.5
Solution 1
(i) $(35 x+28) \div(5 x+4)$
$\frac{35 x+28}{5 x+4}$
$⇒\frac{7(5 x+4)}{5 x+4}$
$⇒\quad 7$
(ii) $7 p^{2} q^{2}(9 r-21) \div 63 p q(r-3)$
⇒ $\frac{7 p^{2} q^{2}(9 r-27)}{63 p q(r-3)}$
⇒ $\frac{7 \cdot p^{2} q^{ 2} \cdot q(r- 3)}{63 pq(x - 3)}$
⇒ pq
Solution 2
(i) $6(2 x+7)(5 x-3) \div 3(5 x-3)$
⇒$\frac{6(2 x+7)(5 x-3)}{ 3(5 x-3)}$
⇒$2(2 x+7)$
⇒$\quad 4 x+14$
(ii) $33 p q(p+3)(2 q-5) \div 11 p(2 q-5)$
⇒$\frac{38 p q(p+3)(2 q-8)}{11-pq (2 q-5)}$
⇒3(p+3)
⇒3 p+9
Solution 3
(i) $\left(7 x^{3}-63 x\right) \div 7(x-3)$
⇒ $\frac{7 x^{3}-6 x}{7(x-3)}$
⇒$\frac{7 x\left(x^{2}-9\right)}{7(x-3)}$
⇒$\frac{x\left(x^{2}-3^{2}\right)}{x-3}$
⇒$\frac{x(x+3)(x-3)}{(x-3)}$
⇒ $x(x+3)$
⇒$x^{2}+3x$
(ii) $\left(3 p^{2}+17 p+10\right) \div(p+5)$
⇒$\frac{3 p^{2}+17 p+10}{p+5}$
⇒$\frac{3 p^{2}+15 p+2 p+10}{p+5}$
⇒$\frac{3 P(p+5)+2(p+5)}{p+5}$
⇒$\frac{(p+5)(3 p+2)}{(p+5)}$
⇒ $3 p+2$
(iii) $10 x y\left(14 y^{2}+43 y-21\right) \div 5 x(7 y-3)$
⇒ $\frac{10 y\left(14 y^{2}+43 y-21\right)}{5x(7 y-3)}$
⇒$\frac{2 y\left(14 y^{2}-6 y+49 y-21\right)}{7 y-3}$
⇒$\frac{2 y(2 y(7 y-3)+7(7 y-3))}{7 y-3}$
⇒$\frac{2 y(7 y-3)(2 y+7)}{(7 y-3)}$
⇒2 y(2 y+7)
(iv) $12 p q r\left(6 p^{2}-13 p q+6 q^{2}\right) \div 6 p q(2 p-3 q)$
⇒ $\frac{12pq r\left(6 p^{2}-13 p q+6 q^{2}\right)}{6 pq(2 p-3 q)}$
⇒ $\frac{2r\left(6 p^{2}-9 p q-4 p q+6 q^{2}\right)}{2 p-3 q}$
⇒ $\frac{2 r(3 p(2 p-3 q)-2 q(2 p-3 q))}{2 p-3 q}$
⇒ $\frac{2 r(2 p-3 q)(3 p-2 q)}{(2 p-3 q)}$
⇒ 2(3 p-2 q)
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