ML AGGARWAL CLASS 8 CHAPTER 11 Factorisation Exercise 11.4

 Exercise 11.4

Solution 1

Sol :

(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

Sol :

(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

Sol :

(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

Sol :

(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

Sol :

(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) (3 x+5)

Solution 6

Sol :

(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)

 

Solution 7

Sol :

(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

Sol :

(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

Sol :

(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

Sol :

(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)