ML AGGARWAL CLASS 8 CHAPTER 11 Factorisation Exercise 11.2
Exercise 11.2
Solution 1
Sol :
(i)
x2+xy−x−y⇒x(x+y)−1(x+y)⇒(x+y)(x−1)
(ii)
y2−yz−5y+5z⇒y(y−z)−5(y−z)⇒(y−z)(y−5)
Solution 2
Sol :
(i)
5xy+7y−5y2−7x⇒5xy−5y2+7y−7x.⇒5xy−5y2−7x+7y⇒5y(x−y)−7(x−y)→(x−y)(5y−7)
(ii) 5p2−8pq−10p+16q
5p2−10p−8pq+16q
5p(p-2)-8q(p-2)
(p-2)(5p-8q)
Solution 3
Sol :
(i)
a2b−ab2+3a−3b⇒ab(a−b)+3(a−b)⇒(a−b)(ab+3)
(ii)
x3−3x2+x−3=x2(x−3)+1(x−3)⇒(x−3)(x2+1)
Solution 4
Sol :
(i)
6xy2−3xy−10y+5⇒3xy(2y−1)−5(2y−1)⇒(2y−1)(3xy−5)
(ii) 3ax-6ay-8by+4 bx
3a(x-2 y)+2b(-2y+x)
⇒3a(x−2y)+2b(x−2y)
⇒(x−2y)(3a+2b)
Solution 5
Sol :
(i) x2+xy(1+y)+y3\
⇒x2+xy+xy2+y3
⇒x(x+y)+y2(x+y)
⇒(x+y)(x+y2)
(ii) y2−xy(1−x)−x3
⇒y2−xy+x2y−x3
⇒y(y−x)+x2(y−x)
⇒(y−x)+(y+x2)
Solution 6
Sol :
(i) ab2+(a−1)⋅b−1
ab2+ab−b−1
ab(b+1)-(b+1)
(b+1)(a b-1)
(ii) 2a-4b-xa+2bx
=2(a-2b)-x(a-2b)
=(a-2b)(2-x)
Solution 7
Sol :
(i) 5ph-10qk+2rph-4q rk
=5(p h-2 q k)+2 r(p h-2 q k)
=(ph−2qk)(5+2r)
(ii) x2−x(a+2b)+2ab
=x2−xa−2bx+2ab
=x(x-a)-2 b(x-a)
=(x-a)(x-2 b)
Solution 8
Sol :
(i) ab(x2+y2)−xy(a2+b2)
=abx2+aby2−xya2−xyb2
=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) (ax+by)2+(bx−ay)2
=a2x2+b2y2+2ax⋅by+b2x2+a2y2−2bx.ay
=x2(a2+b2)+y2(a2+b2)
=(a2+b2)(x2+y2)
Solution 9
Sol :
(i)a3+ab(1−2a)−2b2
=a3+ab−2ab−2b2
=a(a2+b)−2b(a2+b)
= (a2+b)(a−2b)
(ii) 3x2y−3xy+12x−12
=3xy(x-1)+12(x-1)
=(x-1)(3 x y+12)
=(x-1)3(x y+4)
=∴3(x−1)(xy+4)
Solution 10
Sol :
(i) a2b+ab2−abc−b2c+axy+bxy
=(a2b+ab2)−(abc+b2c)+(axy+bxy)
=a b(a+b)-b c(a+b)+x y(a+b)
=(a+b)(a b-b c+x y)
(ii) ax2−bx2+ay2−by2+az2−bz2
=x2(a−b)+y2(a−b)+x2(a−b)
=(a−b)(x2+y2+z2)
Solution 11
Sol :
(i) x−1−(x−1)2+ax−a
=1(x−1)−(x−1)2+a(x−1)
=(x−1)(1−(x−1)+a)
=(x-1)(1-x+1+a)
=(x-1)(2-x+a)
(ii) ax+a2x+aby+by−(ax+by)2
=ax+by+a2x+aby−(ax+by)2
=1(ax+by)+a(ax+by)−(ax+by)2
=(a+b y)(1+a-a x-b y)
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