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RS Aggarwal Solutions Class 9 Maths Chapter 3 Factorisation of Polynomials
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Question 1:
Factorize:
9x2 + 12xy
ANSWER:
We have:
9x2+12xy=3x(3x+4y)
Question 2:
Factorize:
18x2y − 24xyz
ANSWER:
We have:
18x2y–24xyz=6xy(3y–4z)
Question 2:
Factorize:
18x2y − 24xyz
ANSWER:
We have:
18x2y–24xyz
=6xy(3y–4z)
Question 3:
Factorize:
27a3b3 − 45a4b2
ANSWER:
We have:
27a3b3−45a4b2 = 9a3b2(3b−5a)
Question 4:
Factorize:
2a(x + y) − 3b(x + y)
ANSWER:
We have:
2a(x+y)−3b(x+y) = (x+y)(2a−3b)
Question 5:
Factorize:
2x(p2 + q2) + 4y(p2 + q2)
ANSWER:
We have:
2x(p2+q2)+4y(p2+q2) = 2[x(p2+q2)+2y(p2+q2)] = 2(p2+q2)(x+2y)
Question 6:
Factorize:
x(a − 5) + y(5 − a)
ANSWER:
We have:
x(a−5)+y(5−a)=x(a−5)−y(a−5)xa-5+y5-a=xa-5-ya-5
=(a−5)(x−y)
Question 7:
Factorize:
4(a + b) − 6(a + b)2
ANSWER:
We have:
4(a+b)−6(a+b)2=2(a+b)[2−3(a+b)]4a+b-6a+b2=2a+b2-3a+b
=2(a+b)(2−3a−3b)
Question 8:
Factorize:
8(3a − 2b)2 − 10(3a − 2b)
ANSWER:
We have:
8(3a−2b)2−10(3a−2b)=2(3a−2b)[4(3a−2b)−5]83a-2b2-103a-2b=23a-2b43a-2b-5
=2(3a−2b)(12a−8b−5)
Question 9:
Factorize:
x(x + y)3 − 3x2y(x + y)
ANSWER:
We have:
x(x+y)3−3x2y(x+y)=x(x+y)[(x+y)2−3xy]xx+y3-3x2yx+y=xx+yx+y2-3xy
=x(x+y)[x2+y2+2xy−3xy]=x(x+y)(x2+y2−xy)=xx+yx2+y2+2xy-3xy=xx+yx2+y2-xy
Question 10:
Factorize:
x3 + 2x2 + 5x + 10
ANSWER:
We have:
x3+2x2+5x+10=(x3+2x2)+(5x+10)x3+2×2+5x+10=x3+2×2+5x+10
=x2(x+2)+5(x+2)=(x+2)(x2+5)=x2x+2+5x+2=x+2×2+5
Question 11:
Factorize:
x2 + xy − 2xz − 2yz
ANSWER:
We have:
x2+xy−2xz−2yz=(x2+xy)−(2xz+2yz) =x(x+y)−2z(x+y) =(x+y)(x−2z)x2+xy-2xz-2yz=x2+xy-2xz+2yz
=xx+y-2zx+y =x+yx-2z
Question 12:
Factorize:
a3b − a2b + 5ab − 5b
ANSWER:
We have:
a3b−a2b+5ab−5b=b(a3−a2+5a−5) =b[(a3−a2)+(5a−5)]a3b-a2b+5ab-5b=ba3-a2+5a-5 =ba3-a2+5a-5
=b[a2(a−1)+5(a−1)]=b(a−1)(a2+5)=ba2a-1+5a-1=ba-1a2+5
Question 13:
Factorize:
8 − 4a − 2a3 + a4
ANSWER:
We have:
8−4a−2a3+a4= (8−4a)−(2a3−a4) = 4(2−a)− a3(2−a) = (2−a) (4 − a3)
Question 14:
Factorize:
x3 − 2x2y + 3xy2 − 6y3
ANSWER:
We have:
x3−2x2y+3xy2−6y3=(x3−2x2y)+(3xy2−6y3)x3-2x2y+3xy2-6y3=x3-2x2y+3xy2-6y3
=x2(x−2y)+3y2(x−2y)=(x−2y)(x2+3y2)
Question 15:
Factorize:
px − 5q + pq − 5x
ANSWER:
We have:
px−5q+pq−5x=(px−5x)+(pq−5q)px-5q+pq-5x=px-5x+pq-5q
=x(p−5)+q(p−5)=(p−5)(x+q)
Question 16:
Factorize:
x2 + y − xy − x
ANSWER:
We have:
x2+y−xy−x=(x2−xy)−(x−y)x2+y-xy-x=x2-xy-x-y
=x(x−y)−1(x−y)=(x−y)(x−1)
Question 17:
Factorize:
(3a − 1)2 − 6a + 2
ANSWER:
We have:
(3a−1)2−6a+2=(3a−1)2−2(3a−1)3a-12-6a+2=3a-12-23a-1
=(3a−1)[(3a−1)−2]=(3a−1)(3a−1−2)=(3a−1)(3a−3)=3(3a−1)(a−1)
Question 18:
Factorize:
(2x − 3)2 − 8x + 12
ANSWER:
We have:
(2x−3)2−8x+12=(2x−3)2−4(2x−3)2x-32-8x+12=2x-32-42x-3
=(2x−3)[(2x−3)−4]=(2x−3)(2x−3−4)=(2x−3)(2x−7)
Question 19:
Factorize:
a3 + a − 3a2 − 3
ANSWER:
We have:
a3+a−3a2−3=(a3−3a2)+(a−3)a3+a-3a2-3=a3-3a2+a-3
=a2(a−3)+1(a−3)=(a−3)(a2+1)
Question 20:
Factorize:
3ax − 6ay − 8by + 4bx
ANSWER:
We have:
3ax−6ay−8by+4bx=(3ax−6ay)+(4bx−8by)3ax-6ay-8by+4bx=3ax-6ay+4bx-8by
=3a(x−2y)+4b(x−2y)=(x−2y)(3a+4b)
Question 21:
Factorize:
abx2 + a2x + b2x + ab
ANSWER:
We have:
abx2+a2x+b2x+ab=(abx2+b2x)+(a2x+ab)abx2+a2x+b2x+ab=abx2+b2x+a2x+ab
=bx(ax+b)+a(ax+b)=(ax+b)(bx+a)
Question 22:
Factorize:
x3 − x2 + ax + x − a − 1
ANSWER:
We have:
x3−x2+ax+x−a−1=(x3−x2)+(ax−a)+(x−1)x3-x2+ax+x-a-1=x3-x2+ax-a+x-1
=x2(x−1)+a(x−1)+1(x−1)=(x−1)(x2+a+1)=x2x-1+ax-1+1x-1=x-1×2+a+1
Question 23:
Factorize:
2x + 4y − 8xy − 1
ANSWER:
We have:
2x+4y−8xy−1=(2x−8xy)−(1−4y)2x+4y-8xy-1=2x-8xy-1-4y
=2x(1−4y)−1(1−4y)=(1−4y)(2x−1)
Question 24:
Factorize:
ab(x2 + y2) − xy(a2 + b2)
ANSWER:
We have:
ab(x2+y2)−xy(a2+b2)=abx2+aby2−a2xy−b2xyabx2+y2-xya2+b2=abx2+aby2-a2xy-b2xy
=(abx2−a2xy)−(b2xy−aby2)=ax(bx−ay)−by(bx−ay)=(bx−ay)(ax−by)
Question 25:
Factorize:
a2 + ab(b + 1) + b3
ANSWER:
We have:
a2+ab(b+1)+b3=a2+ab2+ab+b3a2+abb+1+b3=a2+ab2+ab+b3
=(a2+ab2)+(ab+b3)=a(a+b2)+b(a+b2)=(a+b2)(a+b)
Question 26:
Factorize:
a3 + ab(1 − 2a) − 2b2
ANSWER:
We have:
a3+ab(1−2a)−2b2=a3+ab−2a2b−2b2a3+ab1-2a-2b2=a3+ab-2a2b-2b2
=(a3−2a2b)+(ab−2b2)=a2(a−2b)+b(a−2b)=(a−2b)(a2+b)
Question 27:
Factorize:
2a2 + bc − 2ab − ac2
ANSWER:
We have:
2a2+bc−2ab−ac=(2a2−2ab)−(ac−bc)2a2+bc-2ab-ac=2a2-2ab-ac-bc
=2a(a−b)−c(a−b)=(a−b)(2a−c)
Question 28:
Factorize:
(ax + by)2 + (bx − ay)2
ANSWER:
We have:
(ax+by)2+(bx−ay)2=[(ax)2+2×ax×by+(by)2]+[(bx)2−2×bx×ay+(ay)2]ax+by2+bx-ay2
=ax2+2×ax×by+by2+bx2-2×bx×ay+ay2
=a2x2+2abxy+b2y2+b2x2−2abxy+a2y2
=a2x2+b2y2+b2x2+a2y2=(a2x2+b2x2)+(a2y2+b2y2)
=x2(a2+b2)+y2(a2+b2)=(a2+b2)(x2+y2)
Question 29:
Factorize:
a(a + b − c) − bc
ANSWER:
We have:
a(a+b−c)−bc=a2+ab−ac−bcaa+b-c-bc=a2+ab-ac-bc
=(a2−ac)+(ab−bc)=a(a−c)+b(a−c)=(a−c)(a+b)
Question 30:
Factorize:
a(a − 2b − c) + 2bc
ANSWER:
We have:
a(a−2b−c)+2bc=a2−2ab−ac+2bcaa-2b-c+2bc=a2-2ab-ac+2bc
=(a2−2ab)−(ac−2bc)=a(a−2b)−c(a−2b)=(a−2b)(a−c)
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