# RS Aggarwal Solutions Class 9 Maths Chapter 3 Factorisation of Polynomials (Updated For 2023)

RS Aggarwal Solutions Class 9 Maths Chapter 3 Factorisation of Polynomials: If you are a Class 9 student preparing for Maths exam, then you must try the RS Aggarwal Solutions Class 9 Maths. All the solutions are easy to understand, reliable and very well-explained, all thanks to the subject matter experts.

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RS Aggarwal Solutions Class 9 Maths Chapter 3 Factorisation of Polynomials

## Access Solutions Of RS Aggarwal Solutions Class 9 Maths Chapter 3 Factorisation of Polynomials

Factorize:
9x2 + 12xy

We have:

Factorize:
18x2y − 24xyz

We have:

Factorize:
18x2y − 24xyz

We have:

#### Question 3:

Factorize:
27a3b3 − 45a4b2

We have:
27a3b345a4b2 = 9a3b2(3b5a)

#### Question 4:

Factorize:
2a(x + y) − 3b(x + y)

We have:
2a(x+y)3b(x+y) = (x+y)(2a3b)

#### Question 5:

Factorize:
2x(p2 + q2) + 4y(p2 + q2)

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)

We have:
x(a5)+y(5a)=x(a5)y(a5)xa-5+y5-a=xa-5-ya-5
=(a5)(xy)

#### Question 7:

Factorize:
4(a + b) − 6(a + b)2

We have:
4(a+b)6(a+b)2=2(a+b)[23(a+b)]4a+b-6a+b2=2a+b2-3a+b
=2(a+b)(23a3b)

#### Question 8:

Factorize:
8(3a − 2b)2 − 10(3a − 2b)

We have:
8(3a2b)210(3a2b)=2(3a2b)[4(3a2b)5]83a-2b2-103a-2b=23a-2b43a-2b-5
=2(3a2b)(12a8b5)

#### Question 9:

Factorize:
x(x + y)3 − 3x2y(x + y)

We have:
x(x+y)33x2y(x+y)=x(x+y)[(x+y)23xy]xx+y3-3x2yx+y=xx+yx+y2-3xy
=x(x+y)[x2+y2+2xy3xy]=x(x+y)(x2+y2xy)=xx+yx2+y2+2xy-3xy=xx+yx2+y2-xy

#### Question 10:

Factorize:
x3 + 2x2 + 5x + 10

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

We have:
x2+xy2xz2yz=(x2+xy)(2xz+2yz)         =x(x+y)2z(x+y)         =(x+y)(x2z)x2+xy-2xz-2yz=x2+xy-2xz+2yz

=xx+y-2zx+y                           =x+yx-2z

#### Question 12:

Factorize:
a3b − a2b + 5ab − 5b

We have:
a3ba2b+5ab5b=b(a3a2+5a5)           =b[(a3a2)+(5a5)]a3b-a2b+5ab-5b=ba3-a2+5a-5          =ba3-a2+5a-5

=b[a2(a1)+5(a1)]=b(a1)(a2+5)=ba2a-1+5a-1=ba-1a2+5

#### Question 13:

Factorize:
8 − 4a − 2a3 + a4

We have:
84a2a3+a4= (84a)(2a3a4)                         = 4(2a) a3(2a)                         = (2a) (4  a3)

#### Question 14:

Factorize:
x3 − 2x2y + 3xy2 − 6y3

We have:
x32x2y+3xy26y3=(x32x2y)+(3xy26y3)x3-2x2y+3xy2-6y3=x3-2x2y+3xy2-6y3
=x2(x2y)+3y2(x2y)=(x2y)(x2+3y2)

#### Question 15:

Factorize:
px − 5q pq − 5x

We have:
px5q+pq5x=(px5x)+(pq5q)px-5q+pq-5x=px-5x+pq-5q
=x(p5)+q(p5)=(p5)(x+q)

#### Question 16:

Factorize:
x2 + y − xy − x

We have:
x2+yxyx=(x2xy)(xy)x2+y-xy-x=x2-xy-x-y
=x(xy)1(xy)=(xy)(x1)

#### Question 17:

Factorize:
(3a − 1)2 − 6a + 2

We have:
(3a1)26a+2=(3a1)22(3a1)3a-12-6a+2=3a-12-23a-1
=(3a1)[(3a1)2]=(3a1)(3a12)=(3a1)(3a3)=3(3a1)(a1)

#### Question 18:

Factorize:
(2x − 3)2 − 8x + 12

We have:
(2x3)28x+12=(2x3)24(2x3)2x-32-8x+12=2x-32-42x-3
=(2x3)[(2x3)4]=(2x3)(2x34)=(2x3)(2x7)

#### Question 19:

Factorize:
a3 + a − 3a2 − 3

We have:
a3+a3a23=(a33a2)+(a3)a3+a-3a2-3=a3-3a2+a-3
=a2(a3)+1(a3)=(a3)(a2+1)

#### Question 20:

Factorize:
3ax − 6ay − 8by + 4bx

We have:
3ax6ay8by+4bx=(3ax6ay)+(4bx8by)3ax-6ay-8by+4bx=3ax-6ay+4bx-8by
=3a(x2y)+4b(x2y)=(x2y)(3a+4b)

#### Question 21:

Factorize:
abx2 + a2x + b2x + ab

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

We have:
x3x2+ax+xa1=(x3x2)+(axa)+(x1)x3-x2+ax+x-a-1=x3-x2+ax-a+x-1
=x2(x1)+a(x1)+1(x1)=(x1)(x2+a+1)=x2x-1+ax-1+1x-1=x-1×2+a+1

#### Question 23:

Factorize:
2x + 4y − 8xy − 1

We have:
2x+4y8xy1=(2x8xy)(14y)2x+4y-8xy-1=2x-8xy-1-4y
=2x(14y)1(14y)=(14y)(2x1)

#### Question 24:

Factorize:
ab(x2 + y2) − xy(a2 + b2)

We have:
ab(x2+y2)xy(a2+b2)=abx2+aby2a2xyb2xyabx2+y2-xya2+b2=abx2+aby2-a2xy-b2xy
=(abx2a2xy)(b2xyaby2)=ax(bxay)by(bxay)=(bxay)(axby)

#### Question 25:

Factorize:
a2 + ab(b + 1) + b3

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

We have:
a3+ab(12a)2b2=a3+ab2a2b2b2a3+ab1-2a-2b2=a3+ab-2a2b-2b2
=(a32a2b)+(ab2b2)=a2(a2b)+b(a2b)=(a2b)(a2+b)

#### Question 27:

Factorize:
2a2 + bc − 2ab − ac2

We have:
2a2+bc2abac=(2a22ab)(acbc)2a2+bc-2ab-ac=2a2-2ab-ac-bc
=2a(ab)c(ab)=(ab)(2ac)

#### Question 28:

Factorize:
(ax + by)2 + (bx − ay)2

We have:
(ax+by)2+(bxay)2=[(ax)2+2×ax×by+(by)2]+[(bx)22×bx×ay+(ay)2]ax+by2+bx-ay2

=ax2+2×ax×by+by2+bx2-2×bx×ay+ay2
=a2x2+2abxy+b2y2+b2x22abxy+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

We have:
a(a+bc)bc=a2+abacbcaa+b-c-bc=a2+ab-ac-bc
=(a2ac)+(abbc)=a(ac)+b(ac)=(ac)(a+b)

#### Question 30:

Factorize:
a(a − 2b − c) + 2bc

We have:
a(a2bc)+2bc=a22abac+2bcaa-2b-c+2bc=a2-2ab-ac+2bc
=(a22ab)(ac2bc)=a(a2b)c(a2b)=(a2b)(ac)

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