RS Aggarwal Solutions: Factorisation of Polynomials- 1 | Mathematics (Maths) Class 9 PDF Download
RS Aggarwal Solutions: Exercise 3A - Factorisation of Polynomials
Q.1. Factorize: 9x2 + 12xy
Ans. We have: 9x2+12xy
= 3x(3x+4y)
Q.2. Factorize: 18x2y − 24xyz
Ans. We have: 18x2y-24xyz
= 6xy(3y-4z)
Q.3. Factorize: 27a3b3 − 45a4b2
Ans. We have: 27a3b3-45a4b2
= 9a3b2(3b-5a)
Q.4. Factorize: 2a(x + y) − 3b(x + y)
Ans. We have: 2a(x+y)-3b(x+y)
=(x+y)(2a-3b)
Q.5. Factorize: 2x(p2 + q2) + 4y(p2 + q2)
Ans. We have: 2x(p2+q2)+4y(p2+q2)
= 2[x(p2+q2)+2y(p2+q2)]
= 2(p2+q2)(x+2y)
Q.6. Factorize: x(a − 5) + y(5 − a)
Ans. We have:
x(a-5)+y(5-a)
= x(a-5)-y(a-5)
=(a-5)(x-y)
Q.7. Factorize: 4(a + b) − 6(a + b)2
Ans. We have: 4(a+b)-6(a+b)2
=2(a+b)[2-3(a+b)]
=2(a+b)(2-3a-3b)
Q.8. Factorize: 8(3a − 2b)2 − 10(3a − 2b)
Ans. We have: 8(3a-2b)2-10(3a-2b)=2(3a-2b)[4(3a-2b)-5]
=2(3a-2b)(12a-8b-5)
Q.9. Factorize: x(x + y)3 − 3x2y(x + y)
Ans. We have: x(x+y)3-3x2y(x+y)=x(x+y)[(x+y)2-3xy]
=x(x+y)[x2+y2+2xy-3xy]
=x(x+y)(x2+y2-xy)
Q.10. Factorize: x3 + 2x2 + 5x + 10
Ans. We have: x3+2x2+5x+10=(x3+2x2)+(5x+10)
=x2(x+2)+5(x+2)
=(x+2)(x2+5)
Q.11. Factorize: x2 + xy − 2xz − 2yz
Ans. We have: x2+xy-2xz-2yz=(x2+xy)-(2xz+2yz)
=x(x+y)-2z(x+y)
=(x+y)(x-2z)
Q.12. Factorize: a3b − a2b + 5ab − 5b
Ans. We have: a3b-a2b+5ab-5b=b(a3-a2+5a-5)
=b[(a3-a2)+(5a-5)]
=b[a2(a-1)+5(a-1)]
=b(a-1)(a2+5)
Q.13. Factorize: 8 − 4a − 2a3 + a4
Ans. We have: 8-4a-2a3+a4= (8-4a)-(2a3-a4)
= 4(2-a)- a3(2-a)
= (2-a) (4 - a3)
Q.14. Factorize: x3 − 2x2y + 3xy2 − 6y3
Ans. We have: x3-2x2y+3xy2-6y3
=(x3-2x2y)+(3xy2-6y3)
=x2(x-2y)+3y2(x-2y)
=(x-2y)(x2+3y2)
Q.15. Factorize: px − 5q + pq − 5x
Ans. We have: px-5q+pq-5x
=(px-5x)+(pq-5q)
=x(p-5)+q(p-5)
=(p-5)(x+q)
Q.16. Factorize: x2 + y − xy − x
Ans. We have: x2+y-xy-x=(x2-xy)-(x-y)
=x(x-y)-1(x-y)
=(x-y)(x-1)
Q.17. Factorize: (3a − 1)2 − 6a + 2
Ans. We have: (3a-1)2-6a+2=(3a-1)2-2(3a-1)
=(3a-1)[(3a-1)-2]
=(3a-1)(3a-1-2)
=(3a-1)(3a-3)
=3(3a-1)(a-1)
Q.18. Factorize: (2x − 3)2 − 8x + 12
Ans. We have: (2x-3)2-8x+12
=(2x-3)2-4(2x-3)
=(2x-3)[(2x-3)-4]
=(2x-3)(2x-3-4)
=(2x-3)(2x-7)
Q.19. Factorize: a3 + a − 3a2 − 3
Ans. We have: a3+a-3a2-3
=(a3-3a2)+(a-3)
=a2(a-3)+1(a-3)
=(a-3)(a2+1)
Q.20. Factorize: 3ax − 6ay − 8by + 4bx
Ans. We have: 3ax-6ay-8by+4bx
=(3ax-6ay)+(4bx-8by)
=3a(x-2y)+4b(x-2y)
=(x-2y)(3a+4b)
Q.21. Factorize: abx2 + a2x + b2x + ab
Ans. We have: abx2+a2x+b2x+ab=(abx2+b2x)+(a2x+ab)
=bx(ax+b)+a(ax+b)=(ax+b)(bx+a)
Q.22. Factorize: x3 − x2 + ax + x − a − 1
Ans. We have: 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)
Q.23. Factorize: 2x + 4y − 8xy − 1
Ans. We have: 2x+4y−8xy−1=(2x−8xy)−(1−4y)
=2x(1−4y)−1(1−4y)
=(1−4y)(2x−1)
Q.24. Factorize: ab(x2 + y2) − xy(a2 + b2)
Ans. We have: ab(x2+y2)−xy(a2+b2)
=abx2+aby2−a2xy−b2xy
=(abx2−a2xy)−(b2xy−aby2)
=ax(bx−ay)−by(bx−ay)
=(bx−ay)(ax−by)
Q.25. Factorize: a2 + ab(b + 1) + b3
Ans. We have: a2+ab(b+1)+b3=a2+ab2+ab+b3
=(a2+ab2)+(ab+b3)
=a(a+b2)+b(a+b2)
=(a+b2)(a+b)
Q.26. Factorize: a3 + ab(1 − 2a) − 2b2
Ans. We have: a3+ab(1−2a)−2b2=a3+ab−2a2b−2b2
=(a3−2a2b)+(ab−2b2)
=a2(a−2b)+b(a−2b)
=(a−2b)(a2+b)
Q.27. Factorize: 2a2 + bc − 2ab − ac2
Ans. We have: 2a2+bc−2ab−ac=(2a2−2ab)−(ac−bc)
=2a(a−b)−c(a−b)
=(a−b)(2a−c)
Q.28. Factorize: (ax + by)2 + (bx − ay)2
Ans. We have: (ax+by)2+(bx−ay)2
=[(ax)2+2×ax×by+(by)2]+[(bx)2−2×bx×ay+(ay)2]
=a2x2+2abxy+b2y2+b2x2−2abxy+a2y2
=a2x2+b2y2+b2x2+a2y2
=(a2x2+b2x2)+(a2y2+b2y2)
=x2(a2+b2)+y2(a2+b2)
=(a2+b2)(x2+y2)
Q.29. Factorize: a(a + b − c) − bc
Ans. We have: a(a+b−c)−bc
=a2+ab−ac−bc
=(a2 - ac) + (ab-bc)
=a(a−c)+b(a−c)
=(a−c)(a+b)
Q.30. Factorize: a(a − 2b − c) + 2bc
Ans. We have: a(a−2b−c)+2bc=a2−2ab−ac+2bc
=(a2−2ab)−(ac−2bc)
=a(a−2b)−c(a−2b)
=(a−2b)(a−c)
Q.31. Factorize: a2x2 + (ax2 + 1)x + a
Ans. We have: a2x2+(ax2+1)x+a=(ax2+1)x+(a2x2+a)
=x(ax2+1)+a(ax2+1)
=(ax2+1)(x+a)
Q.32. Factorize: ab(x2 + 1) + x(a2 + b2)
Ans. We have: ab(x2+1)+x(a2+b2)
=abx2+ab+a2x+b2x
=(abx2+a2x)+(b2x+ab)
=ax(bx+a)+b(bx+a)
=(bx+a)(ax+b)
Q.33. Factorize: x2 − (a + b)x + ab
Ans. We have: x2−(a+b)x+ab
=x2−ax−bx+ab
=x(x−a)−b(x−a)
=(x−a)(x−b)
Q.34. Factorize: x2+1/x2−2−3x+3/x
Ans. We have: x2+1/x2−2−3x+3x/4
= x2−2+1/x2−3x+3/x
=(x)2−2×x×1/x+(1/x)2−3(x−1/x)
=(x−1/x)2−3(x−1/x)
=(x−1/x)(x−1/x−3)
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