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)