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Class 9 Exam  >  Class 9 Notes  >  Extra Documents & Tests for Class 9  >  RS Aggarwal Solutions: Exercise 3F - Factorisation of Polynomials

RS Aggarwal Solutions: Exercise 3F - Factorisation of Polynomials | Extra Documents & Tests for Class 9 PDF Download

Q.1. Factorize: x3 + 27
Ans. x3+27=(x)3+(3)3
=(x+3)(x2−3x+32)
=(x+3)(x2−3x+9)

Q.2. Factorise: 27a3 + 64b3
Ans. We know that x3+y3
=(x+y)(x2+y2−xy)x3+y3
Given: 27a3 + 64b3
x = 3a, y = 4b
27a3 + 64b3=(3a+4b)(9a2+16b2−12ab)

Q.3. Factorize: 125a3+1/8
Ans. 125a3+1/8=(5a)3+(1/2)3
=(5a+1/2)[(5a)2−5a×1/2+(1/2)2]
=(5a+1/2)(25a2−5a/2+1/4)

Q.4. Factorize: 216x3+1/125
Ans. 216x3+1/125=(6x)3+(1/5)3
=(6x+1/5)[(6x)2−6x×1/5+(1/5)2]
=(6x+1/5)(36x2−6x/5+1/25)

Q.5. Factorize: 16x4 + 54x
Ans. 16x4+54x
=2x(8x3+27)
=2x[(2x)3+(3)3]
=2x(2x+3)[(2x)2−2x×3+32]
=2x(2x+3)(4x2−6x+9)

Q.6. Factorize: 7a3 + 56b3
Ans. 7a3+56b3=7(a3+8b3)
=7[(a)3+(2b)3]
=7(a+2b)[a2−a×2b+(2b)2]
=7(a+2b)(a2−2ab+4b2)

Q.7. Factorize: x5 + x2
Ans. x5+x2=x2(x3+1)
=x2(x3+13)
=x2(x+1)(x2−x×1+12)
=x2(x+1)(x2−x+1)

Q.8. Factorize: a3 + 0.008
Ans. a3+0.008=a3+(0.2)3
=(a+0.2)[a2−a×(0.2)+(0.2)2]
=(a+0.2)(a2−0.2a+0.04)

Q.9. Factorise: 1 – 27a3
Ans. 1−27a3=13−(3a)3
=(1−3a)[12+1×3x+(3a)2]
=(1−3a)(1+3a+9a2)

Q.10. Factorize: 64a3 − 343
Ans. 64a3−343=(4a)3−(7)3
=(4a−7) (16a2+4a×7+72)
=(4a−7) (16a2+28a+49)

Q.11. Factorize: x3 − 512
Ans. x3−512 =x3−83
=(x−8) (x2+8x+82)
=(x−8) (x2+8x+64)

Q.12. Factorize: a3 − 0.064
Ans. a3−0.064=(a)3−(0.4)3
=(a−0.4) [a2+a× (0.4) +(0.4)2]
=(a−0.4) (a2+0.4a+0.16)

Q.13. Factorize: 8x3−1/27y3
Ans. 8x3−127y3=(2x)3−(1/3y)3
=(2x−1/3y) [(2x)2+2x×1/3y+(1/3y)2]
=(2x−1/3y) (4x2+2x/3y+1/9y2)

Q.14. Factorise: x3/216−8y3
Ans. We know a3−b3=(a−b) (a2+b2+ab)
We have,
x3/216−8y3=(x/6)3−(2y)3
So, a=x/6, b=2y
x3/216−8y3=(x/6−2y) ((x/6)2+x/6×2y+(2y)2)
=(x/6−2y) (x2/36+xy/3+4y2)

Q.15. Factorize: x − 8xy3
Ans. x−8xy3=x(1−8y3)
=x[13−(2y)3]
=x(1−2y)(12+1×2y+(2y)2)
=x(1−2y)(1+2y+4y2)

Q.16. Factorise: 32x4 – 500x
Ans. 32x4 – 500x=4x(8x3−125) =4x((2x)3−53)
we know
a3−b3=(a−b) (a2+b2+ab)
a=2x, b=5
32x4 – 500x=4x((2x)3−53)
=4x(2x−5) (4x2+25+10x)

Q.17. Factorize: 3a7b − 81a4b4
Ans. 3a7b−81a4b4=3a4b(a3−27b3)
=3a4b[a3−(3b)3]
=3a4b(a−3b)[a2+a×3b+(3b)2]
=3a4b(a−3b)(a2+3ab+9b2)

Q.18. Factorise: x4 y4 – xy
Ans. Using the identity
a3−b3=(a−b)(a2+b2+ab)
x4 y4–xy=xy(x3y3−1)
=xy(xy−1)(x2y2+1+xy)

Q.19. Factorise: 8x2 y3 – x5
Ans. 8x2y3–x5=x2(8y3−x3)
=x2(2y−x)(4y2+x2+2xy)

Q.20. Factorise:1029 – 3x3
Ans. 1029–3x3
=3(343−x3)
=3(73−x3)
=3(7−x)(49+x2+7x)

Q.21. Factorize: x6 − 729
Ans. x6−729=(x2)3−(9)3
=[x2−9][(x2)2+x2×9+92]
=[x2−32](x4+9x2+81)
=(x+3)(x−3)(x4+18x2+81−9x2)
=(x+3)(x−3)[(x2)2+2×x2×9+92−9x2]
=(x+3)(x−3)[(x2+9)2−(3x)2]
=(x+3)(x−3)(x2+9+3x)(x2+9−3x)
=(x+3)(x−3)(x2+3x+9)(x2−3x+9)

Q.22. Factorise: x9 – y9
Ans. x9–y9=(x3)3−(y3)3
we know
a3−b3=(a−b)(a2+b2+ab)
a=x3,b=y3
So,x9–y9=(x3)3−(y3)3
=(x3−y3)(x6+y6+x3y3)
=(x−y)(x2+y2+xy)(x6+y6+x3y3)

Q.23. Factorize: (a + b)3 − (a − b)3
Ans. (a + b)3−(a−b)3=[(a+b)−(a−b)][(a+b)2+(a+b)(a−b)+(a−b)2]
=(a+b−a+b)[a2+2ab+b2+a2−b2+a2−2ab+b2]
=2b(3a2+b2)

Q.24. Factorize: 8a3 − b3 − 4ax + 2bx
Ans. 8a3−b3−4ax+2bx=[(2a)3−(b)3]−2x(2a−b)
=(2a−b)[(2a)2+2ab+b2]−2x(2a−b)
=(2a−b)(4a2+2ab+b2)−2x(2a−b)
=(2a−b)(4a2+2ab+b2−2x)

Q.25. Factorize: a3 + 3a2b + 3ab2 + b3 − 8
Ans. a3+3a2b+3ab2+b3−8=(a3+b3+3a2b+3ab2)−8
=[a3+b3+3ab(a+b)]−8
=(a+b)3−23
=(a+b−2)[(a+b)2+2(a+b)+22]
=(a+b−2)[(a+b)2+2(a+b)+4]

Q.26. Factorize: a3−1/a3−2a+2/a
Ans. a3−1/a3−2a+2/a=(a3−1/a3)−2(a−1/a)
=[(a)3−(1/a)3]−2(a−1/a)
=(a−1/a)[a2+a×1/a+(1/a)2]−2(a−1/a)
=(a−1/a)(a2+1+1/a2)−2(a−1/a)
=(a−1/a)(a2+1+1/a2−2)
=(a−1/a)(a2−1+1/a2)

Q.27. Factorize: 2a3 + 16b3 − 5a − 10b
Ans. 2a3+16b3−5a−10b=2[a3+8b3]−5(a+2b)
=2[a3+(2b)3]−5(a+2b)
=2(a+2b)[a2−a×2b+(2b)2]−5(a+2b)
=2(a+2b)(a2−2ab+4b2)−5(a+2b)
=(a+2b)[2(a2−2ab+4b2)−5]

Q.28. Factorise: a6 + b6
Ans. a6+b6=(a2)3+(b2)3
=(a2+b2)[(a2)2−a2b2+(b2)2]
=(a2+b2)(a4−a2b2+b4)

Q.29. Factorise: a12 – b12
Ans. a12 – b12
=(a6+b6)(a6−b6)
=[(a2)3+(b2)3][(a3)2−(b3)2]
=[(a2+b2)(a4+b4−a2b2)][(a3−b3)(a3+b3)]
=[(a2+b2)(a4+b4−a2b2)][(a−b)(a2+b2+ab)(a+b)(a2+b2−ab)]
=(a−b)(a2+b2+ab)(a+b)(a2+b2−ab)(a2+b2)(a4+b4−a2b2)

Q.30. Factorise: x6 – 7x3 – 8
Ans. Let x3=y
So, the equation becomes
y2−7y−8=y2−8y+y−8
=y(y−8)+(y−8)
=(y−8)(y+1)
=(x3−8)(x3+1)
=(x−2)(x2+4+2x)(x+1)(x2+1−x)

Q.31. Factorise: x3 – 3x2 + 3x + 7
Ans. x3 – 3x2 + 3x + 7
=x3–3x2+3x+7
=x3–3x2+3x+8−1
=x3–3x2+3x−1+8
=(x3–3x2+3x−1)+8
=(x−1)3+23
=(x−1+2)[(x−1)2+4−2(x−1)]
=(x+1)[x2+1−2x+4−2x+2]
=(x+1)(x2−4x+7)

Q.32. Factorise: (x +1)3 + (x – 1)3
Ans. (x +1)3 + (x – 1)3
=(x+1+x−1)[(x+1)2+(x−1)2−(x−1)(x+1)]
=(2x)[(x+1)2+(x−1)2−(x2−1)]
=2x(x2+1+2x+x2+1−2x−x2+1)
=2x(x2+3)

Q.33. Factorise: (2a +1)3 + (a – 1)3
Ans. (2a +1)3 + (a – 1)3
=(2a+1+a−1)[(2a+1)2+(a−1)2−(2a+1)(a−1)]
=(3a)[4a2+1+4a+a2+1−2a−2a2+2a−a+1]
=3a[3a2+3a+3]
=9a(a2+a+1)

Q.34. Factorise: 8(x +y)3 – 27(x – y)3
Ans. 8(x +y)3 – 27(x – y)3
=[2(x+y)]3−[3(x−y)]3
=(2x+2y−3x+3y)[4(x+y)2+9(x−y)2+6(x2−y2)]
=(−x+5y)[4(x2+y2+2xy)+9(x2+y2−2xy)+6(x2−y2)]
=(−x+5y)[4x2+4y2+8xy+9x2+9y2−18xy+6x2−6y2]
=(−x+5y)(19x2+7y2−10xy)

Q.35. Factorise: (x +2)3 + (x – 2)3
Ans. (x +2)3 + (x – 2)3
=(x+2+x−2)[(x+2)2+(x−2)2−(x2−4)]
=2x(x2+4+4x+x2+4−4x−x2+4)
=2x(x2+12)

Q.36. Factorise: (x + 2)3 – (x – 2)3
Ans. (x + 2)3 – (x – 2)3
=(x+2−x+2)[(x+2)2+(x−2)2+(x2−4)]
=4[x2+4+4x+x2+4−4x+x2−4]
=4(3x2+4)

Q.37. Prove that (0.85×0.85×0.85+0.15×0.15×0.15)/(0.85×0.85−0.85×0.15+0.15×0.15)=1.
Ans. LHS: (0.85×0.85×0.85+0.15×0.15×0.15)/(0.85×0.85−0.85×0.15+0.15×0.15)
=(0.85)3+(0.15)3/(0.85)2−0.85×0.15+(0.15)2
We know a3+b3=(a+b)(a2+b2−ab)
Here a=0.85,b=0.15
(0.85)3+(0.15)3/(0.85)2−0.85×0.15+(0.15)2
=(0.85+0.15)((0.85)2−0.85×0.15+(0.15)2)(0.85)2−0.85×0.15+(0.15)2=0.85+0.15=1:RHS
Thus, LHS = RHS

Q.38. Prove that (59×59×59−9×9×9)/(59×59+59×9+9×9)=50.
Ans. (59×59×59−9×9×9)/(59×59+59×9+9×9)
=(59)3−93/592+59×9+92
We know a3+b3=(a+b)(a2+b2−ab)
Here a=59,b=9
So, ((59−9)(592+92+59×9))/(592+92+59×9)=59−9=50:RHS
Thus, LHS=RHS

The document RS Aggarwal Solutions: Exercise 3F - Factorisation of Polynomials | Extra Documents & Tests for Class 9 is a part of the Class 9 Course Extra Documents & Tests for Class 9.
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FAQs on RS Aggarwal Solutions: Exercise 3F - Factorisation of Polynomials - Extra Documents & Tests for Class 9

1. How can I factorise a polynomial using the method of common factors?
Ans. To factorise a polynomial using the method of common factors, you need to find the factor that is common to all the terms of the polynomial and then divide each term by this factor. Repeat this process until you have factored out all common factors.
2. Can all polynomials be factorised completely?
Ans. Not all polynomials can be factorised completely. Some polynomials may have irreducible factors that cannot be further factorised. However, most polynomials can be factorised using various methods such as common factors, grouping, or special products.
3. Is there a specific order in which I should factorise a polynomial?
Ans. There is no specific order in which you must factorise a polynomial. You can choose to factorise based on common factors, grouping, or any other method that you find suitable for the given polynomial. The goal is to simplify the polynomial as much as possible.
4. What is the difference between factorising a polynomial and simplifying a polynomial?
Ans. Factorising a polynomial involves expressing it as a product of its factors, while simplifying a polynomial involves reducing it to its simplest form by combining like terms or performing operations such as addition, subtraction, multiplication, and division.
5. Are there any shortcuts or tricks to factorise polynomials quickly?
Ans. While there are no specific shortcuts or tricks to factorise all polynomials quickly, practicing different factorisation methods and familiarising yourself with common patterns in polynomials can help you factorise them more efficiently. Additionally, using factorisation formulas for special cases can also speed up the process.
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