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: 64a− 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 y– 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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