RS Aggarwal Solutions: Exercise 3E - Factorisation of Polynomials | Extra Documents & Tests for Class 9 PDF Download
Q.1. Expand
(i) (3x + 2)3
(ii) (3a+1/4b)3
(iii) (1+2/3 a)3
Ans. (i) (3x+2)3=(3x)3+3×(3x)2x2+3×3x×(2)2+(2)3
=27x3+54x2+36x+8
(ii) (3a+1/4b)3=(3a)3+(1/4b)3+3(3a)2(1/4b)+3(3a)(14b)2 =27a3+1/64b3+27a2/4b+9a/16b2
(iii) (1+2/3a)3=(2/3a)3+3×(2/3a)2×1+3a2/3a×(1)2+(1)3 =827a3+43a2+2a+1iii 1+23a3=23a3+3×23a2×1+3a23a×12+13 =8/27a3+43a2+2a+1
Q.2. Expand
(i) (5a – 3b)3
(ii) (3x−5/x)3
(iii) (4/5a−2)3
Ans. (i) (5a−3b)3=(5a)3−(3b)3−3(5a)2(3b)+3(5a)(3b)2 =125a3−27b3−225a2b+135ab2
(ii) (3x−5/x)3=(3x)3−(5/x)3−3(3x)2(5/x)+3(3x)(5/x)2
=27x3−125/x3−135x+225/x
(iii) (4/5a−2)3=(4/5a)3−(2)3−3(4/5a)2(2)+3(4/5a)(2)2
=64/125a3−8−96/25a2+48/5a
Q.3. Factorise 8a3+27b3+36a2b+54ab2
Ans. 8a3+27b3+36a2b+54ab2
=(2a)3+(3b)3+3(2a)2(3b)+3(2a)(3b)2
=(2a+3b)3
Hence, factorisation of 8a3+27b3+36a2b+54ab2
Q.4. Factorise 64a3−27b3−144a2b+108ab2
Ans. 64a3−27b3−144a2b+108ab2
=(4a)3−(3b)3−3(4a)2(3b)+3(4a)(3b)2
=(4a-3b)3
Hence, factorisation of 64a3−27b3−144a2b+108ab2
Q.5. Factorise 1+27/125 a3+9a/5+27a2/24
Ans. 1+27/125 a3+9a/5+27a2/25
=(1)3+(3/5a)3+3(1)2(3/5 a)+3(1)(3/5a)2
Hence, factorisation of 1+27/125 a3+9a/5+27a2/25 is (1+3/5a)3
Q.6. Factorise 125x3−27y3−225x2y+135xy2
Ans. 125x3−27y3−225x2y+135xy2
=(5x)3−(3y)3−3(5x)2(3y)+3(5x)(3y)2
=(5x-3y)3
Hence, factorisation of 125x3−27y3−225x2y+135xy2 is (5x−3y)3
Q.7. Factorise: a3x3−3a2bx2+3ab2x−b3
Ans. a3x3−3a2bx2+3ab2x−b3
=(ax)3−(b)3−3(ax)2(b)+3(ax)(b)2
= (ax-b)3
Hence, factorisation of a3x3−3a2bx2+3ab2x−b3 is (ax−b)3.
Q.8. Factorise: 64/125a3−96/25a2+48/5a−8
Ans. 64/125 a3−96/25 a2+48/5 a−8
=(4/5 a)3−(2)3−3(4/5 a)2(2)+3(4/5 a)(2)2
=(4/5 a-2)3
Hence, factorisation of 64/125 a3−96/25 a2+48/5 a is (4/5a−2)3 is (4/5a-2)3.
Q.9. Factorise a3 – 12a(a – 4) – 64
Ans. a3−12a(a−4)−64
=a3−12a2+48a−64
=(a)3−(4)3−3(a)2(4)+3(a)(4)2
= (a-4)3
Hence, factorisation of a3 – 12a(a – 4) – 64 is (a−4)3.
Q.10. Evaluate
(i) (103)3
(ii) (99)3
Ans. (i) (103)3=(100+3)3
=(100)3+(3)3+3(100)2(3)+3(100)(3)2
=1000000+27+90000+2700
=1092727
(ii) (99)3=(100−1)3
=(100)3−(1)3−3(100)2(1)+3(100)(1)2
=1000000−1−30000+300
=1000300−30001
=970299
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