Class 9 Exam > Class 9 Notes > Extra Documents & Tests for Class 9 > RS Aggarwal Solutions: Exercise 3D - Factorisation of Polynomials
RS Aggarwal Solutions: Exercise 3D - Factorisation of Polynomials | Extra Documents & Tests for Class 9 PDF Download
Q.1. Expand:
(i) (a + 2b + 5c)2
(ii) (2b − b + c)2
(iii) (a − 2b − 3c)2
Ans. (i) (a+2b+5c)2
=(a)2 + (2b)2 +(5c)2+2(a)(2b)+2(2b)(5c)+2(5c)(a) =a2+4b2+25c2+4ab+20bc+10ac
(ii) (2a−b+c)2=[(2a)+(−b)+(c)]2
=(2a)2+(−b)2+(c)2+2(2a)(−b)+2(−b)(c)+4(a)(c)
=4a2+b2+c2−4ab−2bc+4ac
(iii) (a−2b−3c)2=[a+(−2b)+(−3c)]2
=(a)2+(−2b)2+(−3c)2+2(a)(−2b)+2(−2b)(−3c)+2(a)(−3c)
=a2+4b2+9c2−4ab+12bc−6ac
Q.2. Expand:
(i) (2a − 5b − 7c)2
(ii) (−3a + 4b − 5c)2
(iii) (12a−14a+2)2
Ans. (i) (2a−5b−7c)2
=[(2a)+(−5b)+(−7c)]2
=(2a)2+(−5b)2+(−7c)2+2(2a)(−5b)+2(−5b)(−7c)+2(2a)(−7c)
=4a2+25b2+49c2−20ab+70bc−28ac
(ii) (−3a+4b−5c)2=[(−3a)+(4b)+(−5c)]2
=(−3a)2+(4b)2+(−5c)2+2(−3a)(4b)+2(4b)(−5c)+2(−3a)(−5c)
=9a2+16b2+25c2−24ab−40bc+30ac
(iii) (1/2a−1/4b+2)2=[(a/2)+(−b/4)+(2)]2
=(a2)2+(−b4)2+(2)2+2(a/2)(−b/4)+2(−b/4)(2)+2(a/2)(2)
=a2/4+b2/16+4−ab/4−b+2a
Q.3. Factorize: 4x2 + 9y2 + 16z2 + 12xy − 24yz − 16xz.
Ans. We have: 4x2+9y2+16z2+12xy−24yz−16xz
=(2x)2+(3y)2+(−4z)2+2(2x)(3y)+2(3y)(−4z)+2(−4z)(2x)
=[(2x)+(3y)+(−4z)]2
=(2x+3y−4z)2
Q. 4. Factorize: 9x2 + 16y2 + 4z2 − 24xy + 16yz − 12xz
Ans. We have: 9x2+16y2+4z2−24xy+16yz−12xz
=(−3x)2+(4y)2+(2z)2+2(−3x)(4y)+2(4y)(2z)+2(2z)(−3x)
=[(−3x)+(4y)+(2z)]2
=(−3x+4y+2z)2
Q.5. Factorize: 25x2 + 4y2 + 9z2 − 20xy − 12yz + 30xz.
Ans. We have: 25x2+4y2+9z2−20xy−12yz+30xz
=(5x)2+(−2y)2+(3z)2+2(5x)(−2y)+2(−2y)(3z)+2(3z)(5x)
=[(5x)+(−2y)+(3z)]2
=(5x−2y+3z)2
Q.6. 16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz
Ans. 16x2+4y2+9z2−16xy−12yz+24xz
=(4x)2+(−2y)2+(3z)2+2(4x)(−2y)+2(−2y)(3z)+2(3z)(4x)
=(4x−2y+3z)2 [using a2+b2+c2+2ab+2bc+2ca=(a+b+c)2]
Hence, 16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz = (4x−2y+3z)2
Q.7. Evaluate
(i) (99)2
(ii) (995)2
(iii) (107)2
Ans. (i) (99)2=(100−1)2
=[(100)+(−1)]2
= (100)2+2×(100)×(−1)+(−1)2
=10000−200+1
=9801
(ii) (995)2=(1000−5)2
=[(1000)+(−5)]2
=(1000)2+2×(1000)×(−5)+(−5)2
=1000000−10000+25
=990025
(iii) (107)2=(100+7)2
=(100)2+2×(100)×(7)+(7)2
=10000+1400+49
=11449
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