i) √3x^2 -9x +6√3
=> √3x^2 - 6x - 3x + 6√3
=> √3x(x-2√3) -3(x-2√3)
=> (√3x-3)(x-2√3)
ii)
5−(3a^2 −2a)(6−3a^2 +2a) = 5−(3a^2 −2a)[6−(3a^2 −2a)] = 5−y(6−y)(Assuming3a^2 −2a=y)
=5−6y+y
2
=y
−6y+5
−5y−y+5
=y(y−5)−1(y−5)
=(y−5)(y−1)
=(3a
−2a−5)(3a
−2a−1)(∵3a
−2a=y)
+3a−5a−5)(3a
−3a+a−1)
=[3a(a+1)−5(a+1)][3a(a−1)+1(a−1)]
=(3a−5)(a−1)(a+1)(3a+1)
Hence, the factorization of 5−(3a^2−2a)(6−3a^2 +2a) is (3a−5)(a−1)(a+1)(3a+1).
Hope it helps
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Answers & Comments
i) √3x^2 -9x +6√3
=> √3x^2 - 6x - 3x + 6√3
=> √3x(x-2√3) -3(x-2√3)
=> (√3x-3)(x-2√3)
ii)
5−(3a^2 −2a)(6−3a^2 +2a) = 5−(3a^2 −2a)[6−(3a^2 −2a)] = 5−y(6−y)(Assuming3a^2 −2a=y)
=5−6y+y
2
=y
2
−6y+5
=y
2
−5y−y+5
=y(y−5)−1(y−5)
=(y−5)(y−1)
=(3a
2
−2a−5)(3a
2
−2a−1)(∵3a
2
−2a=y)
=(3a
2
+3a−5a−5)(3a
2
−3a+a−1)
=[3a(a+1)−5(a+1)][3a(a−1)+1(a−1)]
=(3a−5)(a−1)(a+1)(3a+1)
Hence, the factorization of 5−(3a^2−2a)(6−3a^2 +2a) is (3a−5)(a−1)(a+1)(3a+1).
Hope it helps