Factorise 50a^2-2(b-c)^2
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![Q) \: 50 {a}^{2} - 2(b - c) {}^{2} \\ \\ = 2[(25 {a}^{2} ) - (b - c){}^{2} ] \\ \\ = 2[(5a) {}^{2} - (b - c) {}^{2} ] \\ \\ = 2[(5a + b - c)(5a - b - c)]](https://tex.z-dn.net/?f=Q%29+%5C%3A+50+%7Ba%7D%5E%7B2%7D+-+2%28b+-+c%29+%7B%7D%5E%7B2%7D+%5C%5C+%5C%5C+%3D+2%5B%2825+%7Ba%7D%5E%7B2%7D+%29+-+%28b+-+c%29%7B%7D%5E%7B2%7D+%5D+%5C%5C+%5C%5C+%3D+2%5B%285a%29+%7B%7D%5E%7B2%7D+-+%28b+-+c%29+%7B%7D%5E%7B2%7D+%5D+%5C%5C+%5C%5C+%3D+2%5B%285a+%2B+b+-+c%29%285a+-+b+-+c%29%5D "Q) \: 50 {a}^{2} - 2(b - c) {}^{2} \\ \\ = 2[(25 {a}^{2} ) - (b - c){}^{2} ] \\ \\ = 2[(5a) {}^{2} - (b - c) {}^{2} ] \\ \\ = 2[(5a + b - c)(5a - b - c)]")
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is it right or wrong
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\begin{lgathered}Q) \: 50 {a}^{2} - 2(b - c) {}^{2} \\ \\ = 2[(25 {a}^{2} ) - (b - c){}^{2} ] \\ \\ = 2[(5a) {}^{2} - (b - c) {}^{2} ] \\ \\ = 2[(5a + b - c)(5a - b - c)]\end{lgathered}Q)50a2−2(b−c)2=2[(25a2)−(b−c)2]=2[(5a)2−(b−c)2]=2[(5a+b−c)(5a−b−c)]
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\begin{lgathered}Q) \: 50 {a}^{2} - 2(b - c) {}^{2} \\ \\ = 2[(25 {a}^{2} ) - (b - c){}^{2} ] \\ \\ = 2[(5a) {}^{2} - (b - c) {}^{2} ] \\ \\ = 2[(5a + b - c)(5a - b - c)]\end{lgathered}Q)50a2−2(b−c)2=2[(25a2)−(b−c)2]=2[(5a)2−(b−c)2]=2[(5a+b−c)(5a−b−c)]
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