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(a) Factorise : `x^(2) - (z-5)x - 5z` (b) Factorise : `x^(2) + x - y + y^(2) - 2xy` |
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Answer» (a) `x^(2) - (z-5)x - 5z` = `x^(2) - xz + 5x - 5z = x(x-z) + 5(x-z) = (x-z) (x + 5)` `therefore x^(2) - (z-5)x - 5z = (x-z) (x + 5)` (b) `x^(2) + x - y + y^(2) - 2xy` `= x^(2) + y^(2) - 2xy + x - y = (x-y)^(2) + 1(x-y) = (x-y) (x+y + 1)` `therefore x^(2) + x - y + y^(2) - 2xy = (x-y) (x+y + 1)` |
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