How do you factor the trinomial #x^2 - 2xy - 48y^2#?

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Answer 1 · 2018-06-03T06:43:35

#x^2-2xy-48y^2 = (x-8y)(x+6y)#

Given:

#x^2-2xy-48y^2#

Note that this is a homogeneous quadratic in two variables. So we can factor it in a similar way to factoring a quadratic in one variable.

Find a pair of factors of #48# which differ by #2#.

The pair #8, 6# works in that #8 * 6 = 48# and #8 - 6 = 2#.

Hence we find:

#x^2-2xy-48y^2 = (x-8y)(x+6y)#