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:

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

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