How Do You Factor Completely, 2x^2-3xy-2y^2?

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Vin Ginge answered
2x2 - 3xy - 2y2 = (2x + y)(x - 2y)

When dealing with quadratic expressions the standard form is ax2 + bx + c.
The factor format is (mx + n)(px + q). In this question a = 2, b = -3y, c = -2y2.
The technicalities can be covered later but basically permutate the product of the factors of a with those of c so that the sum generates the value of b (= -3y).
As a = 2 the factors are 1,2.
As c = -2y2 the factors are -y,2y,  (the reverse also applies eg -2y,y)
The permutations are 1 x -y, 2 x 2y, sum 3y: 1 x 2y, 2 x -y, sum 0.  Neither combination produces the sum -3y but the first pairing produces 3y. So, by working with the 'reverse' factors we have 1 x y, 2 x -2y, sum -3y which is the desired result.

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