How do you factor 28x2+5xy−12y2?

1 Answer
George C.
Oct 10, 2015

Use the quadratic formula to find factors:

28x2+5xy−12y2=(4x+3y)(7x−4y)

Explanation:

If you divide this polynomial by y2 then you get a quadratic in xy:

28x2+5xy−12y2y2=28(xy)2+5(xy)−12

So if we let t=xy we get a quadratic in t

28t2+5t−12

which is of the form at2+bt+c, with a=28, b=5 and c=−12.

This has zeros for values of t given by the quadratic formula:

t=−b±√b2−4ac2a=−5±√52−(4×28×−12)2⋅28

=−5±√25+134456=−5±√136956=−5±3756

That is t=−4256=−34 and t=3256=47

Hence:

28t2+5t−12=(4t+3)(7t−4)

So

28(xy)2+5(xy)−12=(4(xy)+3)(7(xy)−4)

Then multiply through by y2 to find:

28x2+5xy−12y2=(4x+3y)(7x−4y)

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