factoring y^2 + xy - 2x^2

mauricev

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How can I factor y^2 + xy - 2x^2? It doesn't follow the traditional quadratic form, ax^2+bx+c. Cymath somehow manages to use the quadratic formula to factor it, but that seems unnecessarily complex. Mathpapa just ignores it.
 
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Oops, it should read y^2 + xy + 2x^2.
That expression does not factor nicely.

If you treat y as constant, you could apply the Quadratic Formula, to obtain a factorization in this form:

A*(x - r1)*(x - r2)

where r1 and r2 are the solutions coming out of the Quadratic Formula and A is a constant. However, there will be imaginary numbers involved.

It may be that your materials contain a typographical error.
 
It's y^2+xy-2x^2.
Okay -- this one factors nicely.

If you're not yet comfortable using the Quadratic Formula, you could try guess-and-check.

I'm going to assume that you already understand why the factorization needs this form:

(y + M)(y + N)

where the product M*N is -2x^2.

So, you could start substituting factors of -2x^2 for symbols M and N, and then multiply out the factorization to see if it works. If it does not work, then try a different pair of factors for -2x^2.

Alternatively, you could multiply out (y + M)(y + N) first:

y^2 + (M+N)*y - M*N

From this we see that M+N must equal x and M*N must equal -2x^2. Now, think back to how you factored quadratic trinomials with A=1 that had numerical coefficients for B and C (instead of symbolic ones, like x and -2x^2). You look for two quantities whose produce is C and whose sum is B, yes?

In other words, find two factors of -2x^2 that add to make x. Those two factors are M and N.



Have you learned Polynomial Long Division? That's another method for guessing one factor (like x+y, for example) and then using division to find the other factor (the check is whether you get a remainder of zero).



If I didn't want to guess and check, I would probably use the Quadratic Formula. It's fairly straightforward, here, because the Discriminant (B^2-4AC) turns out to be a squared expression (i.e., the radical sign goes away).

Treat y as the variable, and treat x as constant. Then:

A = 1

B = x

C = -2x^2

The factorization will take the form (y - r1)(y - r2) where r1 and r2 are the roots that you get from the Quadratic Formula.

Whichever way you decide to go, please show your work, if you need more help. :cool:
 
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Gosh,sorry. I have transcribed it wrong twice. It's y^2+xy-2x^2.
They were supposed to cover how to factor quadratics before they assigned homework on it! :shock:

To learn the general process, try here. Once you've learned the general terms and techniques, note that the method for factoring y2 + y - 2 will work just as well for factoring y2 + xy - 2x2. (Hint: You'll need factors of -2 that add to +1.) ;)
 
They were supposed to cover how to factor quadratics before they assigned homework on it! :shock:

To learn the general process, try here. Once you've learned the general terms and techniques, note that the method for factoring y2 + y - 2 will work just as well for factoring y2 + xy - 2x2. (Hint: You'll need factors of -2 that add to +1.) ;)

The method described at this link, the box method, works very nicely in this case. Thanks for that link and thanks to those who also answered.
 
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