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Mathematics 23 Online
OpenStudy (anonymous):

How many integer values are there for k for which 4x2 + kxy – 9y2 is factorable?

OpenStudy (ranga):

To factor the expression we need to split the middle term into two parts. Let k = p + q where k, p, q are all integers. 4x^2 + kxy - 9y^2 = 4x^2 + pxy + qxy - 9y^2 = (4x^2 + pxy) + (qxy - 9y^2) = x(4x + py) + y(qx - 9y) In order to factor we must have (4x + py) = (qx - 9y) Or q = 4 and p = -9 Therefore, k = p + q = -9 + 4 = -5 That is one integer.

OpenStudy (anonymous):

that is really confusing... can you maybe dumb the instructions down a bit?

OpenStudy (ranga):

I myself am not 100% sure if this is the way to proceed. But there is at least one integer found this way. Let me think if there may be other ways.

OpenStudy (anonymous):

I heard the answer was 9 if that helps

OpenStudy (ranga):

wow!

OpenStudy (anonymous):

what?

OpenStudy (anonymous):

I still need help figuring out how to get 9

OpenStudy (anonymous):

i got k = 0, ±35, ±16, ±9, ±5

OpenStudy (ranga):

What method did you use @souwing ?

OpenStudy (anonymous):

the same way you did actually. Given (4x^2 + pxy) + (qxy - 9y^2), and p+q = k 4x (x + (p/4)y) + qy (y - (9/q)y) so to be factorable, p/4 = -9/q pq = -36 so the solutions are the integer factors of -36. I.e -1 and 36 or 1 and -36 -2 and 18 or 2 and -18 etc... since k = q + p, just add up the factors

OpenStudy (whpalmer4):

the k=0 case is easy — the middle terms drop out leaving you with a difference of squares which is of course factorable

OpenStudy (whpalmer4):

easy in hindsight, that is :-)

OpenStudy (whpalmer4):

In[57]:= eq = Out[49] Out[57]= 4 x^2 + k x y - 9 y^2 In[58]:= Factor[eq /. k -> 0] Out[58]= (2 x - 3 y) (2 x + 3 y) In[59]:= Factor[eq /. k -> 35] Out[59]= (4 x - y) (x + 9 y) In[60]:= Factor[eq /. k -> -35] Out[60]= (x - 9 y) (4 x + y) In[62]:= Factor[eq /. k -> 16] Out[62]= (2 x - y) (2 x + 9 y) In[63]:= Factor[eq /. k -> -16] Out[63]= (2 x - 9 y) (2 x + y) In[64]:= Factor[eq /. k -> 9] Out[64]= (4 x - 3 y) (x + 3 y) In[65]:= Factor[eq /. k -> -9] Out[65]= (x - 3 y) (4 x + 3 y) In[66]:= Factor[eq /. k -> 5] Out[66]= (x - y) (4 x + 9 y) In[67]:= Factor[eq /. k -> -5] Out[67]= (4 x - 9 y) (x + y)

OpenStudy (whpalmer4):

if you had a need to see all of the factorings, you're all set :-)

OpenStudy (ranga):

sfaigan: If you need a simpler version without the p and q: The original expression is: 4x2 + kxy – 9y2 If you multiply the coefficients of the first and last terms you get 4 * -9 = -36. We need to split the middle term coefficient k such that its product is -36 much like the AC method of factoring a quadratic. -36 can be factored as: (-1)*36 or 1*(-36) (-2)*18 or 2*(-18) (-3)*12 or 3*(-12) (-4)*9 or 4*(-9) (-6)*6 or 6*(-6) And the middle coefficient k will be the sum of the above factors. k = 0, ±5, ±9, ±16, ±35 A total of 9 factors.

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