polynomial factoring

**skoker** · June 12th 2011, 10:40 PM

this one is giving me trouble.

$3p^2-50y^2-25yp+7p+2$

I try some normal things and get nowhere...

$(3p+5y)(p-y10)+7p+2$
$(3p+1)(p+2)-25yp-50y^2$

so I try to make perfect squares and fold them together as difference of squares. but to do this given the coefficients I have to start adding, subtracting and splitting terms.

$(p+3)^2-(p+5y)^2 + 3p^2 +p -7 -15yp -25y^2$

at this point Im not sure what to do. I think I could keep going for ever like this and I see no pattern to fold everything together.

I know it can be factored by the CAS into 2 factors with 3 terms each.

**Unknown008** · June 13th 2011, 12:57 AM

I don't know if that will work... but here's my 2 cents.

Since you know there are two factors with 3 terms, the most likely 'form' would be:

$(ap + by + c)(qp + ry + s)$

Which gives, after expansion and simplification; $aqp^2 + (ar+bq)py + bry^2 + (cr+bs)y + (cq+as)p + cs$

Which gives us six equations:

$aq = 3$

$ar+bq = -25$

$br = -50$

$cr + bs = 0$

$cq +as = 7$

$cs = 2$

Oh well, that appears to be long.

I got ar = 5 using the first three equations...

**skoker** · June 13th 2011, 12:22 PM

@Unknown008
I am not sure I get this. does this show the factors? or are you going backward trying to reconstruct the equation and find the coefficient combinations. is there a name for this technique?

I can't have this dirty little wolfram robot taunting me with his nice factor!

I have to do it! so I use quadratic formula...

$\begin{aligned}&3p^2+7p+2-25py-50y^2\\&3p^2+(7-25y)p-50y^2+2\\&\frac{-(7-25y)\pm \sqrt{(7-25y)^2-(4)(3)(2-50y^2)}}{(2)(3)}\\&\frac{-7+25y\pm \sqrt{25(7y-1)^2}}{6}\\&\frac{-7+25y+[5(7y-1)]}{6} \quad ,\quad \frac{-7+25y-[5(7y-1)]}{6}\\&\frac{-2-10y}{6} \quad ,\quad \frac{-12+60y}{6}\\&[p-\Big(\frac{-2-10y}{6}\Big)][p-\Big(\frac{-12+60y}{6}\Big)]\\&(6p+10y+2)(p+2-10y)\\&(3p+5y+1)(p+2-10y)=0\end{aligned}$

it seems you can use this for full conic form with all the terms for rotation and translation. probably other forms also but I have not tried.

**Unknown008** · June 13th 2011, 09:27 PM

I really didn't think of it that way

Nicely done, yes.

And no, I don't know how is this method I tried called, but yes, it's basically going backward trying to reconstruct the polynomial.

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