1
$\begingroup$

A polynomial $2x^2+mxy+3y^2-5y-2$ Find the value of $m$ much that $p(xy)$ can be factorized into two linear factors

$\endgroup$
  • $\begingroup$ Shouldn't it be $p(x,y)$, not just $p(x)$? $\endgroup$ – Gregory Grant May 24 '15 at 15:31
1
$\begingroup$

Putting $m = 7$ you obtain \begin{align*} (x+3y+1)(y+2x-2) &= xy+2x^2-2x+3y^2+6xy-6y+y+2x-2 \\ &= 2x^2 + 7xy +3y^2-5y-2 \end{align*}

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ of course doing it that way I figured it out too ma can be 7 or -7 but the real question is what is the standard way of doing it $\endgroup$ – bulbasaur May 24 '15 at 15:58
0
$\begingroup$

Solve $p(x,y) = 0$ for $x$. Roots are $q_1(y)$ and $q_2(y)$. You will obtain a representation $p = (x - q_1)(x - q_2)$. Now find $m$ that makes $q$ into linear expressions.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.