Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm currently doing some homework, but I'm COMPLETELY stuck on one problem. I need to factor the following trinomial:


How can I solve this problem? I have no idea what to do because of the different variables.

share|improve this question
Why don't you add the work what you would do if $5x^2 + 7x+2$ is the given problem? –  user62089 Jun 5 '13 at 19:44
PS. Once a question is answered in a way that you understand, choose the best solution and click the check mark next to it; this will mark it as accepted, i.e. say that the question is "answered" as far as you're concerned. If you do this (1) you get rep (2) the person gets rep (precious rep) (3) people will be more likely to answer your future questions (4) it's polite. You should go back to your old questions and accept the answers. –  Douglas B. Staple Jun 5 '13 at 20:29
add comment

6 Answers

up vote 1 down vote accepted



We can use a method called grouping to factor this equation.

  • Start by multiplying $5x^2\cdot2y^2$ which equals $10x^2y^2.$

  • Now look for two numbers which multiply to $10x^2y^2$ and add to $7xy$.

  • In this case, $2xy$ & $5xy$ are the two numbers which we want because $2xy\cdot5xy=10x^2y^2$ and $2xy+5xy=7xy$.

  • We can now split the $7xy$ in the original expression:


We will now factor out any common factors.


The final factorization of ${5x^2+7xy+2y^2}$:


share|improve this answer
add comment

Figure it out as follows.

Firstly, realize that the answer has to be something like $(ax + by)(cx + dy)$ in order to get those $x^2$ and $y^2$ terms. But then one of $a$ or $c$ has to be $1$, and the other one has to be $5$, or you'll never get $5x^2$. You can make a similar argument to figure out what $b$ and $d$ might be. In the end there's only four possibilities; two of them give the right answer, and two of them don't.

share|improve this answer
add comment

There are various ways of looking at this homogeneous form.

Suppose you can factorize $5z^2+7z+2=(az+b)(cz+d)$ then the homogeneous form is factorised as follows, taking $z=\cfrac xy$$$5x^2+7xy+2y^2=y^2\left(5(\frac xy)^2+7\frac xy+2\right)=y^2(5z^2+7z+2)=y^2(az+b)(cz+d)$$Now allocate a factor $y$ to each bracket$$=(azy+by)(czy+dy)=(ax+by)(cx+dy)$$

So the factorisation is essentially the obvious one you know, and can be obtained by setting $y=1$, for example. The homogeneous form also admits the possibility $y=0$ and can be seen as extending the original factorisation "to infinity" but sadly not beyond. For this reason homogeneous polynomials (and "projective" structures of various kinds) become significant in Algebraic Geometry - they avoid special cases at infinity.

The basic arithmetic of the factorisation remains the same.

share|improve this answer
add comment

Hint: Substitute $x=-y$ and see what you get.

share|improve this answer
add comment

Do you know the cross product method for factoring?

It is the same process as if the question was this instead: $5x^2 +7x+2$.

share|improve this answer
add comment

By the AC method we can reduce to factoring a monic (leading coefficient $= 1).$

$$\begin{eqnarray} f\, &=&\ \ \: \color{#c00}2y^2\ + 7xy\ \ \ +\ \ \ 5x^2 \\ \Rightarrow\ \color{#c00}2f\, &=&\, (\color{#c00}2y)^2 + 7x(\color{#c00}2y)+ \color{#c00}2\cdot 5\ x^2, \ \ {\rm let}\ \ Y = 2y \\ &=&\quad Y^2 + 7x\ Y + (2x)(5x)\\ &=&\,\ \ (Y + 2x)(Y + 5x) \\ &=&\,\ (2y+2x)(2y+5x)\\ \Rightarrow\ f\, &=&\ \ \ \, (y\ +\ x)(2y+5x)\end{eqnarray}$$

Remark $\ $ Due to unique factorization, this method will always succeed (see the above link for further details).

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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