Factoring Quadratic Trinomials

Question

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

$$5x^2+7xy+2y^2$$

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

Why don't you add the work what you would do if $5x^2 + 7x+2$ is the given problem? — user62089, Jun 05 '13 at 19:44
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score 3 · Answer 1 · answered Jun 05 '13 at 19:55

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.

score 3 · Answer 2 · answered Jun 05 '13 at 20:05

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.

score 1 · Answer 3 · answered Jun 05 '13 at 19:43

1

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

answered Jun 05 '13 at 19:43

response

5,071

score 1 · Answer 4 · answered Jun 05 '13 at 19:44

1

Do you know the cross product method for factoring?

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

answered Jun 05 '13 at 19:44

Sujaan Kunalan

10,944

score 1 · Answer 5 · edited Apr 13 '17 at 12:21

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).

score 1 · Accepted Answer · edited Jun 06 '13 at 00:00

Given:

$\boxed{5x^2+7xy+2y^2}$

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:

$$(5x^2+5xy)+(2xy+2y^2)$$

We will now factor out any common factors.

$$=5x\left(\frac{5x^2+5xy}{5x}\right)+2y\left(\frac{2xy+2y^2}{2y}\right)\\=5x(x+y)+2y(x+y)\\=(5x+2y)(x+y)$$

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

$$(5x+2y)(x+y)$$

Factoring Quadratic Trinomials

6 Answers6