# Factoring Polynomials with four terms and two variables

I've been working on this for hours and cannot figure it out.

When I search, I find factorization techniques that I already know but don't seem to be able to apply here, or that are for polynomials that don't have the same form.

I am beginning to wonder if this can even be factored.

$$4x^2 + 4x - 9y^2 -1$$

The most I can figure to factor is:

$$4x(x+1) - 9y^2 -1$$

Mahalo for the help, I am really trying to understand this.

• What do you want to do with the polynomial when you are done factoring it? – George V. Williams Jan 20 '13 at 2:00
• What are you trying to achieve? Are you trying to get a form like $k(x - a)(x - b)(y - c)(y - d) = 0$? Because that is impossible. – Herng Yi Jan 20 '13 at 2:01
• $1\cdot(4x^2+4x-9y^2-1)$. – David Mitra Jan 20 '13 at 2:01
• But, more seriously, I don't think your expression factors, at least not nicely or non-trivially. (Note "factor" means to write as a product of two or more expressions.) – David Mitra Jan 20 '13 at 2:07
• This polynomial does not factor in any interesting way. The closely related polynomial $4x^2+4x-9y^2+1$ does. Maybe that is the polynomial you are actually interested in. Impossible to know unless one knows where the problem came from. – André Nicolas Jan 20 '13 at 2:07

While the expression cannot be factored (it cannot be represented strictly as the product of factors, we can proceed to manipulate it in a way that involves two factors, with an added (subtracted) constant: \begin{align} 4x^2 + 4x - 9y^2 - 1 &= 4x^2 + 4x + \color{red}{1} - 9y^2 - 1 \color{red}{-1} \\ &= (2x+1)^2 - (9y^2 + 2) \\ &= (2x+1)^2 - (3y)^2 -2 \\ &= (2x + 1 +3y)(2x + 1 - 3y) -2 \\ &= (2x + 3y +1)(2x - 3y + 1) -2 \;? \end{align}

I'm "simply" simplifying the expression you gave (which is not an equation). If you meant $$4x^2 + 4x - 9y^2 - 1 = 0$$ you could write:

$$4x^2 + 4x - 9y^2 - 1=0$$ $$\iff (2x+1)^2 - (3y)^2 = 2$$ $$\iff (2x + 1 +3y)(2x + 1 - 3y) = 2$$ $$\iff (2x + 3y +1)(2x - 3y + 1) = 2$$

• Why did you add and minus 1? – jason Jan 20 '13 at 2:11
• "expression" $+1 + -1 =$ "expression $+ 0$ so it does not alter the expression, it amounts to adding zero to it. I did so to complete the square: I noted that $4x^2 + 4x + 1 = (2x+1)^2$, then to compensate for adding $1$, I subtracted $1$ so as not to change the value of the expression. – Namaste Jan 20 '13 at 2:13
• That's brilliant, thank you so much. What made you think to do that? Is that technique useful in other situations? I'm going to see if I can find something to read about completing the square. – jason Jan 20 '13 at 2:19
• I linked you to "complete the square" (above comment) at Wikipedia]. See also Difference of two squares – Namaste Jan 20 '13 at 2:21
• @amWhy: In changing the order of terms, you switched a "-" to a "+" in front of $3y$. – Clayton Jan 20 '13 at 2:22

The answers that currently are showing seem contradictory. @Jason has accepted one answer which does not factor the given polynomial, and @jathd has shown that the polynomial can not be factored. But the most conclusive answer is the one given by @Clayton. He completes the square and shows that the zeros of the polynomial (in the plane) constitute a hyperbola. If the polynomial had had a factorization, its zeros would have lain on two intersecting lines (because the total degree is two!). So the geometric argument is really quickest and most determinative.

The moral to be taken from this story is that polynomials in two variables rarely factor unless they are homogeneous (i.e. have all monomials of the same degree).

• Thank you for your comment. Notes from those more experienced and knowledgeable in mathematics will help me reach the level of understanding I a working towards. – jason Jan 20 '13 at 20:13

Your polynomial $f(x,y)=4x^2+4x-9y^2-1$ cannot be factored (over $\mathbf R$).

Here's a proof. If it could, you could write $f(x,y)=p(x,y)q(x,y)$. Look at the degree of $p$ in $x$: if it's $0$, then all the $x$ is in $q$, meaning $p(x,y)=p(y)$ doesn't depend on $x$. Then $p$ has some root $y_0$ (say in $\mathbf C$), and for all $x$ we would have $f(x,y_0)=0$, which is absurd.

If the degree of $p$ in $x$ is $2$, then all the $x$ is in $p$, and by the same argument, we reach a contradiction, so it has to be $1$. Using another similar argument, we see that the degrees in $y$ of $p$ and $q$ are also both $1$. In other words, you can write $p(x,y)=ax+by+c$ and $q(x,y)=a'x+b'y+c'$, so $f(x,y)$ is a product of two linear polynomials.

Let's put $x=0$ everywhere: we get $-9y^2-1$ for $f(0,y)$, which doesn't have any root in $\mathbf R$, and $(by+c)(b'y+c')$ for $p(0,y)q(0,y)$ which does have roots in $\mathbf R$, because $b,b'\neq0$ (since their product is the coefficient of $y^2$ in $f(x,y)$, and said coefficient is $\neq0$).

• Thank you for the reply, but I'm don't know enough yet to know what you are talking about. Hopefully soon! – jason Jan 20 '13 at 2:35

If it has anything to do with the equation of an ellipse or something similar, we can also complete the square for each variable; we have \begin{align}4(x^2+x+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2)-9y^2-1&=4\left(x+\frac{1}{2}\right)^2-9y^2-2. \end{align}

• @N.S.: Oops, you are correct. It's been edited. – Clayton Jan 20 '13 at 2:19

$$4x^2 + 4x - 9y^2 - 1 = \left( 1+2x+ \sqrt{2+9y^{2}}\right) \left( 1+2x-\sqrt{2+9y^{2}}\right),$$ courtesy of my HP 50g calculator.