Factoring Polynomials with four terms and two variables

Question

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.

Answer 1

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:

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

Answer 2

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}

Answer 3

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 $\neq0$).

Answer 4

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).