Can you factor quadratic equations where two variables are not divisible?

Question

The question is whether $9x^2+16y^2z^4$ can be factored. Many algebra calculators online are saying that they can't be factored whilst I think that we can factor it using the difference of two squares to solve it, can it be factored?

Thank you

Answer 1

You cannot use a difference of squares (at least with real numbers) thus it is irreducible over $\mathbb{R}$. Differences of squares come in the form $a^2 - b^2$, but $9x^2+16y^2z^4$ does not fit this form since it has a plus sign, not a minus sign.

But if you allow for imaginary numbers, you can do the following:

$$9x^2+16y^2z^4 = 9x^2-(-16y^2z^4)=(3x)^2-(4iyz^2)^2$$

Which gives you the following factors:

$$(3x-4iyz^2)(3x+4iyz^2)$$

Thus it is not irreducible over $\mathbb{C}$. This gives you the two solutions hinted by the Fundamental Theorem of Algebra given by the equation $9x^2+16y^2z^4 = 0$.