Factorizing Equations

Question

I am solving some questions given as assignment and I am having trouble understanding why the method I am using is not working.

The question is Factorize the following: $$ a^2+4b^2\label{eq:i}\tag{i}$$ $$9a^2+16b^\label{eq:ii}\tag{ii}2$$ $$3x^2+3y^2\label{eq:iii}\tag{iii}$$

I solved the first question using the method,

$$a^2+4b^2=0$$ $$a^2=-4b^2$$ $$a=\sqrt{-4b^2}$$ $$a=\pm2bi$$ $$a^2+4b^2=(a+2bi)(a-2bi)$$

But the same method doesn't works for the remaining two questions.

For $9a^2+16b^2$ I get $$9a^2+16b^2=0$$ $$9a^2=-16b^2$$ $$a^2=\frac{-16b^2}{9}$$ $$a=\pm\frac{4bi}{3}$$ $$(a+\frac{4b^2}{3})(a-\frac{4b^2}{3})$$ where the correct answer is $$(3a+4bi)(3a-4bi)$$

For $3x^2+3y^2$ using the same method, I get $$(x+iy)(x-iy)$$ where the correct answer is $$3(x+yi)(x-yi)$$.

I noticed that this method works if when I don't simplify the L.H.S of $\eqref{eq:ii}$ I get the correct answer $$9a^2+16b^2=0$$ $$9a^2=-16b^2$$ $$3a=\pm4bi$$ $$(3a+4bi)(3a-4bi)$$

My questions are:

1. Is this the correct method to factorize these type of equations?
2. If not then why does this method works for specific cases? Was it just a coincidence? Or is there more to it?
3. Also why did it work for $\eqref{eq:ii}$ when I didn't simplify the L.H.S of the equation?

Answer 1

I don't fully understand what you are doing, but here's how I would do it.

First, notice you are factorizing equations, so you are free to multiply or divide your expression by any constant. This means that if I take the equation

$$A a^2 + B b^2 = 0 \; ,$$

then the solution will be the same as the one for the equation

$$a^2 + \frac{B}{A} b^2 = 0 \; .$$

Next, note that

$$ (a+ x b) (a- x b) = a^2- x^2b^2 \; .$$

This can be seen the easiest by labeling $x b \rightarrow b'$ or any other variable.

Therefore the factorization will be

$$A a^2 + B b^2 = 0 \; \rightarrow \left(a+\sqrt{-\frac{B}{A}} b\right)\left(a-\sqrt{-\frac{B}{A}} b\right) = 0$$

So for the two examples you would get

$$a^2+4b^2 =0 \rightarrow (a-2ib)(a+2ib)=0 \\ 9a^2+16b^2 = 0 \rightarrow \left(a-\frac{4}{3}ib\right)\left(a+\frac{4}{3}ib\right)=0 \\ 3x^2+3y^2 =0 \rightarrow (x-iy)(x+iy)=0 $$

Again, the only difference would be how you are handling the division/multiplication with a constant, which you are free to do if you are solving equations where your polynomial is equal to zero. You might alternatively try the approach

$$ A x^2 + B y^2 = (\sqrt{A} x + \sqrt{B} i y)(\sqrt{A} x - \sqrt{B} i y) \; ,$$

which will give the prefactors that I think you expect to get, but then again, for equations, it's the same thing.

Answer 2

You should first factorize common factors if there is anything, e.g.: $3x^2+3y^2=3(x^2+y^2)$. Now, all the questions are sums of squares, for example $9a^2+16b^2=x^2+y^2$. This is factored as $(x+iy)(x-iy)$ for example, if you are allowed to factor using complex numbers.