Completely factoring the expression of 2 Numbers only?

Question

This may sound like stupid question, I have already searched the forums and found nothing but, I have this problem: $$12x^2 - 48$$ I have already factored it to $12(x^2 - 4)$ but that isn't the end of the problem. I have been looking at the examples in the book and the example they did is: $$3 - 12x^2=3(1-4x^2)=3(1^2-(2x)^2)=3(1+2x)(1-2x)$$ How from $1-4x^2$ to $1^2-(2x)^2$?

Answer 1

Since $2x\times 2x = 4x^2$ and $2x \times 2x = (2x)\times (2x) =(2x)^2$ because adding the brackets here changes nothing.

So $1-4x^2=1-(2x)^2$. With problems like this you should looking for the difference of two squares hidden within.

For example, to factorise the general quadratic polynomial you can proceed as follows:

$$p(x)=ax^2+bx+c$$ multiply by $4a$ $$4ap(x)=4a^2x^2+4abx+4ac=(2ax+b)^2-(b^2-4ac)$$ and treat as the difference of two squares to recover the usual quadratic formula.

Answer 2

To get $1^2-(2x)^2$ from $1-4x^2$, notice that $1^2=1$ and $(2x)^2=4x^2$.

The difference of two squares, $a^2-b^2$, is the product of $a-b$ and $a+b$. For your question, let $a=x$ and $b=2$.