Factor the expression a^2 - 4x^2 + 12x - 9.

Step-by-step solution

1. 1

   Count the variables before grouping. There are four terms but only one contains $a$; the other three are all in $x$. That asymmetry is the hint: group the three $x$-terms together and see whether they form something recognisable.

2. 2

   Group with a leading minus sign. Factor $-1$ out of the last three terms — every sign inside flips:

   $a^2-4x^2+12x-9=a^2-\left(4x^2-12x+9\right)$

   Forgetting to flip the sign of $+12x$ and $-9$ here is the usual mistake.

3. 3

   Recognise the perfect square. For $4x^2-12x+9$, check $(2x)^2=4x^2$, $3^2=9$, and $2\cdot 2x\cdot 3=12x$, which matches the middle term, so

   $4x^2-12x+9=(2x-3)^2$

   The expression is now $a^2-(2x-3)^2$.

4. 4

   Apply the difference of squares. With $A=a$ and $B=2x-3$, the identity $A^2-B^2=(A-B)(A+B)$ gives

   $a^2-(2x-3)^2=\bigl(a-(2x-3)\bigr)\bigl(a+(2x-3)\bigr)$

5. 5

   Clear the inner brackets. Distributing the signs:

   $(a-2x+3)(a+2x-3)$

   Neither factor can be broken down further over the integers, so the factorization is complete.

6. 6

   Expand to verify. $(a-2x+3)(a+2x-3)=a^2+a(2x-3)-a(2x-3)-(2x-3)^2=a^2-(4x^2-12x+9)=a^2-4x^2+12x-9$. $\checkmark$