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Dividing by 2 (the greatest common factor) is not required to solve the equation, but it makes the algebra much easier and reduces the chance of mistakes. Here’s why:

1. Simplifies the coefficients
   The original quadratic is

   $$2x^{2} - 8x + 6 = 0 .$$

   All three terms share a factor of 2. If we factor that 2 out of the left‑hand side we get

   $$2\bigl(x^{2} - 4x + 3\bigr) = 0 .$$

   Because the factor 2 is never zero, the equation is equivalent to

   $$x^{2} - 4x + 3 = 0 .$$

   Now the leading coefficient is 1, so the constant term and the linear coefficient are small integers (3 and –4). That makes the next step—finding two numbers that multiply to 3 and add to –4—straightforward.

2. Reduces arithmetic errors
   If we tried to factor the original form directly, we would need two numbers whose product is $2 \times 6 = 12$ (the product of the leading coefficient and the constant) and whose sum is $-8$. The possible pairs (±1, ±12), (±2, ±6), (±3, ±4) are more numerous, and it’s easy to pick the wrong sign or pair. After dividing by 2, the product we need is just +3, and the only integer pair that works is $-1$ and $-3$.

3. Keeps the equation equivalent
   Dividing every term (or factoring out the GCF) does not change the set of solutions because we are multiplying the entire left side by a non‑zero constant. The equation

   $$2x^{2} - 8x + 6 = 0$$

   has exactly the same solutions as

   $$x^{2} - 4x + 3 = 0 .$$

   (If we divided only one term, the equation would no longer be equivalent and we would get the wrong answer—that’s a common mistake.)

4. Makes the zero‑product step clearer
   After the simplification we can factor directly:

   $$x^{2} - 4x + 3 = (x-1)(x-3) = 0 ,$$

   leading immediately to $x = 1$ or $x = 3$. If we kept the factor 2 outside, we would have

   $$2(x-1)(x-3) = 0 ,$$

   and we would still set each factor $(x-1)$ and $(x-3)$ to zero, but the extra “2” is unnecessary work.

Bottom line:
Dividing by the GCF (or factoring it out) is a convenient algebraic shortcut that leaves the same solutions while giving you smaller, easier‑to‑handle numbers. Just remember to apply the division to every term; otherwise the equation changes and you’ll get incorrect results.