Tewtor.ai

When we have

every term contains the factor 2:

* $2x^{2}=2\cdot x^{2}$
* $-8x=2\cdot(-4x)$
* $6=2\cdot3$

Because 2 is a common factor of all three terms, we can factor it out (or divide the entire equation by 2). This does two useful things:

1. Simplifies the numbers – the coefficients become smaller, so the arithmetic in the next steps is easier and less error‑prone.

2. Leaves the equation equivalent – dividing every term by the same non‑zero number does not change the set of solutions. If

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

   then the only way the product can be zero is for the bracketed factor to be zero (the factor 2 can never be zero). So solving

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

   gives exactly the same solutions as the original equation.

If we divided only one term (for example, turning $2x^{2}-8x+6$ into $x^{2}-8x+6$), we would be changing the equation’s balance and would obtain incorrect solutions. The GCF must be applied to all terms, or we can think of it as factoring the 2 out of the whole expression.

After factoring out the 2 we get a simpler quadratic that can be factored or solved with the quadratic formula, leading to the solutions $x=1$ and $x=3$.