Tewtor.ai

Dividing by 2 is simply a way to simplify the equation before we try to factor it.

1. Why we look for a common factor

The original quadratic is

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

All three terms (the two $x$‑terms and the constant) share the factor 2:

* $2x^{2}=2\cdot x^{2}$
* $-8x=2\cdot(-4x)$
* $6=2\cdot 3$

When every term contains the same factor, we can divide the entire equation by that factor without changing the set of solutions. Algebraically:

If $a\neq0$, then

$$a\,A = a\,B \;\Longrightarrow\; A = B .$$

Here $a=2$, $A = x^{2}-4x+3$, and $B = 0$ (because $2\cdot0=0$). So

$$\frac{2x^{2}-8x+6}{2}=0 \quad\Longrightarrow\quad x^{2}-4x+3=0 .$$

2. What we gain

• Smaller numbers – working with $1$, $-4$, and $3$ is easier than with $2$, $-8$, and $6$.
• Easier factoring – we need two numbers that multiply to $+3$ and add to $-4$. Those are $-1$ and $-3$, which are obvious once the coefficients are small.
• Reduced chance of arithmetic errors – fewer large multiplications or sign mistakes.

3. What would happen if we didn’t divide?

You could still solve the quadratic, but you’d have to factor a “scaled‑up” version:

$$2x^{2}-8x+6 = 2\bigl(x^{2}-4x+3\bigr) .$$

Factoring the part in parentheses gives the same factors, and the outer 2 does not affect the roots because $2\neq0$. So the solutions are unchanged, but the intermediate steps are messier.

4. Alternative: quadratic formula (no division needed)

If you prefer not to factor, you can apply the quadratic formula directly to the original coefficients $a=2$, $b=-8$, $c=6$:

$$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} =\frac{-(-8)\pm\sqrt{(-8)^{2}-4(2)(6)}}{2(2)} =\frac{8\pm\sqrt{64-48}}{4} =\frac{8\pm\sqrt{16}}{4} =\frac{8\pm4}{4}.$$

Thus $x=\dfrac{8+4}{4}=3$ or $x=\dfrac{8-4}{4}=1$, the same answers we obtained after dividing by 2.

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Bottom line: Dividing by the greatest common factor is a convenient “pre‑processing” step that makes the numbers smaller and the factoring step straightforward, while preserving all solutions. It’s optional, but it often saves time and reduces errors.

Let me know if you’d like to see another example or practice more problems!