Tewtor.ai

When we solve a quadratic (or any algebraic equation), we want to keep the equation balanced—the left‑hand side (LHS) and right‑hand side (RHS) must stay equal after every operation.

The original equation is

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

All three terms on the left share a common factor of 2:

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

Because the same number multiplies every term, we can divide the entire equation by that number without changing the set of solutions. Doing so simplifies the coefficients, making the next steps (factoring or using the quadratic formula) easier and less error‑prone.

---

What happens if we don’t divide the whole equation?

If we divided only one term, say the constant term $6$ by 2, we would get

$$2x^{2} - 8x + 3 = 0,$$

which is not equivalent to the original equation. The left‑hand side is now smaller, so the solutions to this new equation would be different (and generally incorrect for the original problem).

---

Why dividing by the GCF is useful

1. Simplifies numbers – Smaller coefficients are easier to work with, especially when factoring.
2. Reduces fractions – If we tried to factor or apply the quadratic formula with the original coefficients, we would encounter fractions (e.g., $\frac{-b}{2a}$). Removing the GCF often eliminates those fractions.
3. Keeps the equation balanced – Dividing every term on both sides by the same non‑zero number preserves the equality, guaranteeing that any solution we find still satisfies the original equation.

---

Applying it to the problem

Dividing every term by the GCF 2:

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

Now the quadratic has leading coefficient 1, making it straightforward to factor:

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

so $x=1$ or $x=3$.

---

Bottom line: We divide by 2 (the greatest common factor) to simplify the equation while preserving all its solutions. Always apply the division (or any other operation) to every term on both sides of the equation.

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