Tewtor.ai

When we solve a quadratic (or any algebraic equation), the principle of equality tells us that whatever operation we perform must be applied to every term on both sides of the equation.

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1. Why look for a common factor?

The original equation is

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

All three coefficients (2, –8, 6) share the factor 2. Factoring out that greatest common factor (GCF) simplifies the numbers we have to work with, making the later steps—especially factoring the quadratic—much easier.

If we keep the 2 in front, we would have to find two numbers that multiply to

$$\frac{c}{a}= \frac{6}{2}=3$$

and add to

$$\frac{b}{a}= \frac{-8}{2}= -4 .$$

That works, but the arithmetic is a little messier because the leading coefficient $a$ is not 1. By dividing the whole equation by 2, we turn the quadratic into a monic form (leading coefficient = 1), so we only need numbers that multiply to 3 and add to –4—straightforward integers.

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2. Why must we divide the entire equation?

Dividing only one term (e.g., turning $2x^{2}$ into $x^{2}$ but leaving $-8x$ and $+6$ unchanged) would change the balance of the equation. Algebraically:

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

If we “divide the first term by 2” we would get

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

which is not equivalent to the original equation; the set of solutions would be different (in fact, that new equation has no real roots).

To keep the equality true, we must perform the same operation on every term on the left‑hand side (and on the right‑hand side, which is 0). Since 0 divided by 2 is still 0, the right side stays 0.

So we write

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

Now the equation is exactly equivalent to the original one; it has the same solution set.

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3. What if we didn’t divide?

You could solve the original equation without factoring out the 2, using the quadratic formula:

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

which also gives $x = 1$ or $x = 3$.
Dividing by the GCF simply makes the arithmetic easier and lets us factor by inspection instead of using the formula.

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4. Quick checklist when you see a common factor

1. Identify the greatest common factor of all coefficients.
2. Divide every term (including the constant on the right‑hand side) by that factor.
3. Proceed with factoring, completing the square, or applying the quadratic formula on the simplified equation.

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Bottom line: We divided by 2 because it was a common factor of all terms, and we applied the division to the whole equation to keep the equality valid. This reduces the numbers we work with and makes the factoring step straightforward.

Let me know if you’d like to see another example or try a similar problem on your own!