Hyung-eun is solving the quadratic equation x2−4x−12=0 by grouping. Her work is shown below. At which step does Hyung-eun first make an error?%0D%0A%0D%0Ax2−4x−12=0 %0D%0A%0D%0AStep1: x2+(−6x)+2x−12=0 %0D%0A%0D%0AStep 2: (x2+(−6x))+(2x−12)=0 %0D%0A%0D%0AStep 3: x(x+(−6))+2(x−6)=0 %0D%0A%0D%0Ax(x−6)+2(x−6)=0 %0D%0A%0D%0AStep 4: (x+6)(x−2)=0 %0D%0A%0D%0AStep 5: x+6=0 or x−2=0 %0D%0A%0D%0Ax=−6 or x=2 %0D%0A%0D%0A(1 point)%0D%0AResponses%0D%0A%0D%0AStep 3%0D%0AStep 3%0D%0A%0D%0AStep 4%0D%0AStep 4%0D%0A%0D%0AStep 2%0D%0AStep 2%0D%0A%0D%0AStep 1

Let's analyze Hyung-eun's work step by step to identify any errors:

**Step 1:** \( x^2 + (-6x) + 2x - 12 = 0 \)

This step involves rewriting \(-4x\) in the quadratic equation \(x^2 - 4x - 12 = 0\) as \((-6x) + 2x\). This is correct, as \(-6x + 2x = -4x\).

**Step 2:** \( (x^2 + (-6x)) + (2x - 12) = 0 \)

This step groups the terms into two parts. This is also correct.

**Step 3:** \( x(x + (-6)) + 2(x - 6) = 0 \)

At this step, Hyung-eun is factoring the two groups. Let's break it down:

- From the first part \(x^2 + (-6x)\), she factors out an \(x\) to get \(x(x - 6)\), which should be \(x(x - 6)\) (not \(x + (-6)\)).
- The second part \(2x - 12\) factors out as \(2(x - 6)\), which is correct.

However, it should have been \(x(x - 6) + 2(x - 6) = 0\) leading to step 4.

So the error occurs in Step 3 when she incorrectly writes \(x(x + (-6))\) instead of \(x(x - 6)\).

Thus, the first error occurs in **Step 3**.