Question

Hyung-eun is solving the quadratic equation \(x^2-4x-12=0\) by grouping. Her work is shown
below. At which step does Hyung-eun first make an error?
\(x^2-4x-12=0\)
Step1: \(x^2+\left(-6x\right)+2x-12=0\)
Step 2: \(\left(x^2+\left(-6x\right)\right)+\left(2x-12\right)=0\)
Step 3: \(x\left(x+\left(-6\right)\right)+2\left(x-6\right)=0\)
\(x\left(x-6\right)+2\left(x-6\right)=0\)
Step 4: \(\left(x+6\right)\left(x-2\right)=0\)
Step 5: \(x+6=0\) or \(x-2=0\)
\(x=-6\) or \(x=2\)
(1 point)
Step 4
Step 2
Step 3
Step 1

Answer 1

To determine at which step Hyung-eun first makes an error, let's analyze her steps closely.

The original quadratic equation is:

\[
x^2 - 4x - 12 = 0
\]

### Step 1:

Hyung-eun rewrites the equation as:

\[
x^2 + (-6x) + 2x - 12 = 0
\]

This is incorrect because \(-6 + 2\) does not equal \(-4\). The correct representation would be to split \(-4x\) correctly as \(-6x + 2x\), which is valid, but summing gives \(-4x\) as she intended. So this step is consistent with her goal.

### Step 2:

She now rewrites the equation as:

\[
\left(x^2 + (-6x)\right) + \left(2x-12\right) = 0
\]

This grouping is also valid.

### Step 3:

Next, she factors:

\[
x\left(x - 6\right) + 2\left(x - 6\right) = 0
\]

This step is also valid as she correctly factors out \(x\) from the first group and \(2\) from the second group.

### Step 4:

She combines the terms from Step 3:

\[
\left(x - 6\right)\left(x + 2\right)=0
\]

This is her first error. The correct factorization should be:

\[
\left(x - 6\right)\left(x + 2\right)
\]

Sophisticatedly, both factors should yield \(x^2 - 4x - 12\) when expanded. However, she instead mistakenly combined the groups, impacting the resulting factors.

### Conclusion

The first step at which she makes an error is **Step 4**, where she incorrectly performs the factorization process.

Thus, the answer is:

**Step 4**.