Hyung-eun is solving the quadratic equation \(x^2-4x-12=0\) by grouping. Her work is shown

Question

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.