Question

Hyung-eun is solving the quadratic equation x2−4x−12=0
x
2
−
4
x
−
12
=
0
by grouping. Her work is shown below. At which step does Hyung-eun first make an error?

x2−4x−12=0
x
2
−
4
x
−
12
=
0

Step1: x2+(−6x)+2x−12=0
x
2
+
(
−
6
x
)
+
2
x
−
12
=
0

Step 2: (x2+(−6x))+(2x−12)=0
(
x
2
+
(
−
6
x
)
)
+
(
2
x
−
12
)
=
0

Step 3: x(x+(−6))+2(x−6)=0
x
(
x
+
(
−
6
)
)
+
2
(
x
−
6
)
=
0

x(x−6)+2(x−6)=0
x
(
x
−
6
)
+
2
(
x
−
6
)
=
0

Step 4: (x+6)(x−2)=0
(
x
+
6
)
(
x
−
2
)
=
0

Step 5: x+6=0
x
+
6
=
0
or x−2=0
x
−
2
=
0

x=−6
x
=
−
6
or x=2
x
=
2

(1 point)
Responses

Step 4
Step 4

Step 3
Step 3

Step 1
Step 1

Step 2
Step 2
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Item
1
, Fully Attempted.
Item
2
, Fully Attempted.
Item
3
, Fully Attempted.
Item
4
, Unattempted.
Item
5
, Unattempted.

Answer 1

Let's analyze Hyung-eun's steps in solving the quadratic equation \( x^2 - 4x - 12 = 0 \) by grouping.

1. **Original equation**:
\( x^2 - 4x - 12 = 0 \)

2. **Step 1**:
She rewrites it as \( x^2 + (-6x) + 2x - 12 = 0 \).
**This step looks accurate; however, the grouping itself is incorrect. The correct grouping to factor would involve products that equal -12 and sum to -4, which would be -6 and +2.** Hydung-eun's grouping appears to be aiming to break down -4x incorrectly.

3. **Step 2**:
Grouping becomes: \( (x^2 + (-6x)) + (2x - 12) = 0 \).
**This step is also correct as per her grouping. But again, the phrasing for choosing -6x with 2x for grouping doesn't lead to the correct factorization.**

4. **Step 3**:
She factors out the common terms: \( x(x + (-6)) + 2(x - 6) = 0 \).
**This is where the error starts to manifest because of how she factored the previous terms. The correct factorization from the step before would lead to \( (x - 6) + 2(x + 6) \). Since she factored incorrectly, the next steps will be affected.**

5. **Step 4**:
She writes \( (x + 6)(x - 2) = 0 \).
**This is incorrect as a result of the incorrect factoring in the previous step. She should've had a different pairing leading to correct factors.**

6. **Step 5**:
Solving gives \( x + 6 = 0 \) or \( x - 2 = 0 \), leading to \( x = -6 \) or \( x = 2 \), which is not even the correct pairs and terms originating from the original equation.

### Conclusion
The first significant error occurs in **Step 1**, where the quadratic is incorrectly grouped. The correct grouping of \( -4x \) should capitalize on the sum-product relationship leading to the right pairs for factorization.

So, the answer is **Step 1** as the first place where an error is made.