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 2
Step 2

Step 3
Step 3

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



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Answer 1

To identify where Hyung-eun first makes an error, let's examine each step in detail:

1. **Step 1**: The equation is rewritten as \( x^2 + (-6x) + 2x - 12 = 0 \).
- This step is incorrect because combining -6x and 2x should result in -4x, not leaving -6x as a separate term.
- The correct decomposition should be \( x^2 - 6x + 2x - 12 = 0 \), which is just the original equation but doesn't provide a correct grouping.

2. **Step 2**: The grouping shows \( (x^2 + (-6x)) + (2x - 12) = 0 \).
- This step is based on the incorrect expression from Step 1, so the problems continue.

3. **Step 3**: Then Hyung-eun factored out \( x \) and \( 2 \), leading to \( x(x + (-6)) + 2(x - 6) = 0 \).
- Given the errors from prior steps it further propagates.

4. **Step 4**: Attempts to factor as \( (x + 6)(x - 2) = 0 \).
- This step is incorrect since the factors were derived from an incorrect earlier step.

Based on this, Hyung-eun's first mistake occurs in **Step 1**, where the expression is not correctly grouped or simplified.

Therefore, the answer is:

**Step 1**