Question

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?

x2−4x−12=0


Step1: x2+(−6x)+2x−12=0


Step 2: (x2+(−6x))+(2x−12)=0


Step 3: x(x+(−6))+2(x−6)=0


x(x−6)+2(x−6)=0


Step 4: (x+6)(x−2)=0


Step 5: x+6=0
or x−2=0


x=−6
or x=2


(1 point)
Responses

Step 1
Step 1

Step 2
Step 2

Step 3
Step 3

Step 4
Step 4

Answer 1

To analyze Hyung-eun's steps, we need to verify the correctness of each step starting from the original equation \(x^2 - 4x - 12 = 0\).

1. **Step 1**: \(x^2 + (-6x) + 2x - 12 = 0\)
- This step incorrectly decomposes \(-4x\) into \(-6x + 2x\). The decomposition of \(-4x\) into two numbers should sum to \(-4\) and multiply to \(-12\). The correct pair is \(-6\) and \(2\), so this step is acceptable.

2. **Step 2**: \((x^2 + (-6x)) + (2x - 12) = 0\)
- In this step, she just grouped the terms, which is correct as well.

3. **Step 3**: \(x(x + (-6)) + 2(x - 6) = 0\)
- Here, Hyung-eun attempts to factor the grouped expressions. The factorization of the first group is correct, but the second group should factor to \(2(x - 6)\), which is inconsistent with her claim that it factors to \(2(x - 6)\) since \(2x - 12\) actually factors to \(2(x - 6)\). This looks correct, but let's carefully analyze the expression: it should be \(x(x - 6) + 2(x + 6) = 0\) instead.

4. **Step 4**: \((x + 6)(x - 2) = 0\)
- By applying the grouping incorrectly, the expression shouldn't yield \((x + 6)(x - 2)\).

So, the first error occurs in **Step 3**, as Hyung-eun inaccurately progressed from the correct form of the factored expression from the previous step.

The final response is:
**Step 3**