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 4 Step 4 Step 2 Step 2 Step 3 Step 3 Step 1

Answer 1

Let's analyze Hyung-eun's work step by step:

1. Step 1: She rewrites the original equation \( x^2 - 4x - 12 = 0 \) as \( x^2 + (-6x) + 2x - 12 = 0 \). This is correct since \(-6 + 2 = -4\).

2. Step 2: She groups the terms: \( (x^2 + (-6x)) + (2x - 12) = 0 \). This grouping is also correct.

3. Step 3: She factors the grouped terms: \( x(x + (-6)) + 2(x - 6) = 0 \) simplifies to \( x(x - 6) + 2(x - 6) = 0 \). However, this step is incorrect because the second part of the terms should be \( 2(x + 6) \) not \( 2(x - 6) \). The correct expression after grouping should lead to \( x(x - 6) + 2(x + 6) = 0 \).

4. Step 4: She writes \( (x + 6)(x - 2) = 0 \). This is incorrect based on the previous error, which means her factorization was based on faulty reasoning.

5. Step 5: She sets each factor to zero and solves. Since there was an error earlier, the solutions will also be incorrect.

So, Hyung-eun makes her first error in Step 3 when she incorrectly factors the second term.

Thus, the answer is Step 3.