Question

Hyung-eun is solving the quadratic equation x^2-4x-12x=0 by grouping. Her work is shown below. At which step does she make her first error
Step1 x^2 + (-6x)+2x -12=0
Step2 (x^2+(-6x)) +(2x-12)=0
Step 3 x(x+(-6))+2(x -6)=0
x(x-6)+2(x-6)=0
Step4 (x+6)(x-2)=0
Step5 x+6=0 or x-2=0
x= -6 or x=2

Answer 1

To find the first error in Hyung-eun's work, let's go through her steps one by one, starting from the original equation she is solving:

The original equation is:
\[ x^2 - 4x - 12 = 0 \]

Let's analyze her steps:

**Step 1:**
\[ x^2 + (-6x) + 2x - 12 = 0 \]

Here we see that she rewrote \(-4x\) as \(-6x + 2x\), which is a correct grouping because \(-6 + 2 = -4\). So this step is valid.

**Step 2:**
\[ (x^2 + (-6x)) + (2x - 12) = 0 \]

This is also correct; she grouped the terms appropriately.

**Step 3:**
\[ x(x + (-6)) + 2(x - 6) = 0 \]

Here we have something to examine closely. The first part, \(x(x + (-6))\), should simply be \(x(x - 6)\). However, she has misapplied the parentheses or signs leading to the expression appearing wrong. Furthermore, \(2(x - 6)\) simplifies to \(2x - 12\) correctly, but the expression should have been factored differently to maintain consistency.

**Correctly done, it should be:**
\[ x(x - 6) + 2(x - 6) = 0 \]

Then both parts could have been factored out to yield:
\[ (x - 6)(x + 2) = 0 \]

**Step 4:**
Then she goes on to factor:
\[ (x + 6)(x - 2) = 0 \]

This factored form is incorrectly arrived at due to the mistakes made earlier in Step 3.

**Step 5:**
\[ x + 6 = 0 \quad \text{or} \quad x - 2 = 0 \]

This resulting solution is also incorrect due to the improper factorization.

### Conclusion:
The first error occurs in **Step 3** when she improperly factored the expression. The correct grouping and factorization should lead to a solution that reflects the original equation's roots accurately.

The correct factorization should reveal roots of 6 and -2, contrasting with her proposed roots of -6 and 2.