Hyung-eun is solving the quadratic equation x2−4x−12=0

Question

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 3
Step 3

Step 2
Step 2

Step 4
Step 4

Step 1

Answer 1

Hyung-eun's work shows an error in Step 1.

In the original equation \(x^2 - 4x - 12 = 0\), the intention is to rewrite \(-4x\) in a way that allows for grouping. However, in step 1, she incorrectly decomposes \(-4x\) into \((-6x) + 2x\), which does not add up to \(-4x\) (it actually adds to \(-4x\), but it doesn’t help in factoring).

A correct grouping would keep \( -4x \) together or find two numbers that multiply to \(-12\) and add to \(-4\), which would be \(-6\) and \(2\) instead of breaking it into incorrect parts.

Thus, the first error occurs in Step 1.