Question

hyung-eun is solving the quadratic equation $x^2 - 4x - 12 = 0$ by grouping. her work is shown below. at which step does hyung-eun first make an error? $x^2 - 4x - 12 = 0$ $x(x - 6) + 2(x - 6) = 0$ step1: $x^2 + (-6x) + 2x - 12 = 0$ step 2: $(x^2 + (-6x)) + (2x - 12) = 0$ step 3: $x(x + (-6)) + 2(x - 6) = 0$ step 4: $(x + 6)(x - 2) = 0$ step 5: $x + 6 = 0$ or $x - 2 = 0$ $x = -6$ or $x = 2$ (1 point) step 2 step 4 step 1 step 3

Explanation:

Step1: Recall factoring by grouping

To factor \(x^2 - 4x - 12 = 0\) by grouping, we need two numbers that multiply to \(-12\) and add to \(-4\). The numbers are \(-6\) and \(2\) (since \(-6\times2=-12\) and \(-6 + 2=-4\)). So we rewrite the middle term: \(x^2-6x + 2x-12 = 0\) (this is Step 1, which is correct).

Step2: Group the terms

\((x^2-6x)+(2x - 12)=0\) (Step 2 is correct as it groups the first two and last two terms).

Step3: Factor out GCF from each group

From the first group \(x^2-6x\), we factor out \(x\) to get \(x(x - 6)\). From the second group \(2x-12\), we factor out \(2\) to get \(2(x - 6)\). So Step 3 should be \(x(x - 6)+2(x - 6)=0\), but in Hyung - eun's Step 3, she has \(x(x+(-6))+2(x - 6)=0\) which is equivalent to the correct factoring of the first group, but when we go to Step 4, she tries to factor \((x + 6)(x - 2)=0\). Let's check the factoring of \(x(x - 6)+2(x - 6)\). We can factor out \((x - 6)\) to get \((x - 6)(x + 2)=0\), not \((x + 6)(x - 2)=0\). So Step 4 is where the first error occurs because she factored the grouped terms incorrectly.

Answer:

Step 4