Let's go through Gemma's work step-by-step to verify where she might have made a mistake:
1. The original function is [tex]\( f(x) = 3x^2 - 6x - 45 \)[/tex].
2. Step 1: Gemma factors out 3 from the quadratic equation:
[tex]\[
f(x) = 3(x^2 - 2x - 15)
\][/tex]
This step is correct.
3. Step 2: Gemma factors the quadratic expression inside the parentheses:
[tex]\[
f(x) = 3(x - 5)(x + 3)
\][/tex]
This is also correct because [tex]\((x - 5)(x + 3)\)[/tex] expands back to [tex]\(x^2 - 2x - 15\)[/tex].
4. Step 3: Setting each factor to zero to find the zeros of the function:
[tex]\[
x - 5 = 0 \quad \text{and} \quad x + 3 = 0
\][/tex]
5. Step 4: Solving the equation gives:
[tex]\[
x = 5 \quad \text{and} \quad x = 3
\][/tex]
Now let's identify the mistake:
- Solving [tex]\( x - 5 = 0 \)[/tex] correctly gives [tex]\( x = 5 \)[/tex].
- However, solving [tex]\( x + 3 = 0 \)[/tex] should result in:
[tex]\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\][/tex]
So, Gemma's mistake is in Step 4, where she solved for [tex]\( x + 3 = 0 \)[/tex] incorrectly. The correct steps to solve [tex]\( x + 3 = 0 \)[/tex] should be to subtract 3 from both sides, leading to [tex]\( x = -3 \)[/tex].
The correct zeros of the function are [tex]\( x = 5 \)[/tex] and [tex]\( x = -3 \)[/tex]. Therefore, the mistake is:
In Step 4, when solving for [tex]\( x + 3 = 0 \)[/tex], Gemma should have subtracted 3 from both sides of the equation to get [tex]\( x = -3 \)[/tex].
