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Gemma completed the following steps to find the zeros of the function [tex]$f(x) = 3x^2 - 6x - 45$[/tex].

Step 1: [tex]f(x) = 3(x^2 - 2x - 15)[/tex]
Step 2: [tex]f(x) = 3(x - 5)(x + 3)[/tex]
Step 3: [tex]x - 5 = 0 \quad ; \quad x + 3 = 0[/tex]
Step 4: [tex]x = 5 \quad ; \quad x = 3[/tex]

What mistakes, if any, did Gemma make?

A. 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].
B. In Step 1, Gemma should have factored out [tex]3x[/tex] instead of just factoring out the 3.
C. In Step 2, when factoring, Gemma should have gotten the answer [tex]f(x) = 3(x + 5)(x - 3)[/tex].
D. Gemma did not make any mistakes.



Answer :

GinnyAnswer
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].

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