Question

question
if $f(x) = x^3 + 6x^2 - x - 30$ and $f(2) = 0$, then find all of the zeros of $f(x)$ algebraically.
answer
attempt 1 out of 2
+ add a zero - no zeros
$x = \square$ submit answer

Explanation:

Step1: Factor the polynomial using (x - 2)

Since \( f(2) = 0 \), \( (x - 2) \) is a factor. Use polynomial division or synthetic division. Let's use synthetic division for \( f(x)=x^3 + 6x^2 - x - 30 \) with root 2:

$$
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$$

So, \( f(x)=(x - 2)(x^2 + 8x + 15) \).

Step2: Factor the quadratic

Factor \( x^2 + 8x + 15 \). We need two numbers that multiply to 15 and add to 8, which are 3 and 5. So, \( x^2 + 8x + 15=(x + 3)(x + 5) \).

Step3: Find the zeros

Set each factor equal to zero:

• \( x - 2 = 0 \) gives \( x = 2 \)
• \( x + 3 = 0 \) gives \( x = -3 \)
• \( x + 5 = 0 \) gives \( x = -5 \)

Answer:

The zeros of \( f(x) \) are \( x = -5 \), \( x = -3 \), and \( x = 2 \). (You can add these one by one in the provided interface, e.g., first add \( x = -5 \), then \( x = -3 \), then \( x = 2 \))