Question

what are the zeros of the polynomial function $x^3 - 3x^2 - x + 3$? choose the correct answer from the choices. the zeros of the function are $x = \square$, $x = \square$, and $x = \square$.

Explanation:

Step1: Factor by grouping

Given the polynomial \( f(x)=x^{3}-3x^{2}-x + 3 \). Group the terms:
\( (x^{3}-3x^{2})+(-x + 3) \)
Factor out the greatest common factor from each group:
\( x^{2}(x - 3)-1(x - 3) \)

Step2: Factor out the common binomial

Now, factor out \( (x - 3) \) from both terms:
\( (x - 3)(x^{2}-1) \)

Step3: Factor the difference of squares

Recall that \( a^{2}-b^{2}=(a - b)(a + b) \). Here, \( x^{2}-1=x^{2}-1^{2} \), so it factors to \( (x - 1)(x + 1) \).
So the polynomial factors completely as \( (x - 3)(x - 1)(x + 1) \)

Step4: Find the zeros

To find the zeros, set \( f(x)=0 \):
\( (x - 3)(x - 1)(x + 1)=0 \)
Using the zero - product property (if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \)), we set each factor equal to zero:

• \( x-3=0\Rightarrow x = 3 \)
• \( x - 1=0\Rightarrow x=1 \)
• \( x + 1=0\Rightarrow x=-1 \)

Answer:

The zeros of the function are \( x=- 1 \), \( x = 1 \), and \( x=3 \)