Question

the table shows the values of function f at different values of x. de
$f(x) = x^5 + x^4 - 27x^3 - x^2 + 146x - 120$

x	f(x)
-5	0
-4	240
-3	0
-1	-240
4	0
5	960

select all the correct answers.
□ $x + 5$
□ $x - 5$
□ $x + 3$
□ $x - 3$

Explanation:

Step1: Recall Factor Theorem

The Factor Theorem states that if \( f(c) = 0 \), then \( (x - c) \) is a factor of \( f(x) \). Conversely, if \( (x + c)=(x - (-c)) \) is a factor, then \( f(-c)=0 \).

Step2: Analyze \( x = -5 \)

Given \( f(-5) = 0 \). Using the Factor Theorem, if \( c=-5 \), then the factor is \( (x - (-5))=x + 5 \). So \( x + 5 \) is a factor.

Step3: Analyze \( x=-3 \)

Given \( f(-3)=0 \). Using the Factor Theorem, if \( c = - 3 \), then the factor is \( (x-(-3))=x + 3 \). So \( x + 3 \) is a factor.

Step4: Analyze \( x = 4 \)

Given \( f(4)=0 \). Using the Factor Theorem, if \( c = 4 \), then the factor is \( (x - 4) \). But we are checking the given options. For \( x = 5 \), \( f(5)=960
eq0 \), so \( x - 5 \) is not a factor. For \( x=3 \), we don't have \( f(3) = 0 \) from the table, so \( x - 3 \) is not a factor.

Answer:

A. \( x + 5 \), C. \( x + 3 \)