Answer :
To find the zeros of the function [tex]\( F(x) = \frac{x(x+7)(x+2)}{(x-4)(x+4)} \)[/tex], we need to determine the values of [tex]\( x \)[/tex] that make [tex]\( F(x) = 0 \)[/tex].
1. Identify the numerator and the denominator:
The numerator of [tex]\( F(x) \)[/tex] is [tex]\( x(x+7)(x+2) \)[/tex].
The denominator of [tex]\( F(x) \)[/tex] is [tex]\( (x-4)(x+4) \)[/tex].
2. Find the zeros of the numerator:
The function [tex]\(\frac{P(x)}{Q(x)} = 0\)[/tex] when [tex]\(P(x) = 0\)[/tex].
For [tex]\(x(x+7)(x+2) = 0\)[/tex]:
[tex]\[ x = 0, \quad x + 7 = 0 \implies x = -7, \quad x + 2 = 0 \implies x = -2 \][/tex]
Therefore, the potential zeros of the function are [tex]\( x = 0, x = -7, x = -2 \)[/tex].
3. Ensure these zeros do not make the denominator zero:
Evaluate the denominator at the potential zeros:
[tex]\[ \text{Denominator} = (x-4)(x+4) \][/tex]
- For [tex]\( x = 0 \)[/tex], the denominator is [tex]\( (0-4)(0+4) = (-4)(4) = -16 \neq 0 \)[/tex], so [tex]\( x = 0 \)[/tex] is valid.
- For [tex]\( x = -7 \)[/tex], the denominator is [tex]\( (-7-4)(-7+4) = (-11)(-3) = 33 \neq 0 \)[/tex], so [tex]\( x = -7 \)[/tex] is valid.
- For [tex]\( x = -2 \)[/tex], the denominator is [tex]\( (-2-4)(-2+4) = (-6)(2) = -12 \neq 0 \)[/tex], so [tex]\( x = -2 \)[/tex] is valid.
4. Exclude values that make the denominator zero:
The values that make the denominator zero are not included as zeros of the whole function.
Solve for:
[tex]\[ (x-4)(x+4) = 0 \implies x-4 = 0 \implies x = 4 \text{ and } x+4 = 0 \implies x = -4 \][/tex]
We exclude [tex]\( x = -4 \)[/tex] and [tex]\( x = 4 \)[/tex] from the potential zeros.
5. List the valid zeros:
After excluding the invalid denominations, the valid zeros of the function are:
[tex]\[ x = 0, x = -7, x = -2 \][/tex]
So, checking the provided options:
- B. [tex]\( -7 \)[/tex]
- D. [tex]\( 0 \)[/tex]
- E. [tex]\( -2 \)[/tex]
These are the correct zeros of the function.
1. Identify the numerator and the denominator:
The numerator of [tex]\( F(x) \)[/tex] is [tex]\( x(x+7)(x+2) \)[/tex].
The denominator of [tex]\( F(x) \)[/tex] is [tex]\( (x-4)(x+4) \)[/tex].
2. Find the zeros of the numerator:
The function [tex]\(\frac{P(x)}{Q(x)} = 0\)[/tex] when [tex]\(P(x) = 0\)[/tex].
For [tex]\(x(x+7)(x+2) = 0\)[/tex]:
[tex]\[ x = 0, \quad x + 7 = 0 \implies x = -7, \quad x + 2 = 0 \implies x = -2 \][/tex]
Therefore, the potential zeros of the function are [tex]\( x = 0, x = -7, x = -2 \)[/tex].
3. Ensure these zeros do not make the denominator zero:
Evaluate the denominator at the potential zeros:
[tex]\[ \text{Denominator} = (x-4)(x+4) \][/tex]
- For [tex]\( x = 0 \)[/tex], the denominator is [tex]\( (0-4)(0+4) = (-4)(4) = -16 \neq 0 \)[/tex], so [tex]\( x = 0 \)[/tex] is valid.
- For [tex]\( x = -7 \)[/tex], the denominator is [tex]\( (-7-4)(-7+4) = (-11)(-3) = 33 \neq 0 \)[/tex], so [tex]\( x = -7 \)[/tex] is valid.
- For [tex]\( x = -2 \)[/tex], the denominator is [tex]\( (-2-4)(-2+4) = (-6)(2) = -12 \neq 0 \)[/tex], so [tex]\( x = -2 \)[/tex] is valid.
4. Exclude values that make the denominator zero:
The values that make the denominator zero are not included as zeros of the whole function.
Solve for:
[tex]\[ (x-4)(x+4) = 0 \implies x-4 = 0 \implies x = 4 \text{ and } x+4 = 0 \implies x = -4 \][/tex]
We exclude [tex]\( x = -4 \)[/tex] and [tex]\( x = 4 \)[/tex] from the potential zeros.
5. List the valid zeros:
After excluding the invalid denominations, the valid zeros of the function are:
[tex]\[ x = 0, x = -7, x = -2 \][/tex]
So, checking the provided options:
- B. [tex]\( -7 \)[/tex]
- D. [tex]\( 0 \)[/tex]
- E. [tex]\( -2 \)[/tex]
These are the correct zeros of the function.
