Question

select the box or boxes that correctly identify the zeros of each polynomial.
$x^2 - 16$ -4 -3 0 3 4
$x^2 - 9$ -4 -3 0 3 4
$x^2 + 7x + 12$ -4 -3 0 3 4
7.
select the box or boxes that correctly identify the zeros of each polynomial.
$x^2 - 25$ -5 -1 1 4 5
$x^2 - 3x - 4$ -5 -1 1 4 5
$x^2 + 6x + 5$ -5 -1 1 4 5
8.

1. given the polynomial $x^2 - 6x + 9$, which option correctly identifies a zero and a sketch of the graph of the function defined by the polynomial?

a. $x = -3$; a downward - opening parabola with its vertex at $x = -3$.
b. $x = 3$; a downward - opening parabola with its vertex at $x = 3$.
c. $x = -3$; an upward - opening parabola with its vertex at $x = -3$.
d. $x = -3$; an upward - opening parabola with its vertex at $x = 3$.

1. given the polynomial $x^2 - 4$, which option correctly identifies a zero and a sketch of the graph of the function defined by the polynomial?

Answer:

Problem 7 (Identifying Zeros of Polynomials)

For \( x^2 - 16 \):

Step 1: Factor the polynomial

We can factor \( x^2 - 16 \) as a difference of squares: \( x^2 - 16=(x - 4)(x + 4) \)

Step 2: Find the zeros

Set each factor equal to zero:

• \( x - 4 = 0\Rightarrow x = 4 \)
• \( x + 4 = 0\Rightarrow x=-4 \)

So the zeros of \( x^2 - 16 \) are \( x = 4 \) and \( x=-4 \)

For \( x^2 - 9 \):

Step 1: Factor the polynomial

Factor \( x^2 - 9 \) as a difference of squares: \( x^2 - 9=(x - 3)(x + 3) \)

Step 2: Find the zeros

Set each factor equal to zero:

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

So the zeros of \( x^2 - 9 \) are \( x = 3 \) and \( x=-3 \)

For \( x^2+7x + 12 \):

Step 1: Factor the polynomial

We need two numbers that multiply to \( 12 \) and add up to \( 7 \). The numbers are \( 3 \) and \( 4 \). So \( x^2+7x + 12=(x + 3)(x + 4) \)

Step 2: Find the zeros

Set each factor equal to zero:

• \( x+3 = 0\Rightarrow x=-3 \)
• \( x + 4=0\Rightarrow x=-4 \)

So the zeros of \( x^2+7x + 12 \) are \( x=-3 \) and \( x=-4 \)

Problem 8 (Identifying Zeros of Polynomials)

For \( x^2 - 25 \):

Step 1: Factor the polynomial

Factor \( x^2 - 25 \) as a difference of squares: \( x^2 - 25=(x - 5)(x + 5) \)

Step 2: Find the zeros

Set each factor equal to zero:

• \( x - 5=0\Rightarrow x = 5 \)
• \( x + 5=0\Rightarrow x=-5 \)

So the zeros of \( x^2 - 25 \) are \( x = 5 \) and \( x=-5 \)

For \( x^2-3x - 4 \):

Step 1: Factor the polynomial

We need two numbers that multiply to \( - 4 \) and add up to \( -3 \). The numbers are \( -4 \) and \( 1 \). So \( x^2-3x - 4=(x - 4)(x+1) \)

Step 2: Find the zeros

Set each factor equal to zero:

• \( x - 4=0\Rightarrow x = 4 \)
• \( x + 1=0\Rightarrow x=-1 \)

So the zeros of \( x^2-3x - 4 \) are \( x = 4 \) and \( x=-1 \)

For \( x^2+6x + 5 \):

Step 1: Factor the polynomial

We need two numbers that multiply to \( 5 \) and add up to \( 6 \). The numbers are \( 1 \) and \( 5 \). So \( x^2+6x + 5=(x + 1)(x + 5) \)

Step 2: Find the zeros

Set each factor equal to zero:

• \( x + 1=0\Rightarrow x=-1 \)
• \( x + 5=0\Rightarrow x=-5 \)

So the zeros of \( x^2+6x + 5 \) are \( x=-1 \) and \( x=-5 \)

Problem 9 (Zero and Graph of \( x^2-6x + 9 \))

Step 1: Factor the polynomial

\( x^2-6x + 9=(x - 3)^2 \)

Step 2: Find the zero

Set \( (x - 3)^2=0\Rightarrow x = 3 \)

Step 3: Analyze the graph

For a quadratic function \( y=ax^2+bx + c \), if \( a>0 \), the parabola opens upward. Here \( a = 1>0 \), so the parabola opens upward. The vertex of a quadratic function in the form \( y=(x - h)^2+k \) is at \( (h,k) \). For \( y=(x - 3)^2 \), the vertex is at \( x = 3 \)
So the correct option is B (since there is a typo in option D, assuming it's \( x = 3 \) instead of \( x=-3 \) in the first part, the correct option should be: B. \( x = 3 \); a downward - opening parabola with its vertex at \( x = 3 \) is wrong, the correct is upward - opening, but among the given options, if we correct the opening direction understanding, the zero is \( x = 3 \) and vertex at \( x = 3 \) with upward opening, but the closest is B with a mistake in opening direction or maybe a typo. If we consider the zero first, the zero is \( x = 3 \), and the parabola \( y=x^2-6x + 9=(x - 3)^2 \) opens upward, but the options have B with downward (maybe a typo) but the zero is \( x = 3 \) and vertex at \( x = 3 \))

Problem 10 (Zero and Graph of \( x^2 - 4 \))

Step 1: Factor the polynomial

\( x^2 - 4=(x - 2)(x + 2) \)

Step 2: Find the zeros

Set each factor equal to zero: \( x - 2=0\Rightarrow x = 2 \) and \( x + 2=0\Rightarrow x=-2 \)

Step 3: Analyze the graph

For \( y=x^2 - 4 \), \( a = 1>0 \), so the parabola opens upward. The vertex of \( y=x^2-4 \) is at \( (0,-4) \)

(Note: Since the options for problem 10 are not given, we can't choose, but we have found the zeros and the nature of the graph)

Final Answers (for the zero - identification parts)

• For \( x^2 - 16 \): Zeros at \( x=-4,4 \)
• For \( x^2 - 9 \): Zeros at \( x=-3,3 \)
• For \( x^2+7x + 12 \): Zeros at \( x=-4,-3 \)
• For \( x^2 - 25 \): Zeros at \( x=-5,5 \)
• For \( x^2-3x - 4 \): Zeros at \( x=-1,4 \)
• For \( x^2+6x + 5 \): Zeros at \( x=-5,-1 \)
• For problem 9: The zero is \( x = 3 \), the parabola \( y=x^2-6x + 9 \) opens upward with vertex at \( x = 3 \)
• For problem 10: Zeros are \( x=-2,2 \), the parabola \( y=x^2 - 4 \) opens upward with vertex at \( (0,-4) \)