Question

1. what are the zeros of the function $f(x) = x^2 - 81$?

select two answer choices.
a. -9
b. -3
c. 0
d. 3
e. 9
3.
select the boxes that show which expressions are equivalent.
$(x + 5)(x + 3)$ $(x + 4)^2 - 1$ $x^2 + 8x + 15$
$x^2 + 4x - 5$
$(x - 1)^2 - 16$
$(x - 3)(x - 1)$
$x^2 + 8x + 16$
$x^2 - 4x + 3$

Explanation:

Question 2

Step1: Recall zero definition

To find zeros of \( f(x) = x^2 - 81 \), set \( f(x) = 0 \), so \( x^2 - 81 = 0 \).

Step2: Solve the equation

Rewrite as \( x^2 = 81 \). Take square roots: \( x = \pm\sqrt{81}=\pm9 \). So zeros are \( -9 \) and \( 9 \).

Step1: Expand \( (x + 5)(x + 3) \)

Using FOIL: \( x^2 + 3x + 5x + 15 = x^2 + 8x + 15 \).

Step2: Expand \( (x + 4)^2 - 1 \)

\( (x^2 + 8x + 16) - 1 = x^2 + 8x + 15 \).

Step3: Check other expressions

• \( x^2 + 4x - 5 \): Different from \( x^2 + 8x + 15 \).
• \( (x - 1)^2 - 16 = x^2 - 2x + 1 - 16 = x^2 - 2x - 15 \): Different.
• \( (x - 3)(x - 1) = x^2 - 4x + 3 \): Different.
• \( x^2 + 8x + 16 \): Different (constant term 16 vs 15).
• \( x^2 - 4x + 3 \): Different.

So equivalent expressions are \( (x + 5)(x + 3) \), \( (x + 4)^2 - 1 \), \( x^2 + 8x + 15 \).

Answer:

A. -9, E. 9

Question 3