Certainly! Let's solve the problem step-by-step to determine how many zeroes the polynomial [tex]\((x-3)^2 - 4\)[/tex] has and find what those zeroes are.
### Step 1: Write the Polynomial in Standard Form
First, we start with the given polynomial:
[tex]\[
(x-3)^2 - 4
\][/tex]
### Step 2: Simplify the Polynomial
Expand the expression:
[tex]\[
(x-3)^2 - 4 = (x-3)(x-3) - 4
\][/tex]
[tex]\[
= x^2 - 6x + 9 - 4
\][/tex]
[tex]\[
= x^2 - 6x + 5
\][/tex]
### Step 3: Set the Polynomial Equal to Zero
To find the zeroes, we need to solve:
[tex]\[
x^2 - 6x + 5 = 0
\][/tex]
### Step 4: Solve the Quadratic Equation
We can factorize the quadratic equation:
[tex]\[
x^2 - 6x + 5 = (x-1)(x-5) = 0
\][/tex]
### Step 5: Set Each Factor to Zero and Solve for x
Solving for [tex]\(x\)[/tex] from each factor:
[tex]\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\][/tex]
[tex]\[
x - 5 = 0 \quad \Rightarrow \quad x = 5
\][/tex]
### Step 6: Conclusion
Thus, the polynomial [tex]\((x-3)^2 - 4\)[/tex] has 2 zeroes, and they are:
[tex]\[
x = 1 \quad \text{and} \quad x = 5
\][/tex]
Therefore, the polynomial [tex]\((x-3)^2 - 4\)[/tex] has two zeroes, which are [tex]\(1\)[/tex] and [tex]\(5\)[/tex].
