Answer :
- Given the zero $3i$ of the polynomial $f(x)=x^3-5x^2+9x-45$, the conjugate $-3i$ is also a zero.
- The quadratic factor corresponding to these zeros is $(x-3i)(x+3i) = x^2+9$.
- Dividing $f(x)$ by $x^2+9$ gives the remaining factor $x-5$.
- The remaining zero is $x=5$, so the zeros are $\boxed{3i, -3i, 5}$.
### Explanation
1. Finding the Conjugate and Quadratic Factor
We are given the polynomial $f(x)=x^3-5 x^2+9 x-45$ and one of its zeros, $3i$. Since the polynomial has real coefficients, the complex conjugate of $3i$, which is $-3i$, must also be a zero. Thus, $(x-3i)$ and $(x+3i)$ are factors of $f(x)$. Therefore, $(x-3i)(x+3i) = x^2 - (3i)^2 = x^2 - (-9) = x^2 + 9$ is a factor of $f(x)$.
2. Dividing the Polynomial
To find the remaining factor, we can divide $f(x)$ by $x^2+9$. Performing polynomial long division, we have:
```
x - 5
x^2+9 | x^3 - 5x^2 + 9x - 45
- (x^3 + 0x^2 + 9x)
------------------
-5x^2 + 0x - 45
- (-5x^2 + 0x - 45)
------------------
0
```
So, $f(x) = (x^2+9)(x-5)$.
3. Finding the Remaining Zeros
The remaining zero is the solution to $x-5=0$, which is $x=5$. Thus, the zeros of $f(x)$ are $3i, -3i,$ and $5$.
4. Final Answer
The zeros of $f(x)=x^3-5 x^2+9 x-45$ are $3i, -3i,$ and $5$.
### Examples
Polynomials are used in various fields such as physics, engineering, and economics. For example, in physics, projectile motion can be modeled using a quadratic polynomial. In engineering, polynomials are used to design curves and surfaces. In economics, cost and revenue functions can be modeled using polynomials to analyze business performance.
- The quadratic factor corresponding to these zeros is $(x-3i)(x+3i) = x^2+9$.
- Dividing $f(x)$ by $x^2+9$ gives the remaining factor $x-5$.
- The remaining zero is $x=5$, so the zeros are $\boxed{3i, -3i, 5}$.
### Explanation
1. Finding the Conjugate and Quadratic Factor
We are given the polynomial $f(x)=x^3-5 x^2+9 x-45$ and one of its zeros, $3i$. Since the polynomial has real coefficients, the complex conjugate of $3i$, which is $-3i$, must also be a zero. Thus, $(x-3i)$ and $(x+3i)$ are factors of $f(x)$. Therefore, $(x-3i)(x+3i) = x^2 - (3i)^2 = x^2 - (-9) = x^2 + 9$ is a factor of $f(x)$.
2. Dividing the Polynomial
To find the remaining factor, we can divide $f(x)$ by $x^2+9$. Performing polynomial long division, we have:
```
x - 5
x^2+9 | x^3 - 5x^2 + 9x - 45
- (x^3 + 0x^2 + 9x)
------------------
-5x^2 + 0x - 45
- (-5x^2 + 0x - 45)
------------------
0
```
So, $f(x) = (x^2+9)(x-5)$.
3. Finding the Remaining Zeros
The remaining zero is the solution to $x-5=0$, which is $x=5$. Thus, the zeros of $f(x)$ are $3i, -3i,$ and $5$.
4. Final Answer
The zeros of $f(x)=x^3-5 x^2+9 x-45$ are $3i, -3i,$ and $5$.
### Examples
Polynomials are used in various fields such as physics, engineering, and economics. For example, in physics, projectile motion can be modeled using a quadratic polynomial. In engineering, polynomials are used to design curves and surfaces. In economics, cost and revenue functions can be modeled using polynomials to analyze business performance.
