Answer :
We are given the polynomial
[tex]$$
f(x) = x^3 - 5x^2 + 9x - 45.
$$[/tex]
Since we know that [tex]$x = 5$[/tex] is a zero of [tex]$f(x)$[/tex], we can factor [tex]$f(x)$[/tex] by dividing it by [tex]$(x-5)$[/tex].
### Step 1. Divide [tex]$f(x)$[/tex] by [tex]$(x-5)$[/tex]
Using synthetic division or long division with divisor [tex]$(x-5)$[/tex], we set up the coefficients of [tex]$f(x)$[/tex]:
[tex]$$
1 \quad -5 \quad 9 \quad -45.
$$[/tex]
Perform synthetic division with [tex]$5$[/tex]:
1. Bring down the first coefficient: [tex]$1$[/tex].
2. Multiply [tex]$1$[/tex] by [tex]$5$[/tex]: [tex]$1 \times 5 = 5$[/tex], and add this to [tex]$-5$[/tex]: [tex]$-5 + 5 = 0$[/tex].
3. Multiply [tex]$0$[/tex] by [tex]$5$[/tex]: [tex]$0 \times 5 = 0$[/tex], and add this to [tex]$9$[/tex]: [tex]$9 + 0 = 9$[/tex].
4. Multiply [tex]$9$[/tex] by [tex]$5$[/tex]: [tex]$9 \times 5 = 45$[/tex], and add this to [tex]$-45$[/tex]: [tex]$-45 + 45 = 0$[/tex].
Thus, the quotient is
[tex]$$
x^2 + 0x + 9 = x^2 + 9,
$$[/tex]
and there is no remainder. Therefore, we have factored [tex]$f(x)$[/tex] as
[tex]$$
f(x) = (x-5)(x^2+9).
$$[/tex]
### Step 2. Factor the Quadratic Factor into Linear Factors
The quadratic factor is
[tex]$$
x^2+9.
$$[/tex]
We notice that this quadratic does not factor over the real numbers because its discriminant is negative. However, over the complex numbers, we write
[tex]$$
x^2 + 9 = 0 \quad \Longrightarrow \quad x^2 = -9 \quad \Longrightarrow \quad x = \pm 3i,
$$[/tex]
where [tex]$i$[/tex] is the imaginary unit. This allows us to factor the quadratic as
[tex]$$
x^2 + 9 = (x - 3i)(x + 3i).
$$[/tex]
### Step 3. Write the Complete Factorization and Solve the Equation
Combining the factors, the complete factorization of the polynomial over the complex numbers is
[tex]$$
f(x) = (x-5)(x-3i)(x+3i).
$$[/tex]
To solve the equation
[tex]$$
x^3 - 5x^2 + 9x - 45 = 0,
$$[/tex]
we set each factor equal to zero:
1. [tex]$$ x - 5 = 0 \quad \Longrightarrow \quad x = 5. $$[/tex]
2. [tex]$$ x - 3i = 0 \quad \Longrightarrow \quad x = 3i. $$[/tex]
3. [tex]$$ x + 3i = 0 \quad \Longrightarrow \quad x = -3i. $$[/tex]
### Final Answers
(a) The factorization of the polynomial into factors of the form [tex]$(x-c)$[/tex] is:
[tex]$$
f(x) = (x-5)(x-3i)(x+3i).
$$[/tex]
(b) The solutions of the equation [tex]$x^3 - 5x^2+ 9x-45=0$[/tex] are:
[tex]$$
x = 5,\quad x = 3i,\quad \text{and} \quad x = -3i.
$$[/tex]
[tex]$$
f(x) = x^3 - 5x^2 + 9x - 45.
$$[/tex]
Since we know that [tex]$x = 5$[/tex] is a zero of [tex]$f(x)$[/tex], we can factor [tex]$f(x)$[/tex] by dividing it by [tex]$(x-5)$[/tex].
### Step 1. Divide [tex]$f(x)$[/tex] by [tex]$(x-5)$[/tex]
Using synthetic division or long division with divisor [tex]$(x-5)$[/tex], we set up the coefficients of [tex]$f(x)$[/tex]:
[tex]$$
1 \quad -5 \quad 9 \quad -45.
$$[/tex]
Perform synthetic division with [tex]$5$[/tex]:
1. Bring down the first coefficient: [tex]$1$[/tex].
2. Multiply [tex]$1$[/tex] by [tex]$5$[/tex]: [tex]$1 \times 5 = 5$[/tex], and add this to [tex]$-5$[/tex]: [tex]$-5 + 5 = 0$[/tex].
3. Multiply [tex]$0$[/tex] by [tex]$5$[/tex]: [tex]$0 \times 5 = 0$[/tex], and add this to [tex]$9$[/tex]: [tex]$9 + 0 = 9$[/tex].
4. Multiply [tex]$9$[/tex] by [tex]$5$[/tex]: [tex]$9 \times 5 = 45$[/tex], and add this to [tex]$-45$[/tex]: [tex]$-45 + 45 = 0$[/tex].
Thus, the quotient is
[tex]$$
x^2 + 0x + 9 = x^2 + 9,
$$[/tex]
and there is no remainder. Therefore, we have factored [tex]$f(x)$[/tex] as
[tex]$$
f(x) = (x-5)(x^2+9).
$$[/tex]
### Step 2. Factor the Quadratic Factor into Linear Factors
The quadratic factor is
[tex]$$
x^2+9.
$$[/tex]
We notice that this quadratic does not factor over the real numbers because its discriminant is negative. However, over the complex numbers, we write
[tex]$$
x^2 + 9 = 0 \quad \Longrightarrow \quad x^2 = -9 \quad \Longrightarrow \quad x = \pm 3i,
$$[/tex]
where [tex]$i$[/tex] is the imaginary unit. This allows us to factor the quadratic as
[tex]$$
x^2 + 9 = (x - 3i)(x + 3i).
$$[/tex]
### Step 3. Write the Complete Factorization and Solve the Equation
Combining the factors, the complete factorization of the polynomial over the complex numbers is
[tex]$$
f(x) = (x-5)(x-3i)(x+3i).
$$[/tex]
To solve the equation
[tex]$$
x^3 - 5x^2 + 9x - 45 = 0,
$$[/tex]
we set each factor equal to zero:
1. [tex]$$ x - 5 = 0 \quad \Longrightarrow \quad x = 5. $$[/tex]
2. [tex]$$ x - 3i = 0 \quad \Longrightarrow \quad x = 3i. $$[/tex]
3. [tex]$$ x + 3i = 0 \quad \Longrightarrow \quad x = -3i. $$[/tex]
### Final Answers
(a) The factorization of the polynomial into factors of the form [tex]$(x-c)$[/tex] is:
[tex]$$
f(x) = (x-5)(x-3i)(x+3i).
$$[/tex]
(b) The solutions of the equation [tex]$x^3 - 5x^2+ 9x-45=0$[/tex] are:
[tex]$$
x = 5,\quad x = 3i,\quad \text{and} \quad x = -3i.
$$[/tex]
