Answer :
We start with the cubic equation
[tex]$$
4x^3 - 5x^2 + 20x - 25 = 0.
$$[/tex]
Step 1. Find a real root
By inspection or by testing possible candidates, we find that
[tex]$$
x = \frac{5}{4}
$$[/tex]
is a solution. This means that when [tex]$x = \frac{5}{4}$[/tex], the equation evaluates to zero.
Step 2. Factor the polynomial
Since [tex]$x = \frac{5}{4}$[/tex] is a root, the polynomial must have a factor corresponding to it. Multiplying the factor by 4 to clear the fraction, we have
[tex]$$
4x - 5 = 0.
$$[/tex]
Thus, the polynomial can be factored as
[tex]$$
4x^3 - 5x^2 + 20x - 25 = (4x - 5)(x^2 + 5).
$$[/tex]
Step 3. Solve the factors
1. Solve [tex]$4x - 5 = 0$[/tex]:
Set
[tex]$$
4x - 5 = 0.
$$[/tex]
Solve for [tex]$x$[/tex]:
[tex]$$
4x = 5 \quad \Longrightarrow \quad x = \frac{5}{4}.
$$[/tex]
2. Solve [tex]$x^2 + 5 = 0$[/tex]:
Set
[tex]$$
x^2 + 5 = 0.
$$[/tex]
Solve for [tex]$x$[/tex]:
[tex]$$
x^2 = -5.
$$[/tex]
Taking the square root of both sides gives
[tex]$$
x = \pm \sqrt{-5} = \pm i\sqrt{5}.
$$[/tex]
Final Answer
The solutions to the equation
[tex]$$
4x^3 - 5x^2 + 20x - 25 = 0
$$[/tex]
are
[tex]$$
x = \frac{5}{4}, \quad x = i\sqrt{5}, \quad \text{and} \quad x = -i\sqrt{5}.
$$[/tex]
[tex]$$
4x^3 - 5x^2 + 20x - 25 = 0.
$$[/tex]
Step 1. Find a real root
By inspection or by testing possible candidates, we find that
[tex]$$
x = \frac{5}{4}
$$[/tex]
is a solution. This means that when [tex]$x = \frac{5}{4}$[/tex], the equation evaluates to zero.
Step 2. Factor the polynomial
Since [tex]$x = \frac{5}{4}$[/tex] is a root, the polynomial must have a factor corresponding to it. Multiplying the factor by 4 to clear the fraction, we have
[tex]$$
4x - 5 = 0.
$$[/tex]
Thus, the polynomial can be factored as
[tex]$$
4x^3 - 5x^2 + 20x - 25 = (4x - 5)(x^2 + 5).
$$[/tex]
Step 3. Solve the factors
1. Solve [tex]$4x - 5 = 0$[/tex]:
Set
[tex]$$
4x - 5 = 0.
$$[/tex]
Solve for [tex]$x$[/tex]:
[tex]$$
4x = 5 \quad \Longrightarrow \quad x = \frac{5}{4}.
$$[/tex]
2. Solve [tex]$x^2 + 5 = 0$[/tex]:
Set
[tex]$$
x^2 + 5 = 0.
$$[/tex]
Solve for [tex]$x$[/tex]:
[tex]$$
x^2 = -5.
$$[/tex]
Taking the square root of both sides gives
[tex]$$
x = \pm \sqrt{-5} = \pm i\sqrt{5}.
$$[/tex]
Final Answer
The solutions to the equation
[tex]$$
4x^3 - 5x^2 + 20x - 25 = 0
$$[/tex]
are
[tex]$$
x = \frac{5}{4}, \quad x = i\sqrt{5}, \quad \text{and} \quad x = -i\sqrt{5}.
$$[/tex]
