We want to solve the equation in the form of P(x)=0, where P(x) is a polynomial function. This means that we want to find zeros of P(x).
P(x) = x^3-2x^2+25x-50=0
We also know that the quotient of P(x) and x-2 equals x^2 + 25. This means that x-2 multiplied by x^2 + 25 gives P(x).
(x^3 - 2x^2 + 25x -50) ÷ (x-2) = x^2+25 ⇕
x^3 - 2x^2 + 25x -50 = (x-2)(x^2 + 25)
The degree of P(x) is 3, so we know by the Fundamental Theorem of Algebra that the polynomial has exactly 3 roots. Let's use the Zero Product Property to find them!
(x-2) (x^2 + 25) = 0 ⇕
x-2 = 0 or x^2 + 25 = 0
The first factor is equal to zero when x = 2. Let's move on to factoring the second expression.
x^2 + 25 = 0
a^2+b^2=(a+bi)(a-bi)
(x+5i)(x-5i) = 0
Now we can use the Zero Product Property again.
(x+5i)(x-5i) = 0 ⇕
x + 5i = 0 or x - 5i = 0
The polynomial x^2 + 25 has 2 complex roots, -5i and 5i. Finally, let's list the roots of P(x).
2, -5i, 5i
