To solve the cubic equation [tex]\( x^3 + 3x^2 - 5x + 25 = 0 \)[/tex], we need to find the roots of this polynomial. Here is a detailed analysis and step-by-step solution:
1. Identify the Polynomial: We start with the polynomial
[tex]\[
P(x) = x^3 + 3x^2 - 5x + 25.
\][/tex]
2. Attempt Rational Root Theorem: The Rational Root Theorem suggests that any rational roots [tex]\( \frac{p}{q} \)[/tex] must be factors of the constant term (25) over factors of the leading coefficient (1). The constant term is 25 and its factors are [tex]\(\pm 1, \pm 5, \pm 25\)[/tex].
3. Test Possible Rational Roots:
- Testing [tex]\( x = 1 \)[/tex]:
[tex]\[
1^3 + 3(1)^2 - 5(1) + 25 = 1 + 3 - 5 + 25 = 24 \quad (\text{not zero})
\][/tex]
- Testing [tex]\( x = -1 \)[/tex]:
[tex]\[
(-1)^3 + 3(-1)^2 - 5(-1) + 25 = -1 + 3 + 5 + 25 = 32 \quad (\text{not zero})
\][/tex]
- Testing [tex]\( x = 5 \)[/tex]:
[tex]\[
5^3 + 3(5)^2 - 5(5) + 25 = 125 + 75 - 25 + 25 = 200 \quad (\text{not zero})
\][/tex]
- Testing [tex]\( x = -5 \)[/tex]:
[tex]\[
(-5)^3 + 3(-5)^2 - 5(-5) + 25 = -125 + 75 + 25 + 25 = 0 \quad (\text{zero})
\][/tex]
Since [tex]\( x = -5 \)[/tex] is a root, we have identified one root of the polynomial.
4. Factor the Polynomial: Since [tex]\( x = -5 \)[/tex] is a root, [tex]\( (x + 5) \)[/tex] must be a factor of [tex]\( P(x) \)[/tex]. We will now divide [tex]\( P(x) \)[/tex] by [tex]\( (x + 5) \)[/tex].
- Polynomial long division:
[tex]\[
\begin{array}{r|rrrr}
x^3 & +3x^2 & -5x & +25 \\
- (x \cdot x^2) & \uparrow x^2 & +5x^2 \\
\hline
& -2x^2 & -5x \\
- (-2x \cdot x) & \uparrow -10x \\
\hline
& 5x & +25 \\
- (5 \cdot x+5) & \uparrow +25 \\
\hline
& 0 &\\
\end{array}
\][/tex]
We find:
[tex]\[
P(x) = (x + 5)(x^2 - 2x + 5)
\][/tex]
5. Solve the Quadratic Equation: Now we need to find the roots of the quadratic equation:
[tex]\[
x^2 - 2x + 5 = 0.
\][/tex]
By using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = 5 \)[/tex]:
[tex]\[
x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i.
\][/tex]
6. Collect the Solutions: We have found all the roots of the polynomial:
- The real root: [tex]\( x = -5 \)[/tex]
- The complex roots: [tex]\( x = 1 + 2i \)[/tex] and [tex]\( x = 1 - 2i \)[/tex]
Therefore, the complete set of solutions to the equation [tex]\( x^3 + 3x^2 - 5x + 25 = 0 \)[/tex] are:
[tex]\[
x = -5, \quad x = 1 + 2i, \quad x = 1 - 2i.
\][/tex]
