Given that $-i+2$ is a zero of $f(x)=x^5-6 x^4+11 x^3-x^2-14 x+5$, find all complex roots using synthetic division. List your possible rational roots also.

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06-11-2026
Mathematics

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Given that $-i+2$ is a zero of $f(x)=x^5-6 x^4+11 x^3-x^2-14 x+5$, find all complex roots using synthetic division. List your possible rational roots also.

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GinnyAnswer GinnyAnswer · Answer 1 · 2026-11-06T13:14:01-05:00

The given polynomial is $f(x) = x^5 - 6x^4 + 11x^3 - x^2 - 14x + 5$.
Given the complex root $2-i$, its conjugate $2+i$ is also a root, leading to the factor $(x^2 - 4x + 5)$.
Dividing the polynomial by this factor gives $q(x) = x^3 - 2x^2 - 2x + 1$.
Factoring $q(x)$ reveals the roots $-1$ and $\frac{3 \pm \sqrt{5}}{2}$.
The full set of roots is $\boxed{{2-i, 2+i, -1, \frac{3+\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}}}$.

Explanation

Analyzing the Given Polynomial and Identifying Roots

We are given the polynomial $f(x) = x^5 - 6x^4 + 11x^3 - x^2 - 14x + 5$. We know that $z_1 = 2 - i$ is a zero. Because the polynomial has real coefficients, its complex roots must occur in conjugate pairs. Thus, $z_2 = 2 + i$ is also a zero of $f(x)$. To find the remaining factors, we first construct the quadratic factor associated with these roots:
\[(x - (2 - i))(x - (2 + i)) = (x - 2 + i)(x - 2 - i) = (x - 2)^2 - i^2 = x^2 - 4x + 4 + 1 = x^2 - 4x + 5.\]

Division and Polynomial Factoring

Using the Rational Root Theorem, the possible rational roots are the factors of the constant term (5) divided by the factors of the leading coefficient (1), which are $\pm 1$ and $\pm 5$.
To divide $f(x) = x^5 - 6x^4 + 11x^3 - x^2 - 14x + 5$ by $x^2 - 4x + 5$, we perform the division:
\[(x^5 - 6x^4 + 11x^3 - x^2 - 14x + 5) \div (x^2 - 4x + 5) = x^3 - 2x^2 - 2x + 1.\]
Thus, the polynomial factors as $f(x) = (x^2 - 4x + 5)(x^3 - 2x^2 - 2x + 1)$. We now need to find the roots of the cubic factor $q(x) = x^3 - 2x^2 - 2x + 1$.

Finding the Remaining Roots

We test the candidate rational roots for $q(x) = x^3 - 2x^2 - 2x + 1$:

For $x = 1$: $q(1) = 1^3 - 2(1)^2 - 2(1) + 1 = 1 - 2 - 2 + 1 = -2 eq 0$.
For $x = -1$: $q(-1) = (-1)^3 - 2(-1)^2 - 2(-1) + 1 = -1 - 2 + 2 + 1 = 0$.

Since $x = -1$ is a root, $(x + 1)$ is a factor of $q(x)$. Dividing $q(x)$ by $(x + 1)$ using synthetic division:
\[(x^3 - 2x^2 - 2x + 1) \div (x + 1) = x^2 - 3x + 1.\]
To find the remaining roots, we solve $x^2 - 3x + 1 = 0$ using the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(1)}}{2(1)} = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}.\]

Conclusion and Summary of Roots

The five roots of the polynomial $f(x) = x^5 - 6x^4 + 11x^3 - x^2 - 14x + 5$ are the two complex roots $2 \pm i$, the rational root $-1$, and the two irrational roots $\frac{3 \pm \sqrt{5}}{2}$.
The full set of roots is $\{2 - i, 2 + i, -1, \frac{3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}\}$.

Examples

In engineering and physics, finding the roots of a polynomial is essential for stability analysis in control systems. For example, if a system's behavior is modeled by a 5th-degree characteristic equation, the roots tell us whether the system is stable or oscillates. Complex roots, like the $2 \pm i$ found here, indicate decaying or growing oscillations, while real roots indicate exponential growth or decay, allowing engineers to design safer and more efficient systems.