Problem 61 - Algebra and Trigonometry

Question

Find all zeros of the polynomial. \(P(x)=x^{5}-3 x^{4}+12 x^{3}-28 x^{2}+27 x-9\)

Step-by-Step Solution

Answer

The zeros are \(x = 1\) (multiplicity 3), \(x = 3i\), and \(x = -3i\).

Step 1: Understand the Polynomial

The polynomial given is \(P(x)=x^{5}-3x^{4}+12x^{3}-28x^{2}+27x-9\). It is a fifth-degree polynomial, which implies it can have up to 5 real roots (or zeros).

Step 2: Use the Rational Root Theorem

The Rational Root Theorem suggests that any rational root is of the form \(\frac{p}{q}\) where \(p\) is a factor of the constant term (-9) and \(q\) is a factor of the leading coefficient (1). The potential rational roots are therefore \(\pm 1, \pm 3, \pm 9\).

Step 3: Test Potential Rational Roots

Plug the potential rational roots into the polynomial to test which, if any, make the polynomial zero. Begin with \(x = 1\):\[P(1) = 1^5 - 3(1)^4 + 12(1)^3 - 28(1)^2 + 27(1) - 9 = 0\] Thus, \(x = 1\) is a root.

Step 4: Use Synthetic Division

Perform synthetic division of \(P(x)\) by \(x - 1\) to reduce the polynomial. The division results in a quotient of \(x^4 - 2x^3 + 10x^2 - 18x + 9\). The remainder is 0, confirming \(x = 1\) is a root.

Step 5: Factor the Reduced Polynomial

Attempt to factor \(x^4 - 2x^3 + 10x^2 - 18x + 9\) further. Apply the Rational Root Theorem again to find potential rational roots: test \(x = 1\), \(x = -1\), \(x = 3\), and others. Found another root: \(x = 1\).

Step 6: Further Divide with Synthetic Division

Applying synthetic division again on the quadratic quotient by \(x - 1\), we have: \(x^3 - x^2 + 9x - 9\).Perform division again to simplify further.

Step 7: Factor Again

Now, factor or apply synthetic division on \(x^3 - x^2 + 9x - 9\) by testing possible roots. Found another root: \(x = 1\), applying synthetic division results in \(x^2 + 9\).

Step 8: Solve the Quadratic Equation

The quadratic \(x^2 + 9\) has no real roots, as the discriminant \((b^2 - 4ac = 0 - 36)\) is negative. The roots are complex: \(x = \pm 3i\).

Step 9: Compilation of All Roots

The roots found are: \(x = 1\) (with multiplicity 3), and the complex roots \(x = 3i\), and \(x = -3i\).

Key Concepts

Fifth-Degree Polynomial

A fifth-degree polynomial is an algebraic expression of the form \[x^5 + ax^4 + bx^3 + cx^2 + dx + e\]where the highest exponent of the variable, in this case, is 5. This indicates that the polynomial can have up to 5 roots in total, counting both real and complex roots. Working with higher-degree polynomials can be more complex than lower-degree ones because they have a larger number of potential roots and may include multiple reducible factors.

While solving a fifth-degree polynomial, the goal is often to break it down into simpler components or factors. This simplifies the problem, allowing us to find individual zeros or roots more easily. It's important to remember that the Fundamental Theorem of Algebra guarantees that a polynomial of degree \(n\) has exactly \(n\) roots in the complex number system, counting multiplicities.

Rational Root Theorem

The Rational Root Theorem is a handy tool when dealing with polynomials, especially when attempting to find rational zeros. It states that if a polynomial has a rational root \(\frac{p}{q}\), then \(p\) must be a factor of the constant term, and \(q\) must be a factor of the leading coefficient. In our exercise, the polynomial is \[x^5 - 3x^4 + 12x^3 - 28x^2 + 27x - 9\]where the constant term is -9 and the leading coefficient is 1.

This gives us potential rational roots of \(\pm 1, \pm 3,\) and \(\pm 9\). Checking these values efficiently helps us identify which of them, if any, are actually roots of the polynomial. In this scenario, testing these values led to confirming \(x = 1\) as a root, simplifying our task to factor the polynomial further.

Synthetic Division

Synthetic division is a simplified form of polynomial division, specifically for dividing by linear expressions of the form \(x - c\). It's quicker and requires less arithmetic than long division, making it ideal when testing potential rational roots from the Rational Root Theorem.

For our fifth-degree polynomial, once \(x = 1\) was identified as a root, synthetic division was employed to divide the polynomial by \(x - 1\). This process works by using coefficients of the polynomial solely, yielding a new polynomial with reduced degree if \(c\) is indeed a root.

In our exercise, applying synthetic division by \(x - 1\) reduced the polynomial to \(x^4 - 2x^3 + 10x^2 - 18x + 9\). Each successful division demonstrates a root and reduces the polynomial further, thus simplifying the search for additional roots.

Complex Roots

Complex roots come into play when a polynomial has roots that cannot be expressed as real numbers. When the polynomial was reduced to \(x^2 + 9\), the discriminant of this quadratic expression was found to be negative, indicating the presence of complex roots.

A negative discriminant signifies that there are no real solutions, but instead complex conjugates. For the quadratic \(x^2 + 9\), the solutions are obtained by solving \[x^2 = -9\]giving us roots \(x = 3i\) and \(x = -3i\), where \(i\) is the imaginary unit representing \(i = \sqrt{-1}\).

In a polynomial of degree 5, having complex roots must be acknowledged as part of the total solution set. Here, the complete solution set includes three real roots (all \(x = 1\) with multiplicity) and two complex roots, \(x = 3i\) and \(x = -3i\). Understanding complex roots and how they operate in pairings is vital for comprehensively solving higher-degree polynomials.