Problem 48 - Algebra and Trigonometry

Question

Find all zeros of the polynomial. \(P(x)=x^{3}+7 x^{2}+18 x+18\)

Step-by-Step Solution

Answer

The zeros are \(x = -3, -2 + i\sqrt{2}, -2 - i\sqrt{2}\).

Step 1: Check Rational Roots

To start, use the Rational Root Theorem to identify potential rational roots of the polynomial. The possible rational roots are the factors of the constant term, 18, divided by the factors of the leading coefficient, which is 1. So the candidates are \(\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18\).

Step 2: Test Rational Roots

Test each possible rational root by substituting it into the polynomial \(P(x)\). After testing, we find that \(x = -3\) is a root since \(P(-3) = 0\).

Step 3: Factor the Polynomial by Division

Perform synthetic division using \(x = -3\) to divide the polynomial and obtain the quotient. Doing so gives the quotient as \(x^2 + 4x + 6\). This means the polynomial can be written as \((x + 3)(x^2 + 4x + 6)\).

Step 4: Find the Roots of the Quadratic

Solve the quadratic equation \(x^2 + 4x + 6 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = 4\), and \(c = 6\). The discriminant \(b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot 6 = 16 - 24 = -8\), which is negative, indicating complex roots.

Step 5: Calculate Complex Roots

Using the quadratic formula, compute the roots as \(x = \frac{-4 \pm \sqrt{-8}}{2}\). Simplify to find the complex roots as \(x = -2 \pm i\sqrt{2}\).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a valuable tool when we're trying to find the zeros of a polynomial. It gives us a list of possible rational zeros without having to factorize the entire polynomial immediately.

Here's how it works: The theorem states that if a polynomial has a rational root \(\frac{p}{q}\), then \(p\) is a factor of the constant term, and \(q\) is a factor of the leading coefficient. In our polynomial \(P(x)=x^{3}+7x^{2}+18x+18\), the constant term is 18, and the leading coefficient is 1.

• Factors of 18 are: \(\pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18\).
• Since the leading coefficient is 1, its factors are simply \(\pm 1\).

Hence, the possible rational roots of the polynomial, according to the Rational Root Theorem, are these factors of 18 divided by 1, which means they remain the same factors of 18. This initial list serves as a starting point to test potential solutions.

Synthetic Division

Synthetic division is a simplified method to divide polynomials, especially useful when testing potential zeros like those found using the Rational Root Theorem.

To use synthetic division, you need a candidate root. From our exercise, we found one rational root: \(x = -3\). We use this value to test if it divides the polynomial evenly and to simplify the polynomial further.

• Write down the coefficients of the polynomial: \(1, 7, 18, 18\).
• Use \(-3\) for the synthetic division.
• Perform the synthetic division process by bringing down the leading coefficient, multiplying and adding stepwise through the remaining coefficients.

The result of our synthetic division gives us \(x^{2} + 4x + 6\) as the quotient, meaning our polynomial can be factored as \((x + 3)(x^2 + 4x + 6)\). This method quickly narrows down the polynomial, making subsequent steps more straightforward.

Quadratic Formula

When a polynomial has a quadratic component, like our factored form \(x^2 + 4x + 6\), the quadratic formula is the key to finding its complex or real roots.

The formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
For the quadratic \(x^2 + 4x + 6\), we identify:

• \(a = 1\)
• \(b = 4\)
• \(c = 6\)

Next, calculate the discriminant \(b^2 - 4ac\). Here, the discriminant is \(16 - 24 = -8\), which is less than zero, indicating that roots are not real numbers, but complex.
Insert the values into the quadratic formula to find the complex roots. This will yield \(x = -2 \pm i\sqrt{2}\).

Complex Roots

Complex roots occur when the discriminant of a quadratic equation is less than zero, implying that no real number solutions exist.

In our exercise case, starting from the quadratic \(x^2 + 4x + 6 = 0\), the discriminant is \(-8\). Such a negative discriminant denotes that the solutions involve imaginary numbers.

The quadratic formula solution gives \(x = \frac{-4 \pm \sqrt{-8}}{2}\). Simplifying further:

• \(\sqrt{-8} = \sqrt{4 \cdot -2} = 2i\sqrt{2}\)
• Hence, the solutions are \(x = -2 \pm i\sqrt{2}\)

These are the complex roots. They consist of a real part and an imaginary part and are typically written in the form \(a \pm bi\) where \(a\) and \(b\) are real numbers, while \(i\) is the imaginary unit, defined by \(i^2 = -1\). This gives us insight into how polynomials can have solutions that extend beyond the real number line.