Question

Use the given zero of \(f\) to find all the zeros of \(f\). $$f(x)=25 x^{3}-55 x^{2}-54 x-18, \frac{1}{5}(-2+\sqrt{2} i)$$

Short answer

The zeros of the polynomial function are \(\frac{1}{5}(-2+\sqrt{2} i)\), \(\frac{1}{5}(-2-\sqrt{2} i)\), and the roots of the remaining quadratic polynomial obtained in Step 3.

Step-by-step solution

Confirm that the given number is a root

Substitute \(\frac{1}{5}(-2+\sqrt{2} i)\) into the polynomial function \(f\). If \(f(\frac{1}{5}(-2+\sqrt{2} i)) = 0\), then it is the root of the polynomial function - if not, there have been made a mistake when providing the root.

Identify the conjugate root

If a polynomial with real coefficients has a complex root, it must also have that complex root's conjugate. So if \(\frac{1}{5}(-2+\sqrt{2} i)\) is a root, then its conjugate \(\frac{1}{5}(-2-\sqrt{2} i)\) is also a root of the function.

Factorize the polynomial

The polynomial \(f\) can be written in terms of its factors. Two of these are \((5x-1+\sqrt{2}i)\) and \((5x-1-\sqrt{2}i)\), corresponding to the roots \(\frac{1}{5}(-2+\sqrt{2} i)\) and \(\frac{1}{5}(-2-\sqrt{2} i)\) respectively. We can multiply these factors out and divide the original polynomial \(f\) by this to find the remaining factor.

Find the remaining root

The remaining factor is a quadratic (reduced) polynomial, we can solve it for \(x\), either by using the quadratic formula or by factoring if possible (only if it has integer coefficients and solutions). The roots from this step are the remaining roots of the original polynomial function \(f(x)\).

Key concepts

Complex Numbers

Complex numbers are an extension of the traditional number system we know, which includes the real numbers. They consist of two parts: a real part and an imaginary part, denoted as \( a + bi \) where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit defined by \( i^2 = -1 \). For example, in the given exercise, \( \frac{1}{5}(-2+\sqrt{2} i) \) is a complex number because it includes both a real component (\( \frac{-2}{5} \)) and an imaginary component (\( \frac{\sqrt{2}}{5} \)).

Understanding complex numbers is crucial because they allow us to solve equations that have no real solutions. In the context of polynomial equations, when the polynomial has complex roots and real coefficients, these roots will always come in conjugate pairs. Once you have a firm handle on this, working with complex numbers becomes much simpler.

Conjugate Roots

In equations with real coefficients, complex roots occur as conjugate pairs. A conjugate of a complex number \( a + bi \) is \( a - bi \). With this, if \( \frac{1}{5}(-2+\sqrt{2} i) \) is a root of a polynomial, then \( \frac{1}{5}(-2-\sqrt{2} i) \) is also a root. These conjugate pairs ensure that when multiplied, they cancel out the imaginary components, resulting in real-coefficient factors.

This property is extremely useful when working with polynomials because it helps in the factorization process. When you have found one complex root of a real polynomial, you automatically also have its conjugate root, which aids in simplifying the polynomial into real factors.

Factorization

Factorization is the process of breaking down a polynomial into simpler polynomial expressions, or factors, whose product will give the original polynomial. In our exercise, factorization involves using the complex roots we've identified. We start by writing each of these complex roots as factors: \((5x-1+\sqrt{2}i)\) and \((5x-1-\sqrt{2}i)\).

These can be multiplied to form a quadratic polynomial which is then separated from the original polynomial to find any remaining factors. The goal is to express \(f(x)\) as a product of these simpler factors, gradually solving smaller polynomial equations that are easier to manage. Factorization simplifies solving equations by reducing degrees and therefore the complexity of the polynomial structure.

Quadratic Formula

The quadratic formula is a valuable tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). The formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), helps find the roots of the polynomial by calculating the solution for \( x \). After factorization in our exercise, any remaining polynomial is likely a quadratic one.

Using the quadratic formula is straightforward: plug in the coefficients \( a \), \( b \), and \( c \) into the formula, and calculate the discriminant \( b^2 - 4ac \). The discriminant will tell you about the nature of the roots. If it's positive, you'll have two distinct real roots; if zero, one real root; and if negative, complex conjugate roots. This technique is especially useful when roots are not easily factorable. It's a powerful and universal method that always works for quadratic polynomials.