Problem 11 - College Algebra and Calculus: An Applied Approach

Question

Find all the zeros of the function and write the polynomial as a product of linear factors. $$f(x)=x^{4}-81$$

Step-by-Step Solution

Answer

The zeros of the function \(f(x)=x^{4}-81\) are \(x=3\), \(x=-3\), \(x=3i\), and \(x=-3i\)

Step 1: Identify the type of polynomial

The given function is \(f(x)=x^{4}-81\). This is a quartic polynomial, i.e. a polynomial of degree 4.

Step 2: Factor the function

Factoring the function using difference of squares formula, we get \(f(x)=(x^{2}-9)\) \(x^{2}+9\)

Step 3: Further factor the function

Factoring the function \(x^{2}-9\) using the difference of squares formula again, we get \(f(x)=(x-3)(x+3)\) \(x^{2}+9\)

Step 4: Find the zeros

Setting \(f(x)=0\), so the zeros of \(f(x)\) are the solutions of the equation \(x-3=0\), \(x+3=0\) and \(x^{2}+9 =0\). Solving these we get \(x=3\), \(x=-3\), and \((x=3i, -3i)\)

Key Concepts

Quartic Polynomials

Quartic polynomials are algebraic expressions of degree four. They have the general form of \(ax^{4} + bx^{3} + cx^{2} + dx + e = 0\), where \(a\), \(b\), \(c\), \(d\), and \(e\) are coefficients and \(a eq 0\). Identifying a polynomial as quartic is important because it dictates the number of possible zeros and the methods used for finding them.

In our exercise, the quartic polynomial presented is \(f(x) = x^{4} - 81\), which can be recognized by its highest degree term, \(x^{4}\). Since there are no other terms except the constant, this particular polynomial can be factored directly using the difference of squares method, which is an efficient way to simplify expressions and reveal its zeros.

Difference of Squares

The difference of squares is a pattern in algebra that you can use to factor certain types of binomials. Specifically, it applies when you have an expression of the form \(a^{2} - b^{2}\), which can be factored into \((a + b)(a - b)\). This identity is pivotal in simplifying and solving polynomial equations as it breaks them down into more manageable linear factors.

For instance, when looking at our exercise's polynomial \(f(x) = x^{4} - 81\), we notice it is a difference of squares where \(a^{2} = x^{4}\) and \(b^{2} = 81\). Following the pattern, we factorize it to \((x^{2} - 9)(x^{2} + 9)\). The first binomial \((x^{2} - 9)\) is also a difference of squares and can be further factored, simplifying our quest to find the polynomial's zeros.

Linear Factors

Linear factors are expressions of the first degree, which means they are in the form of \(ax + b\) where \(a\) and \(b\) are coefficients and \(a\) isn’t zero. When a polynomial is fully factored into linear factors, solving for the zeros—the values of \(x\) for which the polynomial equals zero—becomes straightforward.

In our given function, after applying the difference of squares twice, we arrived at the factorized form \((x - 3)(x + 3)(x^{2} + 9)\). The first two factors are linear and directly give us two of the zeros: \(x = 3\) and \(x = -3\). However, the third factor, \(x^{2} + 9\), cannot be factored into real linear factors because it would require taking the square root of a negative number. Instead, by applying the concept of complex numbers, we find the remaining two zeros: \(x = 3i\) and \(x = -3i\), introducing the concept of imaginary zeros in polynomials.