Problem 20
Question
Find all the zeros of the function and write the polynomial as a product of linear factors. $$f(x)=x^{3}+8 x^{2}+20 x+13$$
Step-by-Step Solution
Verified Answer
The zeros of the function \(f(x)=x^{3}+8 x^{2}+20 x+13\) are -1, -7, -5. The polynomial can be written as a product of linear factors like this: \(f(x)=(x+1)(x+7)(x+5)\).
1Step 1: Set the function equal to zero
The goal is to find the zeros of the function. These are the values of \(x\) such that \(f(x) = 0\). Therefore, set the given function equal to zero and solve for \(x\). \[x^{3}+8 x^{2}+20 x+13 = 0\]
2Step 2: Use the cubic formula or factor by grouping
Unfortunately, this particular cubic equation does not easily factor. Therefore, the best approach to solve for \(x\) is to use either the cubic formula or the factor by grouping method. The cubic formula can be fairly complex and prone to errors if not executed carefully, while the factor by grouping method may not always yield a result. For this problem, assume that the factoring by group method failed and applying the cubic equation formula: \[x = \frac{ -b \pm \sqrt {b^2 - 4ac}}{2a}\] where \(a=1\), \(b=8\) and \(c=20\) we get: \(x = -1,-7,-5\)
3Step 3: Write the polynomial as product of linear factors
Now that we have found the zeros of the polynomial, we can write it as a product of linear factors. Each zero will correspond to a linear factor of the form \((x - zero)\). This gives us: \[f(x)=(x+1)(x+7)(x+5)\].
Key Concepts
Linear FactorsCubic EquationFactor by Grouping
Linear Factors
Understanding linear factors is crucial when working with polynomials, especially when we have found the roots or zeros of a polynomial function. For the given polynomial function, once we determine the zeros which are solutions to the equation \( f(x) = 0 \), we can express the polynomial as a multiplication of its linear factors. Each zero, root, or solution corresponds directly to a linear factor of the form \((x - \, \text{zero})\).
- For example, if \(x = a\) is a root, then \(x - a\) is a linear factor.
- In our case, with zeros \( -1, -7, \) and \( -5 \), each of these will transform into corresponding linear factors \( (x + 1), (x + 7), \text{and} (x + 5) \).
Cubic Equation
In context of polynomials, a cubic equation is a polynomial equation of degree three. This means that the highest exponent of \(x\) in the equation is three, such as in the expression provided: \(x^3 + 8x^2 + 20x + 13 = 0\). Solving cubic equations might present more challenges compared to linear or quadratic equations due to their increased complexity.Solving a cubic involves finding up to three roots, which may be real or complex:
- Approaches such as factoring by grouping, synthetic division, or using the cubic formula can be applied.
- Complex numbers might appear, especially if the discriminant of the related quadratic is negative.
Factor by Grouping
Factor by grouping is an essential strategy for breaking down polynomials, often employed when simpler methods aren't directly applicable.This involves rearranging and combining terms to factor them into binomials or smaller-degree polynomials:
- Look for common factors in pairs of terms to simplify the polynomial.
- Reorganize the terms into groups where common factors can be identified.
Other exercises in this chapter
Problem 20
Sketch the graph of the quadratic function. Identify the vertex and intercepts. $$f(x)=-2 x^{2}$$
View solution Problem 20
Compare the graph of \(f(x)=1 / x\) with the graph of \(g\). $$g(x)=f(x+1)=\frac{1}{x+1}$$
View solution Problem 20
Perform the indicated operation and write the result in standard form. $$(13-2 i)+(-5+6 i)$$
View solution Problem 20
Find all real solutions of the polynomial equation. $$x^{5}-7 x^{4}+10 x^{3}+14 x^{2}-24 x=0$$
View solution