Problem 15
Question
Find all real solutions of the polynomial equation. $$z^{4}-z^{3}-2 z-4=0$$
Step-by-Step Solution
Verified Answer
The real solutions of the polynomial equation \(z^{4}-z^{3}-2 z-4=0\) are \(z = 0\) (multiplicity 2), \(z = 1\), and \(z = -2\).
1Step 1: Factorization
Start by grouping terms: \( (z^4-z^3) - (2z+4) = 0 \). Factoring \(z^3\): from the first group and 2 from the second one yields \( z^3(z-1) - 2(z+2) = 0 \).
2Step 2: Solve for z
Setting the factored equation to zero allows us to find the solutions for \(z\). The way to proceed is to set each factor equal to zero: \(z^3(z-1) = 0 \Rightarrow z = 0 \) or \( z = 1 \), and \( -2(z+2) = 0 \Rightarrow z = -2 \). So, three roots are: \(0, 1 \), and \(-2 \). These are all real numbers.
3Step 3: Check missing root
Given this is a fourth degree polynomial, it should have 4 solutions. We've only found three so far. This means either one solution is missing or one of the solutions is of degree two. Further analysis shows that \(z=0\) is of degree two. We know this because the coefficient of the \(z^3\) term is zero, which indicates that \(z=0\) is a root two times over as cubic equations usually have three solutions if the coefficient of the cube term is not zero. Same applies to higher degree equations.
Key Concepts
FactorizationReal SolutionsRoots of Polynomials
Factorization
Polynomials can often be simplified through factorization, a process which makes it easier to find their roots. In our equation, we began by grouping terms in pairs that share common factors:
- First Group: \( z^4 - z^3 \)
- Second Group: \( -2z - 4 \)
- From \( z^4 - z^3 \), we take out \( z^3 \), leaving \( z^3(z-1) \).
- From \( -2z - 4 \), we take out a \( -2 \), which results in \( -2(z+2) \).
Real Solutions
Finding real solutions of a polynomial involves determining the values of the variable that satisfy the equation. In our context, real solutions for the polynomial \( z^4 - z^3 - 2z - 4 = 0 \) are those that make the entire expression equal zero. In practical terms, here's how it is done after factorization:
- Set \( z^3(z-1) = 0 \) which gives solutions \( z = 0 \) and \( z = 1 \).
- Similarly, set \( -2(z+2) = 0 \) yielding the solution \( z = -2 \).
Roots of Polynomials
The roots of a polynomial are values that transform the polynomial into zero. For a polynomial of degree \( n \), there should ideally be \( n \) roots, accounting for multiplicity. Previously, we identified three roots from the equation: \( 0, 1, \) and \( -2 \).However, since we have a fourth-degree polynomial, it's expected to have four roots. This leads us to the concept of multiplicity, where a particular solution like \( z = 0 \) appears more than once.
- This particular repetition accounts for one of the missing solutions.
- Thus, \( z = 0 \) appears twice, making it a root of multiplicity 2.
Other exercises in this chapter
Problem 15
Find all the zeros of the function and write the polynomial as a product of linear factors. $$h(x)=x^{3}-11 x^{2}-15 x+325$$
View solution Problem 15
Write the complex number in standard form and find its complex conjugate. $$-5 i^{5}$$
View solution Problem 16
Use long division to divide. Divisor \(x-2\) Dividend $$x^{3}-x^{2}+2 x-8$$
View solution Problem 16
Use the graph of \(y=x^{4}\) to sketch the graph of the function. $$f(x)=\frac{1}{2}(x-1)^{4}$$
View solution