Chapter 3: Problem 44
Graphing Polynomials Factor the polynomial and use the factored form to find the zeros. Then sketch the graph. $$P(x)=x^{6}-2 x^{3}+1$$
Short Answer
Expert verified
Zero at \( x = 1 \) with multiplicity 2; bounce at \( x = 1 \).
Step by step solution
01
Set the Polynomial Equal to Zero
To find the zeros of the polynomial, start by setting the polynomial equation equal to zero: \[ x^6 - 2x^3 + 1 = 0 \] This will allow us to solve for the values of \( x \) that make the polynomial equal to zero.
02
Substitute with a New Variable
Observe that \( x^6 \) is \( (x^3)^2 \). Let \( y = x^3 \). Rewrite the equation in terms of \( y \):\[ y^2 - 2y + 1 = 0 \] This makes it a quadratic equation in terms of \( y \).
03
Factor the Quadratic Equation
The equation \( y^2 - 2y + 1 = 0 \) is a perfect square trinomial. Factor it as: \[ (y - 1)^2 = 0 \] This means \( y - 1 = 0 \) or \( y = 1 \).
04
Substitute Back the Original Variable
Given \( y = x^3 \) and now \( y = 1 \), substitute back to find \( x \):\[ x^3 = 1 \] Solve this by taking the cube root of both sides to find \( x = 1 \).
05
Determine the Multiplicity of the Zero
Since the factor \( (y - 1)^2 \) repeated, the zero \( x = 1 \) has a multiplicity of 2. This means at \( x = 1 \), the curve touches the x-axis and bounces back.
06
Sketch the Graph
To sketch the graph of \( P(x) = x^6 - 2x^3 + 1 \), plot the zero at \( x = 1 \) with a bounce (because of the multiplicity of 2). Since \( x^6 \) is the leading term, the end behavior resembles that of \( x^6 \), rising on both ends of the graph as \( x \rightarrow \pm \infty \). The graph does not cross the x-axis anywhere else.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factored Form
A polynomial expressed in factored form makes it much easier to identify the zeros, or roots, of the function. Consider the polynomial given: To express it in factored form, the key step is recognizing parts of the polynomial that can be rewritten in a more manageable way. In our example, we used a substitution method where we let a new variable, such as \( y = x^3 \), simplify our polynomial to a quadratic form.
- This transformation helps to recognize the polynomial as a perfect square trinomial.
Finding Zeros
Zeros of a polynomial are the values of \( x \) that make the polynomial equal to zero. In our exercise, the polynomial is set to zero, giving:\[ x^6 - 2x^3 + 1 = 0 \]The substitution \( y = x^3 \) changes the polynomial to a quadratic form, \( y^2 - 2y + 1 = 0 \). Factoring it to \( (y-1)^2 = 0 \) reveals the zero of the quadratic equation:\( y = 1 \) Since \( y = x^3 \), substituting back gives \( x^3 = 1 \). Solving for \( x \) involves taking the cube root to find \( x = 1 \).
- Thus, \( x = 1 \) is a zero of the original polynomial.
Multiplicity of Roots
The multiplicity of a root refers to the number of times that root appears as a factor in the polynomial. When a root repeats, as seen in this problem,
- its higher multiplicity implies distinctive behavior at that zero.
- At \( x = 1 \), the graph only touches the x-axis and then bounces back.