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Magazine Chapter 4: Problem 12
\(1-12\) . A polynomial \(P\) is given. (a) Find all zeros of \(P\) , real and complex. (b) Factor \(P\) completely. $$ P(x)=x^{6}-7 x^{3}-8 $$
Short Answer
Expert verified
The zeros are \(x = 2, -1, \omega, \omega^2\). Factored form: \((x-2)^2(x+1)^2(x^2+x+1)\).
Step by step solution
01
Identify the polynomial form
Examine the polynomial given: \( P(x) = x^6 - 7x^3 - 8 \). Notice it can be expressed in terms of a substitution by observing it has a quadratic form in terms of \( x^3 \).
02
Substitute variable
Let \( y = x^3 \). Then the polynomial becomes a quadratic: \( y^2 - 7y - 8 \).
03
Solve the quadratic equation
We solve \( y^2 - 7y - 8 = 0 \) using the quadratic formula: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here \( a = 1, \ b = -7, \ c = -8 \).
04
Apply the quadratic formula
Using the formula, we calculate: \[ y = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} = \frac{7 \pm \sqrt{49 + 32}}{2} = \frac{7 \pm \sqrt{81}}{2} \] \( \sqrt{81} = 9 \), so we have solutions \( y = \frac{7 + 9}{2} = 8 \) and \( y = \frac{7 - 9}{2} = -1 \).
05
Back-substitute to find x
For \( y = 8 \), solve \( x^3 = 8 \). The real solution is \( x = 2 \).For \( y = -1 \), solve \( x^3 = -1 \). The real solution is \( x = -1 \). Additionally, there are complex solutions.
06
Find all roots for \( x^3 = -1 \)
The roots of \( x^3 = -1 \) are the cube roots of \( -1 \), which are \( x = -1, \omega, \omega^2 \) where \( \omega = \text{cis}\left(\frac{2\pi}{3}\right)\) and \( \omega^2 = \text{cis}\left(\frac{4\pi}{3}\right) \).
07
Write polynomial as a product of factors
Using the roots found, the polynomial can be factored as:\[ P(x) = (x - 2)(x + 1)(x - \omega)(x - \omega^2)(x - 2)(x + 1) \]
08
Simplify and present the final factorized form
Combine like terms when the polynomial is factored:\[ P(x) = (x - 2)^2 (x + 1)^2 (x^2 + x + 1) \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Zeros of Polynomials
The zeros of a polynomial are the values of the variable that make the polynomial equal to zero. Finding zeros is crucial because they are where the graph of the polynomial function intersects the x-axis. In the example we're dealing with, we have a polynomial in terms of a substitution by setting \( y = x^3 \). This allows us to simplify our polynomial to a quadratic form, \( y^2 - 7y - 8 = 0 \).
- The zeros of this simplified form are determined using the quadratic formula.
- Once the zeros \( y = 8 \) and \( y = -1 \) are found, we substitute back to find the zeros in terms of \( x \).
- For \( y = 8 \), the zero in terms of \( x \) is \( x = 2 \).
- For \( y = -1 \), solving \( x^3 = -1 \) gives us both real and complex roots.
Quadratic Equation
Quadratic equations are polynomials of the form \( ax^2 + bx + c = 0 \). Solving them often involves using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]For our problem, after substituting \( y = x^3 \), the polynomial changes to a quadratic form \( y^2 - 7y - 8 = 0 \). Here's how it's solved:
- Identify the coefficients: here, \( a = 1 \), \( b = -7 \), and \( c = -8 \).
- Plug these into the quadratic formula to find the solutions for \( y \).
- The discriminant \( b^2 - 4ac \) determines the nature of the roots. For this equation, it's \( 81 \), which is a perfect square, giving us two real roots.
Complex Roots
Complex roots occur in pairs and arise from polynomials when the discriminant is negative or when dealing with higher-degree polynomials. In the exercise, when solving \( x^3 = -1 \), we encounter complex solutions. The original polynomial \( P(x) = x^6 - 7x^3 - 8 \) involves these type of roots.
- The complex roots of \( x^3 = -1 \) are \( \omega = \text{cis}\left(\frac{2\pi}{3}\right) \) and \( \omega^2 = \text{cis}\left(\frac{4\pi}{3}\right) \).
- These are derived from Euler's formula, expressing complex numbers on the unit circle.
- These roots contribute significantly to the factorization of the polynomial, revealing more about its algebraic structure.
Cubic Roots
Finding cubic roots essentially means solving equations of degree three, like \( x^3 = c \). In our exercise, once we substitute \( y = x^3 \), finding \( x \) involves looking for roots of these cubic forms:
- For \( y = 8 \), \( x^3 = 8 \) gives the straightforward real root solution \( x = 2 \).
- For \( y = -1 \), \( x^3 = -1 \) requires finding both real and complex roots.
- A key involvement of cubic roots is the cube root of unity properties, seen in the complex roots \( \omega \) and \( \omega^2 \).
