Chapter 4: Problem 63
For each polynomial at least one zero is given. Find all others analytically. $$P(x)=x^{3}-2 x+1 ; 1$$
Short Answer
Expert verified
Roots are 1, \( \frac{-1 + \sqrt{5}}{2} \), and \( \frac{-1 - \sqrt{5}}{2} \).
Step by step solution
01
Verify the given root
We are given that 1 is a root of the polynomial \( P(x) = x^3 - 2x + 1 \). To verify, substitute \( x = 1 \) into the polynomial: \[ P(1) = 1^3 - 2 \times 1 + 1 = 1 - 2 + 1 = 0. \] Since \( P(1) = 0 \), 1 is indeed a root of the polynomial.
02
Perform synthetic division
Since we have verified that 1 is a root of the polynomial, we can use synthetic division to divide \( P(x) \) by \( x - 1 \). The coefficients of \( P(x) \) are [1, 0, -2, 1]. 1. Drop the 1 down: [1, 0, -2, 1]2. Multiply 1 by 1 (the root), add to 0: new row [1, 1]3. Multiply 1 by 1, add to -2: new row [1, 1, -1]4. Multiply 1 by -1, add to 1: new row [1, 1, -1, 0]The quotient is \( x^2 + x - 1 \), with a remainder of 0, confirming \( x - 1 \) is a factor.
03
Find roots of the quadratic factor
The division gives the quadratic \( x^2 + x - 1 = 0 \). To find its roots, use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1, b = 1, c = -1 \).Substitute these values into the formula: \[ x = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times (-1)}}{2 \times 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}. \] Thus, the roots are \( x = \frac{-1 + \sqrt{5}}{2} \) and \( x = \frac{-1 - \sqrt{5}}{2} \).
04
Conclude with all roots
The roots of the polynomial \( P(x) = x^3 - 2x + 1 \) are 1, \( \frac{-1 + \sqrt{5}}{2} \), and \( \frac{-1 - \sqrt{5}}{2} \). These cover all the zeros of the polynomial.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Synthetic Division
Synthetic division is a handy tool for solving polynomials, especially when you know at least one root. It simplifies dividing a polynomial by a linear binomial like \(x - c\), where \(c\) is a known root. Unlike long division, synthetic division is a more streamlined and quicker method. Here's how it works:
- List the coefficients of the polynomial you wish to divide.
- Bring down the leading coefficient as it is.
- Multiply this by the known root \(c\) and write the result under the second coefficient.
- Add down the second column to find a new entry.
- Repeat this process for all columns.
Quadratic Formula
The quadratic formula is a universal solution for quadratic equations in the form \(ax^2 + bx + c = 0\). This formula is derived from completing the square of a quadratic expression and helps find the roots of any quadratic, even when factoring is not straightforward. Here's the formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Each component of the formula plays a role:
- \(-b\) gives the opposite of the linear coefficient, shifting the result one step closer to the roots.
- The discriminant \(b^2 - 4ac\) determines the nature of the roots: real and distinct, real and equal, or complex.
- \(\pm\) means you'll generally get two possible values: one for addition and one for subtraction, indicating the equation's two roots.
- Dividing by \(2a\) scales the roots by doubling the leading coefficient's influence.
Polynomial Factoring
Polynomial factoring involves expressing a polynomial as a product of its roots when possible. It's an efficient way to simplify polynomials and solve equations. Factoring reduces a complex polynomial into manageable linear (or sometimes quadratic) factors:
- If a given polynomial can be factored into simpler polynomials, it is broken down into linear factors for real roots, or irreducible factors for imaginary ones.
- Once factored, setting each factor to zero helps find its roots—points where the polynomial equals zero.
- Factoring also confirms if known roots are accurate, since correct factors ensure no remainder is left after division.