Problem 63 - A Graphical Approach to Precalculus with Limits

Question

For each polynomial at least one zero is given. Find all others analytically. $$P(x)=x^{3}-2 x+1 ; 1$$

Step-by-Step Solution

Answer

Roots are 1, \( \frac{-1 + \sqrt{5}}{2} \), and \( \frac{-1 - \sqrt{5}}{2} \).

Step 1: 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.

Step 2: 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.

Step 3: 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 \( x = \frac{-1 - \sqrt{5}}{2} \).

Step 4: 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.

Key Concepts

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.

Each step is simple, ensuring you work efficiently without writing too much. If the remainder is zero after the process, the divisor binomial actually divides the polynomial completely. The last line you get, if the remainder is zero, is your quotient. In our example, synthetic division allowed us to reduce a cubic polynomial to a quadratic polynomial, making it much easier to handle.

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.

For our polynomial’s quadratic factor \(x^2 + x - 1\), substituting the coefficients \(a = 1\), \(b = 1\), \(c = -1\) into the quadratic formula provided the roots \(\frac{-1 + \sqrt{5}}{2}\) and \(\frac{-1 - \sqrt{5}}{2}\). These represent the points where the polynomial crosses the x-axis.

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.

In the given example, after identifying 1 as a root of \(x^3 - 2x + 1\), synthetic division revealed a quadratic \(x^2 + x - 1\). This factor was then further analyzed using the quadratic formula to find the remaining roots. Defining the complete factors of a polynomial is like peeling layers of complexity, simplifying until only the basic components—roots—are left.