Problem 51 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use synthetic division to show that \(x\) is a solution of the thirddegree polynomial equation, and use the result to factor the polynomial completely. List all the real solutions of the equation. Value of \(x\) \(x=4\) \(x=-6\) \(x=-\frac{3}{2}\) \(x=\frac{1}{3}\) \(x=\sqrt{3}\) \(x=2-\sqrt{5}\) Polynomial Equation $$x^{3}+2 x^{2}-3 x-6=0$$

Step-by-Step Solution

Answer

The real solutions to the equation \(x^{3}+2x^{2}-3x-6=0\) are \(x=-6\), \(x=-\frac{3}{2}\), and \(x=2\). The completely factored form of the polynomial is \(x^{3}+2x^{2}-3x-6 = (x+6)(x+\frac{3}{2})(x-2)\)

Step 1: Setting up the Synthetic Division Table

We take the coefficients of our polynomial \(x^{3}+2x^{2}-3x-6\) which are 1, 2, -3, -6 and set up our synthetic division table. We will then use \(x=4\) first and set it to our synthetic division factor.

Step 2: Using Synthetic Division

Conduct synthetic division: Write down the first coefficient (1), multiply the dividing factor (4) by this coefficient and write the result below the second coefficient (2). Add vertically to get the new second coefficient (6). Repeat this process with each subsequent coefficient. The last number obtained is the remainder.

Step 3: Interpreting the Result of Synthetic Division

We get coefficients 1, 6,21 and 78, with the last coefficient being the remainder and because it is not 0, \(x=4\) is not a root of the original polynomial equation.

Step 4: Repeat Synthetic Division for Other Values

Repeat the process of synthetic division for other given values of \(x\). After going through all the given values, it is found that \(x=-6\) and \(x=-\frac{3}{2}\) yield a remainder of 0 and are therefore roots of the original equation.

Step 5: Factoring the Polynomial

Given that \(x=-6\) and \(x=-\frac{3}{2}\) are roots, we can factor the polynomial to \(x^{3}+2x^{2}-3x-6 = (x+6)(x+3/2)(x-a)\). The value of \(x-a\) can be found by long division of \((x+6)(x+3/2)\) from the original polynomial, yielding \(x-a = x-2\). Thus the fully factored polynomial is \(x^{3}+2x^{2}-3x-6 = (x+6)(x+3/2)(x-2)\)

Key Concepts

Polynomial Equation

A polynomial equation is an expression that consists of variables, coefficients, and exponents. The general form of a polynomial equation is written as: \[ a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 = 0 \] where \( a_n, a_{n-1}, \ldots, a_0 \) are coefficients and \( x \) is the variable. The degree of the polynomial is determined by the highest power of \( x \), which in this case is 3, indicating a cubic polynomial.
For example, in the exercise \( x^3 + 2x^2 - 3x - 6 = 0 \) is a cubic polynomial equation. Our task is to find the roots, or solutions, of this equation using synthetic division. Synthetic division simplifies the process of testing potential solutions (values of \( x \)). If a value results in a remainder of zero when performing synthetic division, that value is a root of the polynomial equation.

Factoring Polynomials

Factoring polynomials involves breaking down the polynomial into simpler, multiplied components, or factors. The goal is to express the original polynomial as a product of its linear or quadratic factors, making it easier to identify the solutions to the equation.
The exercise illustrates factoring by finding roots first through synthetic division. If a root is found, the polynomial can be factored using this root. For example, if \( x = -6 \) is a root, then \( x + 6 \) is a factor.

• Determine if a given \( x \, \) value is a root using synthetic division.
• If it is a root, write one factor as \( (x - \text{root}) \).
• Continuing dividing the polynomial by the discovered factors until fully factored.

For \( x^3 + 2x^2 - 3x - 6 \), we find \( x = -6 \) and \( x = -\frac{3}{2} \) as roots, then factor: \((x+6)(x+\frac{3}{2})(x-a)\). Finally, by division, \( x-a = x-2 \) completes the factorization.

Roots of Polynomial

The roots of a polynomial are the values of \( x \) that make the polynomial equation equal to zero. Identifying roots is crucial as these are the x-values where the polynomial graph intersects the x-axis.
From the exercise, through synthetic division, \( x = -6 \) and \( x = -\frac{3}{2} \) were confirmed as roots. For each valid root, one obtains a factor of the polynomial.

• A root \( r \) transforms the polynomial equation \( f(x) = 0 \) into \((x - r)f'(x) = 0\).
• Continual checking of given values with synthetic division confirms whether each is a valid root.
• All real solutions can be listed, providing a complete solution to the polynomial equation.

In conclusion, a fully factored polynomial gives all roots: \( x = -6, -\frac{3}{2}, 2 \). This process underscores the significance of roots and their role in understanding equation solutions.

Remainder Theorem

The Remainder Theorem is a useful tool in polynomial division, stating that for any polynomial \( f(x) \), the remainder of the division of \( f(x) \) by \( x - c \) is \( f(c) \).

This theorem is instrumental in synthetic division, as it helps rapidly verify if a number is a root of the polynomial. If the remainder is zero after synthetic division, then \( x = c \) is indeed a root.

• The theorem provides a quick method to test potential solutions without complete division.
• Each division step allows for a check of polynomial function values at specific points.
• It simplifies the verification of roots, as seen in our exercise strategy with different values.

In the exercise, employing the Remainder Theorem helped find valid roots (\( x = -6 \) and \( x = -\frac{3}{2} \)). This method substantially reduces the time and calculation effort required in polynomial root finding.