Problem 48 - 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 $$x^{3}-31 x+30=0$$

Step-by-Step Solution

Answer

The solution of the equation \(x^{3}-31 x+30=0\) is \(x=\frac{1}{3}\), \(\frac{-1 + \sqrt{361}}{2}\), and \(\frac{-1 - \sqrt{361}}{2}\).

Step 1: Synthetic Division with \(x=4\)

First, use synthetic division to check if \(4\) is the solution of the given equation. Arrange the digits below the line and start the synthetic division process.\[\begin{array}{c|ccc}4 & 1 & -31 & 30 \ & & 4 & -108 \\hline & 1 & -27 & -78 \\end{array}\]Since the remainder is not zero, \(x=4\) is not a solution.

Step 2: Synthetic Division with \(x=-6\)

Next, use synthetic division to check if \(-6\) is the solution of the given equation.\[\begin{array}{c|ccc}-6 & 1 & -31 & 30 \ & & -6 & 222 \\hline & 1 & -37 & 252 \\end{array}\]Since the remainder is not zero, \(x=-6\) is not a solution.

Step 3: Synthetic Division with \(x=-\frac{3}{2}\)

Next, use synthetic division to check if \(-\frac{3}{2}\) is the solution of the given equation.\[\begin{array}{c|ccc}-\frac{3}{2} & 1 & -31 & 30 \ & & -\frac{3}{2} & 48 \\hline & 1 & -\frac{47}{2} & \frac{60}{1} \\end{array}\]Since the remainder is not zero, \(x=-\frac{3}{2}\) is not a solution.

Step 4: Synthetic Division with \(x=\frac{1}{3}\)

Next, use synthetic division to check if \(\frac{1}{3}\) is the solution of the given equation.\[\begin{array}{c|ccc}\frac{1}{3} & 1 & -31 & 30 \ & & \frac{1}{3} & -10 \\hline & 1 & -\frac{92}{3} & 0 \\end{array}\]Since the result of the division process is zero, we can say that \(x=\frac{1}{3}\) is a solution for the given equation. Now you can use this solution to completely factor the polynomial.

Step 5: Synthetic Division with \(x=\sqrt{3}\)

Again use synthetic division to check if \(\sqrt{3}\) is a solution of the given equation.\[\begin{array}{c|ccc}\sqrt{3} & 1 & -31 & 30 \ & & \sqrt{3} & 0.769 \\hline & 1 & -\sqrt{3} & 0.769 \\end{array}\]Since the remainder is not zero, \(x=\sqrt{3}\) is not a solution.

Step 6: Synthetic Division with \(x=2-\sqrt{5}\)

Lastly, use synthetic division to check if \(2-\sqrt{5}\) is a solution of the equation, we use the polynomial equation \(x^2-4x+1=0\) as an example.\[\begin{array}{c|ccc}2-\sqrt{5} & 1 & -31 & 30 \ & & 2-\sqrt{5} & 30-(2-\sqrt{5}) \\hline & 1 & -33+\sqrt{5} & 0 \\end{array}\]Since the remainder is not zero, \(x=2-\sqrt{5}\) is not a solution.

Step 7: Factoring The Polynomial

Now using the solution found in Step 4, \(x=\frac{1}{3}\), you can factor the original polynomial equation.The equation \(x^{3}-31 x+30=0\) factors to \((x-\frac{1}{3})(x^2 + \frac{1}{3}x -90)\).The solutions to the equation are therefore \(x=\frac{1}{3}\), \(\frac{-1 + \sqrt{361}}{2}\), and \(\frac{-1 - \sqrt{361}}{2}\).

Key Concepts

Polynomial Equation

A polynomial equation consists of a polynomial expression that is equal to zero. In general, a polynomial is an expression that has the form:

• It is a combination of variables, exponents, and coefficients.
• Each term in a polynomial is made up of a variable raised to a power, multiplied by a coefficient.
• Polynomials can have different degrees. The degree is determined by the highest power of the variable in the expression.

For the equation given in the exercise, \(x^3 - 31x + 30 = 0\), it is a third-degree polynomial equation. This means there is a term with \(x\) to the power of 3, making it cubic. When we solve a polynomial equation, we are looking for the values of \(x\) which make the equation true, that is making the whole expression equal to zero.

Factorization

Factorization is the process of breaking down a polynomial into simpler pieces, called factors, that when multiplied together give you the original polynomial. The goal is to express the polynomial in that simpler form.

• Factoring helps in finding the roots or solutions of polynomial equations.
• If a polynomial can be factored completely, you can often find all its solutions by setting each factor to zero.

In the provided exercise, after finding that \(x = \frac{1}{3}\) is a solution, the polynomial \(x^3 - 31x + 30\) was factored into \((x - \frac{1}{3})(x^2 + \frac{1}{3}x - 90)\). By factoring, it makes solving the polynomial equation for the remaining roots more manageable.

Real Solutions

The real solutions of a polynomial equation are the \(x\)-values that satisfy the equation such that all terms result in zero. Real solutions can be visualized on a graph as the points where the graph of the polynomial intersects the \(x\)-axis. Identifying real solutions involves:

• Finding values for which the entire polynomial equates to zero.
• These values can be single numbers called roots or zeros.

In this exercise, one of the real solutions identified was by using synthetic division, where \(x = \frac{1}{3}\) results in zero, confirming it's a root. Other real solutions can be found by solving the quadratic part derived from the factorization \((x^2 + \frac{1}{3}x - 90)\). The equation gets more straightforward once broken down into these simpler components.

Algebraic Solution Steps

Solving polynomial equations algebraically involves a specific set of steps, particularly when using synthetic division. The steps help determine if a given \(x\) value is an actual solution and aid in subsequent factorization. Here's a simplified procedure:

• Perform synthetic division with a trial \(x\)-value, following these substeps:
• Arrange the polynomial coefficients.
• Bring down the leading coefficient to the bottom row.
• Multiply the trial value by this number and add the result to the next coefficient.
• Repeat the process for all coefficients.
• A zero remainder indicates the trial value is a solution.

The exercise showed this method successfully with \(x = \frac{1}{3}\), resulting in a zero remainder, confirming it's a valid root. This method also assists in re-factoring the polynomial, making it easier to find additional real solutions.