Problem 7 - Basic Technical Mathematics with Calculus

Question

Find the remaining roots of the given equations using synthetic division, given the roots indicated. $$x^{3}-5 x^{2}+2 x+8=0 \quad\left(r_{1}=2\right)$$

Step-by-Step Solution

Answer

The remaining roots are \(x = 4\) and \(x = -1\).

Step 1: Set Up Synthetic Division

We will use synthetic division to divide the polynomial \(x^3 - 5x^2 + 2x + 8\) by \(x - r_1\), where \(r_1 = 2\). Write down the coefficients of the polynomial: \(1, -5, 2, 8\). We will use \(2\) (the given root) for synthetic division.

Step 2: Perform Synthetic Division

Write down \(2\) (the given root) on the left. Bring down the leading coefficient \(1\) to the bottom row:\[\begin{array}{r|rrrr}2 & 1 & -5 & 2 & 8 \ & & 2 & -6 & -8 \\hline & 1 & -3 & -4 & 0 \\end{array}\]Multiply \(2\) by \(1\) and write it under \(-5\); add to get \(-3\). Multiply \(2\) by \(-3\) and write it under \(2\); add to get \(-4\). Multiply \(2\) by \(-4\) and place it under \(8\); add to get \(0\). Since the remainder is \(0\), the division is exact, confirming \(x - 2\) is a factor.

Step 3: Identify Quotient Polynomial

The result of the synthetic division, \(1, -3, -4\), represents the coefficients of the quotient polynomial. Thus, the quotient is \(x^2 - 3x - 4\). This is a quadratic equation representing the remaining factors of the original polynomial.

Step 4: Solve Quadratic Equation

To find the remaining roots, solve \(x^2 - 3x - 4 = 0\). We will use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -3\), \(c = -4\).Calculate the discriminant: \(b^2 - 4ac = (-3)^2 - 4 \times 1 \times (-4) = 9 + 16 = 25\).Calculate the roots:\(x = \frac{3 \pm \sqrt{25}}{2}\).Simplify:\(x = \frac{3 \pm 5}{2}\).This gives us roots \(x = 4\) and \(x = -1\).

Step 5: Confirm All Roots

We now have all roots of the equation \(x^3 - 5x^2 + 2x + 8 = 0\) as \(x = 2\), \(x = 4\), and \(x = -1\). These satisfy the original equation.

Key Concepts

Polynomial Roots

Polynomial roots are the values of \(x\) that satisfy the equation when set to zero. In this exercise, the polynomial \(x^3 - 5x^2 + 2x + 8 = 0\) has three roots. These roots are solutions to the polynomial equation that make the entire expression equal to zero.
Roots are fundamental because they give insights into the behavior of the polynomial. For a cubic polynomial like this one, there can be up to three roots.
If roots are real and distinct, each root crosses the x-axis. In contrast, complex roots occur in conjugate pairs and do not intersect the x-axis.

Quadratic Equation

After simplifying our cubic polynomial using synthetic division, we obtain a quadratic equation: \(x^2 - 3x - 4 = 0\). Quadratic equations have the standard form \(ax^2 + bx + c = 0\). Here, \(a = 1\), \(b = -3\), and \(c = -4\).
Solving quadratic equations can provide the remaining roots of the polynomial. There are several methods for this:

• Factoring: Breaks down the equation into simpler components.
• Completing the Square: Rewriting the equation to form an easy-to-solve perfect square.
• Quadratic Formula: A universal method employing all three coefficients.

All these methods ultimately help in finding the values of \(x\) that satisfy the equation.

Discriminant

The discriminant is part of the quadratic formula and is crucial for determining the nature of the roots. It is calculated as \(b^2 - 4ac\). In our case, with coefficients \(b = -3\), \(a = 1\), \(c = -4\), we calculate:
\[ (-3)^2 - 4 \times 1 \times (-4) = 9 + 16 = 25 \]
This value (25) tells you about the roots:

• If the discriminant is positive, as in this case, the equation has two real and distinct roots.
• If it is zero, the equation has one real double (or repeated) root.
• If negative, the roots are complex or imaginary.

Understanding the discriminant provides insight into the solutions without solving the entire equation.

Factorization

Factorization involves breaking down the polynomial into products of simpler polynomials. The goal is to rewrite the equation in a form that reveals each root.
Initially, the given polynomial is not factorized. By employing synthetic division with one root \(x - 2\), we simplify it to a quadratic \(x^2 - 3x - 4\).
Factoring this quadratic further, we seek two numbers whose product is \(-4\) (the constant term) and whose sum is \(-3\) (the coefficient of \(x\)). The numbers \(-4\) and \(1\) fit:

• Expression: \((x - 4)(x + 1)\)

Thus, the roots from factorization are \(x = 4\) and \(x = -1\).
Ultimately, this reveals that all roots of the original cubic equation are \(x = 2\), \(x = 4\), and \(x = -1\), confirming our complete solution.