Problem 55 - Algebra for College Students

Question

Use the rational root theorem, Descartes' rule of signs, and the theorem on bounds as aids in finding all solutions to each equation. $$x^{3}+x+10=0$$

Step-by-Step Solution

Answer

The solutions to the equation are \(x = -2, 1+2i\), and \(1-2i \).

Step 1: List Potential Rational Roots

According to the Rational Root Theorem, any possible rational root of the polynomial equation must be a factor of the constant term divided by a factor of the leading coefficient. For the equation \[x^{3} + x + 10 = 0\], the constant term is 10 and the leading coefficient is 1. The possible rational roots are thus the factors of 10: \[\text{Factors of 10}: \ \text{±1, ±2, ±5, ±10}\].

Step 2: Apply Descartes' Rule of Signs

Descartes' Rule of Signs helps determine the number of positive and negative real roots. For positive roots, count the sign changes in \[x^{3} + x + 10\]: no sign change, which means 0 positive roots. For negative roots, substitute \[x \rightarrow -x\] in the equation: \[-x^{3} - x + 10\]There is 1 sign change (from \[-x^{3} \rightarrow -x\] and \[-x \rightarrow +10\]), meaning 1 negative root.

Step 3: Test Potential Rational Roots

Evaluate the polynomial at each potential rational root to find which one is a solution. Testing \(x = -2\): \[(-2)^{3} + (-2) + 10 = -8 - 2 + 10 = 0\] So, \(x = -2\) is a root of the polynomial.

Step 4: Perform Synthetic Division

Use synthetic division to divide the polynomial \(x^{3} + x + 10\) by \(x + 2\) to find the remaining polynomial: \[1 \quad 0 \quad 1 \quad 10 \ -2 \quad \downarrow \quad -2 \quad 4 \quad -10 \ 1 \quad -2 \quad 3 \quad 0\] So, the quotient is \(x^{2} - 2x + 5\).

Step 5: Solve the Quadratic Equation

Solve the quadratic equation \(x^{2} - 2x + 5 = 0\) using the quadratic formula: \[x = \frac{2 \times 2 \times \bb{-4}}{2 \times 1} = \frac{2 \times \bb{-4}}{2 \times 1} \] find the discriminant \[4 - 20 = -16\]. Since the discriminant is less than 0, the solutions are complex: \[x = 1 \ \text{±2i}.\]

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a useful tool for finding potential rational solutions (roots) to polynomial equations. This theorem states that any rational root of a polynomial equation, with integer coefficients, is a fraction \( \frac{p}{q} \), where:

• \( p \) is a factor of the constant term (the term without a variable)
• \( q \) is a factor of the leading coefficient (the coefficient of the term with the highest degree)

For example, in the polynomial \( x^3 + x + 10 = 0 \), the constant term is 10 and the leading coefficient is 1. So, the possible rational roots are the factors of 10 divided by the factors of 1, which are \( \pm 1, \pm 2, \pm 5, \pm 10 \). This gives us a way to narrow down and test possible solutions systematically.

Descartes' Rule of Signs

Descartes' Rule of Signs is a method to determine the number of positive and negative real roots of a polynomial. It is based on counting the number of sign changes in the polynomial's coefficients.

• For positive roots, look at the polynomial as is, and count the number of times the signs of the coefficients change.
• For negative roots, substitute \( x \) with \( -x \) in the polynomial and count the sign changes in the new polynomial.

For the polynomial \( x^3 + x + 10 \), there are no sign changes, implying zero positive roots. When we substitute \( -x \) for \( x \), we have \( -x^3 - x + 10 \), which shows one sign change (from \( -x \) to 10), indicating one negative root.

Synthetic Division

Synthetic division is a shortcut method for dividing a polynomial by a binomial of the form \( x - c \). It is especially useful for testing potential roots found by the Rational Root Theorem.

To perform synthetic division:

• Write down the coefficients of the polynomial.
• Bring down the leading coefficient.
• Multiply this coefficient by the root being tested (in our case, \( -2 \)), and write the result under the next coefficient.
• Add this result to the next coefficient and repeat the process for the entire polynomial.

For the polynomial \( x^3 + x + 10 \), when testing \( x = -2 \), the synthetic division process confirms that \( -2 \) is indeed a root because it results in a zero remainder. The quotient we obtain is the simplified polynomial \( x^2 - 2x + 5 \).

Quadratic Formula

When a polynomial is reduced to a quadratic equation (of the form \( ax^2 + bx + c = 0 \)), the quadratic formula can be used to find its roots. The quadratic formula is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For the quadratic equation \( x^2 - 2x + 5 = 0 \), we use the formula with \( a = 1 \), \( b = -2 \) and \( c = 5 \).

• First, find the discriminant: \( b^2 - 4ac = (-2)^2 - 4(1)(5) = 4 - 20 = -16 \)
• Since the discriminant is negative (-16), the roots are complex numbers.
• The roots are: \( x = 1 \pm 2i \), where \( i \) is the imaginary unit.

This means the polynomial \( x^3 + x + 10 = 0 \) has one real root (\( x = -2 \)) and two complex roots (\( x = 1 + 2i \) and \( x = 1 - 2i \)).