Problem 13 - Precalculus

Question

Show that each equation has no rational roots. $$x^{3}+x^{2}-x+1=0$$

Step-by-Step Solution

Answer

The equation has no rational roots because neither \( x = 1 \) nor \( x = -1 \) satisfies it.

Step 1: Understanding Rational Root Theorem

The Rational Root Theorem states that if a polynomial has a rational root \( \frac{p}{q} \), then \( p \) must divide the constant term and \( q \) must divide the leading coefficient. For the polynomial \( x^3 + x^2 - x + 1 = 0 \), the constant term is 1 and the leading coefficient is 1. Hence, any rational root must be \( \pm 1 \).

Step 2: Testing Rational Roots

According to the Rational Root Theorem, the potential rational roots are \( x = 1 \) and \( x = -1 \). We substitute these values into the equation \( x^3 + x^2 - x + 1 = 0 \).

Step 3: Substitute \( x = 1 \)

Plug \( x = 1 \) into \( x^3 + x^2 - x + 1 \), we get: \[ 1^3 + 1^2 - 1 + 1 = 1 + 1 - 1 + 1 = 2 \] Since 2 is not 0, \( x = 1 \) is not a root.

Step 4: Substitute \( x = -1 \)

Plug \( x = -1 \) into \( x^3 + x^2 - x + 1 \) , we get: \[-1^3 + (-1)^2 - (-1) + 1 = -1 + 1 + 1 + 1 = 2 \] Since 2 is not 0, \( x = -1 \) is not a root.

Step 5: Conclusion

Neither potential rational root \( x = 1 \) nor \( x = -1 \) satisfies the equation \( x^3 + x^2 - x + 1 = 0 \). Therefore, this polynomial has no rational roots.

Key Concepts

Polynomial Roots

Polynomial roots are the solutions to a polynomial equation where the polynomial equals zero. In other words, if you have a polynomial \( f(x) \), then roots are values of \( x \) that make \( f(x) = 0 \). Understanding this is crucial because finding roots helps in solving polynomial equations.
When working with polynomials, especially third-degree polynomials like \( x^3 + x^2 - x + 1 = 0 \), roots can be integers or fractions. These fractions, if they are roots, are made up of integers—specifically, rational numbers, which leads us to our next topic.

Rational Numbers

Rational numbers are numbers that can be written as fractions with integer numerators and non-zero integer denominators. For instance, \( \frac{3}{4} \) is a rational number while \( \sqrt{2} \) is not because it can't be expressed as a simple fraction of integers.

• They include integers because you can write any integer \( n \) as \( \frac{n}{1} \).
• In the context of polynomial roots, rational numbers are significant because the Rational Root Theorem helps identify possible roots that are rational numbers.

If a rational number is a root, the polynomial will equal zero at that value.

Testing Roots

Testing roots is like investigating which values actually solve the polynomial equation by plugging them into the polynomial. Let's apply this to our polynomial. Using the Rational Root Theorem, potential rational roots of \( x^3 + x^2 - x + 1 = 0 \) were identified as \( x = 1 \) and \( x = -1 \).
When testing these values, we plug them into the polynomial to see if they result in zero:

• Substituting \( x = 1 \) resulted in \( 2 \), not zero, so \( x = 1 \) isn't a root.
• Substituting \( x = -1 \) also resulted in \( 2 \), so \( x = -1 \) isn’t a root either.

Through testing, neither value turned out to be a root of our polynomial, supporting the conclusion of no rational roots.

Leading Coefficient

In any polynomial, the leading coefficient is the non-zero coefficient of the term with the highest degree. For example, in \( x^3 + x^2 - x + 1 \), the leading coefficient is \( 1 \) because \( x^3 \) is the term with the highest power.
The leading coefficient plays a significant role when using the Rational Root Theorem. This theorem asserts that for a polynomial with integer coefficients, any potential rational root \( \frac{p}{q} \) must have \( p \) dividing the constant term and \( q \) dividing the leading coefficient.

• This means if the leading coefficient is \( 1 \), any possible rational root must also be an integer, typically \( \pm 1 \) if the constant term is also \( 1 \).

Understanding the leading coefficient helps limit the possible values to test as potential roots, making problem-solving more efficient.