Question

Find the rational roots, if any, of the following: (a) \(x^{3}-4 x^{2}-x+6=0\) (c) \(x^{3}+\frac{3}{4} x^{2}-\frac{3}{8} x-\frac{1}{8}=0\) (b) \(8 x^{3}+6 x^{2}-3 x-1=0\) (d) \(x^{4}-6 x^{3}+7 \frac{3}{4} x^{2}-\frac{3}{2} x-2=0\)

Short answer

Rational roots are x=1 (a), x=1 (c), x=-1 (b), and x=2 (d).

Step-by-step solution

Understanding the Rational Root Theorem

According to the Rational Root Theorem, any possible rational root of the polynomial \(a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0\) is in the form of \(\frac{p}{q}\), where \(p\) is a factor of the constant term \(a_0\) and \(q\) is a factor of the leading coefficient \(a_n\). We'll apply this theorem to all given polynomials.

Finding Rational Roots for Part (a)

For the polynomial \(x^3 - 4x^2 - x + 6 = 0\), the constant term \(a_0 = 6\) has factors \(\pm 1, \pm 2, \pm 3, \pm 6\), and the leading coefficient \(a_3 = 1\) has factors \(\pm 1\). Possible rational roots are \(\pm 1, \pm 2, \pm 3, \pm 6\). Testing these values: \(x = 1\) satisfies the equation \((1)^3 - 4(1)^2 - 1 + 6 = 0\). Thus, \(x = 1\) is a rational root.

Finding Rational Roots for Part (c)

For the polynomial \(x^3 + \frac{3}{4}x^2 - \frac{3}{8}x - \frac{1}{8} = 0\), the constant term \(a_0 = -\frac{1}{8}\) has rational factors \(\pm \frac{1}{8}, \pm \frac{1}{1}\), and the leading coefficient \(a_3 = 1\) has factors \(\pm 1\). Rational root candidates are \(\pm \frac{1}{8}, \pm 1\). Testing \(x = 1\) confirms it is a root, \((1)^3 + \frac{3}{4}(1)^2 - \frac{3}{8}(1) - \frac{1}{8} = 0\).

Finding Rational Roots for Part (b)

For the polynomial \(8x^3 + 6x^2 - 3x - 1 = 0\), the constant term \(a_0 = -1\) has factors \(\pm 1\), and the leading coefficient \(a_3 = 8\) has factors \(\pm 1, \pm 2, \pm 4, \pm 8\). Rational root candidates are \(\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8}\). Testing \(x = -1\) satisfies the equation, thus \(x = -1\) is a rational root.

Finding Rational Roots for Part (d)

For the polynomial \(x^4 - 6x^3 + \frac{31}{4}x^2 - \frac{3}{2}x - 2 = 0\), the constant term \(a_0 = -2\) has factors \(\pm 1, \pm 2\), and the leading coefficient \(a_4 = 1\) has factors \(\pm 1\). Thus, rational roots could be \(\pm 1, \pm 2\). Testing \(x = 2\) satisfies the equation, \((2)^4 - 6(2)^3 + \frac{31}{4}(2)^2 - \frac{3}{2}(2) - 2 = 0\). Hence, \(x = 2\) is a rational root.

Key concepts

Polynomial Equations

Polynomial equations are expressions involving variables and coefficients, arranged in terms of powers. They form the foundation of algebra and come in different forms depending on the highest power of the variable. For example, equations can be linear, quadratic, cubic, etc., based on whether the highest power is 1, 2, or 3 respectively. In the case of the given exercise, you are dealing with cubic and quartic polynomials, meaning their highest powers are 3 and 4.

Polynomials can be represented as \[ a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0 = 0 \] where each term has a coefficient (\(a_n\)) and a variable raised to a power (\(x^n\)). The goal when solving such polynomials is usually to find the roots or solutions of the equation. These roots are the values of \(x\) that would make the entire polynomial equal to zero.

Factoring Polynomials

Factoring polynomials involves breaking down a larger polynomial into simpler components, often called factors, that can be multiplied to yield the original polynomial. It's like reverse multiplication.

Factoring is especially useful in solving polynomial equations because it can simplify the equation into a product of simpler expressions. For example, the expression \( x^3 - 4x^2 - x + 6 \) can be factored if we find the roots, allowing for easier manipulation.

Polynomials that can be factored into linear expressions provide solutions directly. For example, if \( (x - a)(x - b)(x - c) = 0 \), then \( x = a, x = b, ext{ and } x = c \) are the roots. Sometimes, factoring requires recognizing special patterns like the difference of squares or perfect square trinomials, or it might require using the Rational Root Theorem to find one root.

Rational Roots

The Rational Root Theorem is a very handy tool to find possible rational roots of a polynomial equation. According to the theorem, if the polynomial has any rational root, it will be in the form of \( \frac{p}{q} \) where \( p \) is a factor of the constant term and \( q \) is a factor of the leading coefficient.

For example, in the polynomial \( 8x^3 + 6x^2 - 3x - 1 = 0 \), the constant term is \( -1 \) and the leading coefficient is \( 8 \). Thus, the possible rational roots can be formed using factors of these numbers: \( \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8} \). After testing these, we discovered \( x = -1 \) is a rational root.

Using this theorem helps narrow down potential solutions significantly, making it easier to then verify through substitution or further factoring.

Algebraic Solutions

Finding algebraic solutions to polynomial equations involves solving for the roots or zeroes of the equation, which are the values of the variable that result in a zero value for the polynomial. Algebraic solutions can be rational (like integers or fractions) or irrational (like square roots or other roots).

In algebra, you often leverage techniques like factoring, the use of the Rational Root Theorem, and even synthetic division to solve equations. For instance, after using the Rational Root Theorem to find a rational root in our original equations, that root was then used to simplify the polynomial further by factoring, making it easier to solve for the remaining roots.

It is crucial to always verify your solutions by substituting them back into the original equation to confirm that they satisfy the equation. This ensures not only that the solution is correct but also that no possible solutions have been overlooked.