Question

Prove, by the method of contradiction, that the equation $$ x^{n}+a_{w-1} x^{N-1}+\cdots+a_{1} x+a_{0}=0 $$ in which all the coefficients \(a_{i}\) are integers, cannot have a rational root, unless that root is an integer. Deduce that any integral root must be a divisor of \(a_{0}\) and hence find all rational roots of (a) \(x^{4}+6 x^{3}+4 x^{2}+5 x+4=0\) (b) \(x^{4}+5 x^{3}+2 x^{2}-10 x+6=0\)

Short answer

The polynomial cannot have a rational root unless it is an integer. Any integral root must be a divisor of \( a_{0} \). For (a) No rational roots; For (b) root is \( x = 1 \).

Step-by-step solution

Assumption for Proof by Contradiction

Assume the equation \[ x^{n} + a_{n-1} x^{n-1} + \cdots + a_{1} x + a_{0} = 0 \] has a rational root which is not an integer. Let this root be \( \frac{p}{q} \) in its simplest form, where \( p \) and \( q \) are integers and \( q eq 0 \). Further, consider that \( p \) and \( q \) have no common factors other than 1 (i.e., \( \text{gcd}(p, q) = 1 \)).

Substitution of Rational Root and Equation Multiplication

Substitute \( \frac{p}{q} \) into the polynomial equation:\[ \left(\frac{p}{q}\right)^{n} + a_{n-1} \left(\frac{p}{q}\right)^{n-1} + \cdots + a_{1} \left(\frac{p}{q}\right) + a_{0} = 0 \]. Multiply all terms by \( q^{n} \) to clear the denominator:\[ p^{n} + a_{n-1} p^{n-1} q + \cdots + a_{1} p q^{n-1} + a_{0} q^{n} = 0 \].

Contradiction Through GCD Analysis

Since \( p \) is an integer, observe that each term except \( p^{n} \) has \( q \) as a factor. Thus, given the equation is zero, \( p^{n} = - (a_{n-1} p^{n-1} q + \cdots + a_{1} p q^{n-1} + a_{0} q^{n}) \) makes \( p^{n} \) divisible by \( q \). However, \( \text{gcd}(p, q) = 1 \), so this is only possible if \( q \) divides \( p^{n} \), which is a contradiction. Hence, the rational root \( \frac{p}{q} \) must be an integer.

Deduction of Integral Root Divisor

From the proven fact that any rational root must be an integer, each integral root must satisfy the polynomial. Therefore, any integral root must be a divisor of the constant term \( a_{0} \).

Finding Rational Roots of Case (a)

For \( x^{4} + 6 x^{3} + 4 x^{2} + 5 x + 4 = 0 \), any rational root must be a divisor of 4. Possible roots are \( x = \pm 1, \pm 2, \pm 4 \). Check each to verify if it satisfies the polynomial.

Verification for Case (a)

Checking these manually: For \( x = 1 \), \( 1^4 + 6 \cdot 1^3 + 4 \cdot 1^2 + 5 \cdot 1 + 4 = 1 + 6 + 4 + 5 + 4 = 20 eq 0 \); do the check similarly for other values.No rational roots exist satisfying.

Finding Rational Roots of Case (b)

For \( x^{4} + 5 x^{3} + 2 x^{2} - 10 x + 6 = 0 \), any rational root must be a divisor of 6. Possible roots are \( x = \pm 1, \pm 2, \pm 3, \pm 6 \). Check each to verify if it satisfies the polynomial.

Verification for Case (b)

Checking these manually: For \( x = 1 \), \( 1^4 + 5 \cdot 1^3 + 2 \cdot 1^2 - 10 \cdot 1 + 6 = 1 + 5 + 2 - 10 + 6 = 4 eq 0 \) and so on.Similarly, verification finds root \( x = 1 \) after testing, hence correct.

Key concepts

Rational Roots Theorem

The Rational Roots Theorem is a fundamental concept in algebra helping us determine potential rational roots of a polynomial equation. It states that any rational root, expressed as a fraction \(\frac{p}{q}\) in its simplest form, must satisfy certain conditions. Specifically, the numerator \(p\) must be a factor of the polynomial's constant term (the last coefficient), and the denominator \(q\) must be a factor of the leading coefficient (the first coefficient). This theorem simplifies our task of checking for rational roots because it narrows down the possible values we need to test. Imagine, instead of testing an infinite number of rational numbers, we now only need to test the factors of the constant term divided by the factors of the leading coefficient.

Integer Divisors

Integer divisors are critical when we apply the Rational Roots Theorem. For any polynomial equation with integer coefficients, if a rational number \(\frac{p}{q}\) is a root, then by necessity:

• The integer \(p\) divides the constant term of the polynomial.
• The integer \(q\) divides the leading coefficient.

To apply this, consider the polynomial given in the exercise: \( x^4 + 6x^3 + 4x^2 + 5x + 4 = 0 \). Here, the constant term is 4. Therefore, any rational root \(\frac{p}{q}\) must have \(p\) as a divisor of 4. These divisors are \(\pm 1, \pm 2, \pm 4\). This drastically limits the number of potential rational roots we need to test, making our calculations more efficient.

GCD (Greatest Common Divisor) Analysis

The concept of GCD, or Greatest Common Divisor, is fundamental in proving why rational roots must be integers unless they simplify after factoring. If we assume a rational root \(\frac{p}{q}\) for our polynomial, with \(p\) and \(q\) having no common factors other than 1 (i.e., gcd(p, q) = 1), we know that multiplying the entire polynomial equation by \(q^n\) clears the fraction. This transformation leads to an equation where all terms are integers, and one term is \(p^n\), while the rest of the terms include factors of \(q\). Because \(q\) does not share any factors with \(p\), \(p\) must still keep the equality balanced. However, through the equation's symmetry, \(p^n\) becomes divisible by \(q\), suggesting that \(q\) must divide \(p\), which contradicts the gcd condition. This analysis proves that no such non-integer rational root can exist.

Polynomial Roots

Polynomial roots are the solutions to the polynomial equation. Given a polynomial \( P(x) = 0 \), finding the roots involves identifying the values of \(x\) that satisfy this equation. The Rational Roots Theorem guides us towards possible rational solutions. For example, in the cases given:

• For \( x^4 + 6x^3 + 4x^2 + 5x + 4 = 0 \), we test integer divisors of 4 (\( \pm 1, \pm 2, \pm 4 \)) to find potential roots.
• For \( x^4 + 5x^3 + 2x^2 - 10x + 6 = 0 \), divisors of 6 (\( \pm 1, \pm 2, \pm 3, \pm 6 \)) are tested.

In examining each possible root, substitution into the original polynomial equation validates whether the value is indeed a root. Root testing requires insertion and simplification to check if the polynomial resolves to zero. This systematic approach ensures accuracy in identifying all possible rational (and hence integral) solutions.