In algebra, the rational root theorem (or rational root test, rational zero theorem, rational zero test or p/q theorem) states a constraint on rational solutions of a polynomial equation
a n x n + a n − 1 x n − 1 + ⋯ + a 0 = 0 {\displaystyle a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{0}=0}
with integer coefficients a i ∈ Z {\displaystyle a_{i}\in \mathbb {Z} } and a 0 , a n ≠ 0 {\displaystyle a_{0},a_{n}\neq 0} . Solutions of the equation are also called roots or zeros of the polynomial on the left side. The theorem states that each rational solution x = p q {\displaystyle x={\tfrac {p}{q}}} written in lowest terms (that is, p and q are relatively prime), satisfies:
p is an integer factor of the constant term a0, and q is an integer factor of the leading coefficient an. The rational root theorem is a special case (for a single linear factor) of Gauss's lemma on the factorization of polynomials. The integral root theorem is the special case of the rational root theorem when the leading coefficient is an = 1.
Application The theorem is used to find all rational roots of a polynomial, if any. It gives a finite number of possible fractions which can be checked to see if they are roots. If a rational root x = r is found, a linear polynomial (x – r) can be factored out of the polynomial using polynomial long division, resulting in a polynomial of lower degree whose roots are also roots of the original polynomial.
Cubic equation The general cubic equation
a x 3 + b x 2 + c x + d = 0 {\displaystyle ax^{3}+bx^{2}+cx+d=0}
with integer coefficients has three solutions in the complex plane. If the rational root test finds no rational solutions, then the only way to express the solutions algebraically uses cube roots. But if the test finds a rational solution r, then factoring out (x – r) leaves a quadratic polynomial whose two roots, found with the quadratic formula, are the remaining two roots of the cubic, avoiding cube roots.
Proofs
Elementary proof Let P ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0 {\displaystyle P(x)\ =\ a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}} with a 0 , … , a n ∈ Z , a 0 , a n ≠ 0. {\displaystyle a_{0},\ldots ,a_{n}\in \mathbb {Z} ,a_{0},a_{n}\neq 0.} Suppose P(p/q) = 0 for some coprime p, q ∈ ℤ:
P ( p q ) = a n ( p q ) n + a n − 1 ( p q ) n − 1 + ⋯ + a 1 ( p q ) + a 0 = 0. {\displaystyle P\left({\tfrac {p}{q}}\right)=a_{n}\left({\tfrac {p}{q}}\right)^{n}+a_{n-1}\left({\tfrac {p}{q}}\right)^{n-1}+\cdots +a_{1}\left({\tfrac {p}{q}}\right)+a_{0}=0.}
To clear denominators, multiply both sides by qn:
… excerpt ends here. Continue reading the full article.
