I'm still researching, but, at the moment, I would give credit to Rene Descartes who noted the relationship between the rational roots of a polynomial and the coefficients of the first and last terms. See Google Books link below (bottom of page 243 of text.
http://books.google.com/books?id=5NIDb84jOmcC&pg=PA243&lpg=PA243&dq=history+rational+roots+theorem&source=bl&ots=KYMZS2m4RE&sig=WZS7COaip1KcW4ZfLSHBgcp-Co8&hl=en&ei=3sMmStOACJqutAPFvMHFBg&sa=X&oi=book_result&ct=result&resnum=10#PPA243,M1
Rene' Descartes is credited with founding rational root theorem. He also created the rules of signs to be used with solving equations.
In algebra, the rational root theorem (or rational root test, rational zero theorem or rational zero test) states a constraint on rational solutions (or roots) of a polynomialequationwith integer coefficients.If a0 and an are nonzero, then each rational solution x, when written as a fraction x = p/q in lowest terms (i.e., the greatest common divisor of p and q is 1), satisfiesp is an integer factor of the constant term a0, andq 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 lemmaon the factorization of polynomials. The integral root theorem is a special case of the rational root theorem if the leading coefficient an = 1.
1. Quadratic Formula 2. Rational Root Theorem 3. Zero Product Theorem
2 = 2/1 is rational. Sqrt(2) is not rational.
To determine whether a polynomial equation has imaginary solutions, you must first identify what type of equation it is. If it is a quadratic equation, you can use the quadratic formula to solve for the solutions. If the equation is a cubic or higher order polynomial, you can use the Rational Root Theorem to determine if there are any imaginary solutions. The Rational Root Theorem states that if a polynomial equation has rational solutions, they must be a factor of the constant term divided by a factor of the leading coefficient. If there are no rational solutions, then the equation has imaginary solutions. To use the Rational Root Theorem, first list out all the possible rational solutions. Then, plug each possible rational solution into the equation and see if it is a solution. If there are any solutions, then the equation has imaginary solutions. If not, then there are no imaginary solutions.
No, the square root of 1000 is not rational.
No, the square root of 3 is not rational.
is the square root of 3 rational
The square root of 4 is 2. 2 is a rational number so they square root of 4 is rational.
The square root of 48 is a rational or irrational
No, the square root of 1500 is not a rational number.
It is rational. The root of a perfect square, such as 4, is rational; the root of any positive integer that is not a perfect square is an irrational number.