Laws of signs in math

Answer 1

Descartes' Rule of Signs is a used for finding the number of zeroes of a polynomial. Descartes' Rule of Signs wont tell you where the polynomial's zeroes are, it just tells you how many there are. The rule states that if the terms of a single-variable polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is either equal to the number of sign differences between consecutive nonzero coefficients, or less than it by a multiple of 2. Multiple roots of the same value are counted separately. Another way to say this is The number of positive real roots of a polynomial is bounded by the number of changes of sign in its coefficients. Here are some examples to help understand this definition. If f(x) is a linear polynomial (that is, of degree 1), say, f(x) = mx + b then the only root of f(x) is x = -b / m , which is positive only when m and b are of opposite sign Here is an example of the application to a an equation of degree 3. 3x3+5x2-4x+3=0 is a polynomial of degreee 3. The signs, starting at the sign for 3 are (+,+,-,+) where sign change occurs between the the x2 term and the x term and then the x term and the constant term. So there are two sign changes. + to - and - back to + This means there are at most 2 positive real solutions. It general there is that number minus an even number, but in this case that would be 2-2, or 0. Now look at the same exmaple. We want f(-x). To change polynomial f(x) into f(-x), change the signs of all the odd order terms. So f(-x) in your example is -3x3+5x2+4x+3=0 There is a sign change between the 3 and the 5 and then no more. So only one sign change. This changes sign once, so the number of positive roots of f(-x) is one, so there is one negative root of f(x) Now putting both f(x) and f(-x) together, we see there are either zero or two positive roots, either two or zero (respectively) nonreal roots, and exactly one negative root. We know that the nonreal roots always come in pairs and are conjugates. For example if a+bi is a root, so is a-bi. Then if we had 2 real positive roots, we could only have one nonreal root since there are only 3 roots to an equation of degree 3. This is impossible, since we just said nonreal roots come in pairs. So we must have 1 real root and 2 nonreal roots. That tells us the one real root must be negative since we either have 2 or 0 positive roots. If anyone does not fully understand this, please ask and more examples will be given