When solving third degree polynomials, such as 0 = 2x^3-9x^2-5x-550, is there an easier way of finding the test root?Without having to substitute each factor of 550 / each factor of 2 until the...
When solving third degree polynomials, such as 0 = 2x^3-9x^2-5x-550, is there an easier way of finding the test root?
Without having to substitute each factor of 550 / each factor of 2 until the function equals 0. This can be very time consuming, and we do not have access to a graphing calculator or wolfram alpha.
Two other things you might try/consider:
(1) Descarte's rule of signs canlet you know thenumber of positive or negative roots the polynomial can have. (See link) If the coefficients are all of the same sign, then there are no positive roots. In this case there is one sign change so there is 1 real positive root.
(2) You only need to find a value for x where the function is negative and a value for x where the function is positive. Then you know there is a zero in the interval between these values (e.g. f(-1)=5 and f(1)=-19, then there is a zero between 1 and -1). This only helps narrow the range of the search, and does little if the zero is irrational, but it helps you from having to try all values.
Not really. Graphing is the easiest way. There is a java applet called Geogebra that can graph, and it is free, it runs on PC's, Macs, and Linux. You can get it at geogebra.org. Try that. You could factor out the two, since you are not trying to factor the polynomial, just find roots. I can tell you that this method only finds the rational roots. This cubic does not have any rational roots.
I would add 550 to both sides of the equation.
2x^3 – 9x^2 -5x = 550
I would factor out an x from the left side of the equation.
x (2x^2 – 9x^1 -5) = 550
I would see that -10 + 1 = -9.
x (2x^2 – 10x^1+1x^1 - 5) = 550
I would factor out a 2x from the 1st part of equation and a 1 from 2nd part.
(x) 2x^1 (x^1 – 5) + 1 (x^1 - 5) = 550
I would rewrite the equation.
(x) (2x^1 + 1) (x^1 – 5) = 550