(Trying to) solve Higher order polynomial euations. Some history: The uadratic formula (Dating back to antiuity) allows us to solve any uadratic euation. What about any cubic euation? ax 2 + bx + c = 0 ax 3 + bx 2 + cx + d = 0? In the 1540 s Cardano produced a cubic formula. It is too complicated to actually write down here. See today s extra credit. What about any uartic euation? ax 4 + bx 3 + cx 2 + dx + e = 0? A few decades after Cardano, Ferrari produced a uartic formula More complicated still than the cubic formula. In the 1820 s Galois (age 19) proved that there is no general algebraic formula for the solutions to a degree 5 polynomial. In fact there is no purely algebraic way to solve x 5 x 1 = 0. He went on to die in a duel. This means that, as disheartening as it may feel, we will never get a formulaic solution to a general polynomial euation. The best we can get is tricks that work sometimes. Trick 1 If you find one solution, then you can find a factor and reduce to a simpler polynomial euation. Example. 2x 2 x 2 1 = 0 has x = 1 as a solution. This means that x 1 MUST divide 2x 3 x 2 1. Use polynomial long division to write 2x 3 x 2 1 as (x 1) (something). Now find the remaining two roots 1
2 For you: Find all of the solutions to x 3 + x 2 + x + 1 = 0 given that x = 1 is a solution to this euation. The second trick The Rational roots test. Imagine that 2x 3 + 11x 2 7x 6 = 0 had a rational solution p. We may as well demand that p is in reduced terms. Then: 2 p3 3 11p2 2 7p = 6 Clearing the denominator (by multiplying both sides by 3 ) gives: Notice that p divides the (left or right) pick one hand side! So p must divide the (left or right) pick one. But p was in reduced terms so p must divide. List all of the possible p s Now move everything with no s to the other side of the euation: Now divides the (left or right) pick one hand side! So must divide the (left or right) pick one. But p was in reduced terms so must divide. List all the possible s List all of the possible p s Are any of them actually roots?
3 Theorem (The rational roots test). If the polynomial euation a d x d + a d 1 x d 1 + + a 2 x 2 + a 1 x + a 0 = 0 has a rational solution p then p must divide a 0 and must divide a d. your turn: Use this theorem to find all of the possible rational solutions to 3x 3 +2x+2 = 0. possible values for p: possible values for : possible values for p (remember the possible minus sign) Are any of them actual solutions? Third trick Descartes (The same as the inventor of Cartesian coordinates and responsible for the etymological idea of I think therefore I am ) is credited with the following result saying how many solutions should be positive and negative. Theorem (Descartes Rule of signs). Let p(x) = a d x d + a d 1 x d 1 + + a 2 x 2 + a 1 x + a 0 be a polynomial (written with the powers of x in descending order.) The number of positive solutions to p(x) = 0 is eual to either (1) the number of times the sign of the coefficient a k changes OR (2) that same number minus some even integer. Notice that p( x) = a d ( x) d + a d 1 ( x) d 1 + + a 2 ( x) 2 + a 1 ( x) + a 0 = ( 1) d a d x d + ( 1) d a d 1 x d 1 + + a 2 x 2 a 1 x + a 0 The number of negative solutions to p(x) = 0 is eual to either (1) the number of times the sign of the coefficient ( 1) k a k changes OR (2) that same number minus some even integer. example How many positive solutions might p(x) = 4x 4 2x 2 + x + 1 = 0 have? How many negative solutions might p(x) = 4x 4 2x 2 + x + 1 = 0 have? p( x) =
4 Putting this all together: Try to find the solutions to p(x) = x 4 + 6x 2 12x 18 = 0 Step 1: This is a degree polynomial. What is the greatest that the number of zero s could be? Step 2: Using Descartes Rule of signs, how many positive / negative zero s should there be? Step 3: Use the rational root test to find any possible rational roots. Are any of them actual roots? (split up the work with your groupmates). For each root use polynomial long division (or synthetic division) to pull a linear factor out of the polynomial. REMARK: You might have a root of high multiplicity If you get down to a degree 2 polynomial then use the uadratic formula.
Group work: Try to find the solutions to p(x) = x 5 7x 4 + 19x 3 37x 2 + 60x 36 = 0 Step 1: This is a degree polynomial. What is the greatest that the number of zero s could be? 5 Step 2: Using Descartes Rule of signs, how many positive / negative zero s should there be? Step 3: Use the rational root test to find any possible rational roots. Are any of them actual roots? (split up the work with your groupmates). For each root use polynomial long division (or synthetic division) to pull a linear factor out of the polynomial. REMARK: You might have a root of high multiplicity If you get down to a degree 2 polynomial then use the uadratic formula.