(Trying to) solve Higher order polynomial equations. Featuring a recall of polynomial long division. Some history: The quadratic formula (Dating back to antiquity) allows us to solve any quadratic equation. ax2 + bx + c = 0 What about any cubic equation? ax3 + bx2 + cx + d = 0? In the 1540’s Cardano produced a “cubic formula.” It is too complicated to actually write down here. What about any quartic equation? ax4 + bx3 + cx2 + dx + e = 0? A few decades after Cardano, Ferrari produced a “quartic 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 for “most” degree ≥ 5 polynomials there is no purely algebraic way to solve p(x) = 0. In fact: x5 − x − 1 = 0 has no algebraic solution. Galois died in a duel at age 21. This means that, as disheartening as it may feel, we will never get a formulaic solution to a general polynomial equation. The best we can do is numeric approx- imations and “tricks” that work sometimes. Think of these tricks as analogous to the strategies you use to factor a degree 2 polynomial. Trick 1 If you find one solution, then you can find a factor and reduce to a simpler polynomial equation. Example. 2x3 − x2 − 7x + 6 = 0 has x = 1 as a solution.

This means that x − 1 MUST divide 2x3 − x2 − 7x + 6.

Use polynomial long division to write 2x3 − x2 − 7x + 6 as (x − 1) · (something).

Now find the remaining two roots

1 2

Another: Find all of the solutions to x3 + x2 + x + 1 = 0 given that x = −1 is a solution to this equation.

For you: Find all of the solutions to 2x3 − 10x2 − x + 5 = 0 given that x = 5 is a solution to this equation.

The technique abover requires us to know one solution before we can even begin. The following trick will allow us to find all possible rational solutions. These will serve as initial guesses to be solutions. The second trick The rational roots test. p Imagine that 2x3 + 11x2 − 7x − 6 = 0 had a rational solution . We may as well demand q p that is in reduced terms. q Then: p3 p2 p 2 − 11 − 7 = 6 q3 q2 q Clear the denominator (by multiplying both sides by q3):

Notice that p divides the (left or right)pick one hand side! So p must divide the (left or p right) . But was in reduced terms so p must divide . pick one q List all of the possible p’s 3

Now move everything with no q’s to the other side of the equation:

Now q divides the (left or right)pick one hand side! So q must divide the (left or right)pick one. p But was in reduced terms so q must divide . q List all the possible q’s

p List all of the possible ’s q

Are any of them actually roots? (Different members of the class will try plugging in p different ’s. q

Theorem (The rational roots test). If the polynomial equation d d−1 2 adx + ad−1x + ··· + a2x + a1x + a0 = 0 p has a rational solution ± then p must divide a and q must divide a . q 0 d Use this theorem to find all of the possible rational solutions to 3x3 + 2x + 2 = 0.

possible values for p: possible values for q:

p possible values for (remember the possible minus sign) q

Are any of them actual solutions? 4

Third trick Descartes (The same as the inventor of “Cartesian coordinates” and respon- sible for the epistemological 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 d d−1 2 p(x) = adx + ad−1x + ··· + a2x + a1x + a0 be a polynomial (written with the powers of x in descending order.) • The number of positive solutions to p(x) = 0 is equal to either (1) the number of times the sign of the coefficient ak changes OR (2) that same number minus some even integer. • The number of negative solutions to p(x) = 0 is equal to either (1) the number of times d d d d−1 2 the sign of the coefficient of p(−x) = (−1) adx +(−1) ad−1x +···+a2x −a1x+a0 changes OR (2) that same number minus some even integer. example How many positive solutions might p(x) = 4x4 − 2x2 + x + 1 = 0 have?

How many negative solutions might p(x) = 4x4 − 2x2 + x + 1 = 0 have?

p(−x) = 5

Putting this all together: Find the solutions to p(x) = x5−7x4+19x3−37x2+60x−36 = 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. (You might use a machine for this part.)

For each root use polynomial long division (or synthetic division) to pull a linear factor out of the polynomial.

This only gets us to degree 3. Check for repeated roots and use polynomial long division.

Once you get down to a degree 2 polynomial, use the quadratic formula. 6

Group work: Find the solutions to p(x) = x4 − 3x3 − 5x2 + 13x + 6 = 0. I will ask a similar problem on the next group quiz. 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.

This only gets us down to degree 3. Check for a repeat root and do Polynomial long division.

Once you get down to a degree 2 polynomial then find its roots.