Solving Cubic Polynomials - SHSU

1 Solving Cubic The general solution to the quadratic equationThere are four steps to finding the zeroes of a quadratic First divide by the leading term, making the Then, givenx2+a1x+a0, substitutex=y a12to obtain an equation without the linear term.(This is the depressed equation.)3. Solve then foryas a square root. (Remember to use both signs of the square root.)4. Once this is done, recoverxusing the fact thatx=y example, let s solve2x2+ 7x 15 = , we divide both sides by 2 to create an equation with leading term equal to one:x2+72x 152= replacexbyx=y a12=y 74to obtain:y2=16916 Solve fory:y=134or 134 Then, Solving back forx, we havex=32or method is equivalent to completing the square and is the steps taken in developing the much-memorized quadratic formula. For example, if the original equation is our high school quadratic ax2+bx+c= 0then the first step creates the equationx2+bax+ca= then writex=y b2aand obtain, after simplifying,y2 b2 4ac4a2= 0so thaty= b2 4ac2aand sox= b2a b2 solutions to this quadratic depend heavily on the value ofb2 4ac.

2 We give this a name (thediscriminant) and a symbol ( ) and so discuss the discriminant =b2 the coefficients of the polynomial are integers and is a perfect square integer, we have rational the discriminant is positive, we have real roots. If the discriminant is zero, we have a single root. Ifthe discriminant is negative, we have imaginary note for later that if the discriminant =b2 4acis equal to zero then we have a single root andso our polynomial is a perfect The general solution to the Cubic equationEvery polynomial equation involves two steps to turn the polynomial into a slightly simpler First divide by the leading term, creating amonicpolynomial (in which the highest power ofxhas coefficient one.) This does not change the Then, givenxn+an 1xn 1+an 2xn 2+..a1x+a0, substitutex=y an 1nto obtain an equationwithout the term of degreen 1.(This is thedepressedpolynomial.)

3 Since this step is reversible,solutions to the depressed equation give us solutions to the original are the first steps in Cardano s method of Solving the Cubic . First, we do the two automaticsteps:1. Divide by the leading term, creating a Cubic polynomialx3+a2x2+a1x+a0with leading Then substitutex=y a23to obtain an equation without the term of degree two. This creates anequation of the formx3+Px Q= would rewrite this equation in the formx3+Px= then noticed (!) the followingalgebra identity:(a b)3+ 3ab(a b) =a3 b3.(1)Givenx3+Px=Q, setab=P3anda3 b3= this system by substitution and then assignx:=a example, we might replaceabyP3b(using the first equation) and then substitute into the secondequation to obtain(P3b)3 b3= byb3, we have(P3)3 (b3)2=Qb3which is a quadratic equation this equation as(b3)2+Qb3 (P3)3= 02so thatb3= Q2 (2)where the discriminant is := (P3)3+ (Q2) there are three possible solutions to equation (2).

4 This is where Cardano struggled. The next steps sometimes involved imaginary numbers and hewasn t sure what to do with them. We will press on and finish Cardano s work by jumping ahead toEuler s day and using his understanding of complex Euler s formulaei = cos +isin (3)we will create the element :=e2 i/3= 12+ that 3= 1 and that ( 2)3= 1 so that 1, and 2areallcube roots of 1!Using , we can see thatanyreal number has three cube roots. Given one cube root, ,the otherswill be and 2 .We use this to finish equation 2 by picking on cube root of Q/2 . We call that one particularsolution .The other solutions to the equation (2) areb= andb= 2 .Each solution to (2) gives a value foraso thatab= :=P3 .Then , 2 ,and 2are all solutions to the Cubic s exampleWe work through an example due to Euler: We find all solutions tox3 6x= 4.(4)HereP= 6 andQ= 4 and so the discriminant is = 8 + 4 = 4 so = 2 choose a sign and solve the Cubic equationb3= 2 + 2 + 2iin polar form as 8 (e3 4i).

5 Thenb3= 8e 4ihas a particular solution = 2e 4i= 1 + = 2 = 2 2e 4i= 2e 4i= 2e3 , in cartesian form, = 1 + = ( 1 +i) (1 +i) = 2is one solution to the Cubic there are others!3 Ifx1= is a solution then so arex2= 2 = 2(e3 4ie4 3i e 4ie2 3i) = 2(e25 12i e11 12i) = 2(e 12i+e 12i)andx3= 2= 2(e3 4ie2 3i e 4ie4 3i) = 2(e17 12i e19 12i) = 2(e 7 12i+e7 12i)By Euler s formula,e i+e i= 2 cos .So our answers are equivalent tox2= 2 2 cos( 12) andx3= 2 2 cos(7 12).The value of cos( 12) is not immediate, but we can find it from a trig (2 ) = 2 cos2 1thencos2 =1 + cos 2 12= 12+ 34andx2= 8 12+ 34= 4 + 2 a similar manner we can final answers arex2= 4 + 2 3 andx3= 4 2 example from EulerWe solve Euler s Cubic :x3 6x= 9.(5)Since (a b)3+ 3ab(a b) =a3 b3, we set3ab= 6 anda3 b3= 9.(LetP:= 6;Q:= 9.)Thena= 2/b; ( 2/b)3 b3= 9so 8 b6= 9b3orb6+ 9b3+ 8 = this as a quadratic inb3so that(b3)2+ 9b3+ 8 = 0 = (b3+ 8)(b3+ 1) = 0.

6 (6)Therefore eitherb3= 8 orb3= 8 and presumablyb= 1 andx=a b= 1 thena= 2 andx=a b= 3, so choosing the other factor does not give new have foundonesolution,x= what about other solutions?Set = 2 as a solution to the Cubic equation (6), so that 3= 8 and let = 1 be the correspondingchoice = 2 is also a solution to that Cubic and in this casea= 2= 2is the correspondingchoice a second solution isx=a b= 2+ 2 = ( 12 32i) + ( 1 + 3i) =12( 3 + 3i).If instead we havea= andb= 2then the final solution isx=a b= + 2 2= ( 12+ 32i) + ( 1 3i) =12( 3 3i).So we have found all three solutions:3,12( 3 + 3i).and12( 3 3i). Rational SolutionsWe can avoid the lengthy computations, above, if we are lucky enough to find a rational solution to ourpolynomial. For example, in the last problem, if we had merely stumbled on the rootx= 3, we couldhave divided the Cubic polynomialx3 6x 9 byx 3 and rewritten it asx3 6x 9 = (x 3)(x2+ 3x+ 3).

7 The quadratic equation, applied tox2+ 3x+ 3 would have given us the final two solutions without theextra how do we find these rational solutions when they occur?Ifx=pqis a rational solution to the polynomial equationf(x) = 0 thenqx pis a factor of thepolynomialf(x) and so we can use long division to writef(x) = (qx p)g(x) whereg(x) is a polynomialof smaller teach a version of this method in high school when students learn to solve quadratic equations byfactoring. For example, one might solve the equation 3x2 2x 8 = 0 by factoring the left-hand sideinto (3x+ 4)(x 2), obtain solutionsx= 43andx= 2 and so avoid the quadratic can try this method on Polynomials of higher degree with integer coefficients. Descartes wouldpoint out, in the 1600s, that iff(x) =anxn+an 1xn 1+..a1x+a0has a rootx=p/qthenqmustdivideanandpmust dividea0(Notice how this works on the quadratic 3x2 2x 8.) With then a shortlist of possible rational solutions, Descartes was willing to try these solutions and exhaust the possibilitiesfor a rational modern version of this is to pull out a graphing calculator, graph the polynomial equationy=f(x)and hope that the calculator identifies a nice rational (or even integer!)

8 Example, with Euler s cubicx3 6x 9 , we discover thatx= 3 is a root. When then divide thepolynomial byx 3 to obtain a quadratic polynomial and now we can go ahead and use the method is much faster than the general method, but it requires that we be lucky and stumbleupon a Quartic PolynomialsAfter del Ferro and Tartaglia solved the general Cubic equation (and the result was released to the publicby Cardano), mathematicians then concentrated on the quartic equation. Cardano s student, Ferrari,around 1540, suggested a general method for Solving the depressed the depressed quarticx4+px2+qx+r= 0rewrite the problem so that it appears to involve a square:(x2+p)2=px2 qx+ (p2 r).Now insert a new variableyand expand (x2+p+y)2with the hopes that the right side, (p+ 2y)x2 qx+ (p2 r+ 2py+y2) is a perfect the quadratic formula, the right-hand side will be a perfect square when the discriminant 4(p+2y)(p2 r+ 2py+y2) q2is we first need to solve4(p+ 2y)(p2 r+ 2py+y2) q2= 0(7) equation (7) is a Cubic equation, we know how to do that!

9 In other words, given a quartic, we make it monic, and then turn it into a depressed equation, inserty, and solve the associated Cubic fory. Once this is done, we solve the quadratic equation (x2+p+y)2=(p+ 2y)x2 qx+ (p2 r+ 2py+y2) is lengthy and tedious, but it works in general! (Well, it works if one is willing to believe incomplex numbers.)By 1545, after Cardano published Ferrari s work (with his permission, apparently!) mathematiciansbegan to search for solutions to the general fifth-degree or quintic equation. It would take the work ofAbel, Galois and others to show that in fact a general solution to the quintic is will save that story for another on Cubic equations1. For each of the equations, below, do the appropriate substitution to turn the polynomial on theleft-hand side into a depressed polynomial.(a) 2x2 5x+ 3 = 0(b)x3 6x2+ 2x+ 3 = 0(c)x4+ 8x3+ 3x 12 = 02. Complete your work in equation part (a), above, finding the solutions to the equation by completingthe For parts (b) and (c) in problem 1, guess a rational solution and then use this to find all solutionsto the Find a rational solution for the following Polynomials .

10 Then use this solution to find all the roots.(a) Viete and Cardano-Tartaglia examined this polynomial:x3+ 63x 316.(b) Viete and Cardano-Tartaglia also did this one:x3 63x 162.(c) Euler s Cubic :x3 6x