Download Solving Cubic Polynomials and more Lecture notes Algebra in PDF only on Docsity!
Solving Cubic Polynomials
1.1 The general solution to the quadratic equation
There are four steps to finding the zeroes of a quadratic polynomial.
- First divide by the leading term, making the polynomial monic.
- Then, given x^2 + a 1 x + a 0 , substitute x = y −
a 1 2
to obtain an equation without the linear term. (This is the “depressed” equation.)
- Solve then for y as a square root. (Remember to use both signs of the square root.)
- Once this is done, recover x using the fact that x = y −
a 1 2
For example, let’s solve 2 x^2 + 7x − 15 = 0. First, we divide both sides by 2 to create an equation with leading term equal to one:
x^2 +
x −
Then replace x by x = y − a 1 2
= y −
to obtain:
y^2 =
Solve for y:
y =
or −
Then, solving back for x, we have x =
or − 5.
This 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”
ax^2 + bx + c = 0
then the first step creates the equation
x^2 +
b a
x +
c a
We then write x = y −
b 2 a and obtain, after simplifying,
y^2 − b^2 − 4 ac 4 a^2
so that
y = ±
b^2 − 4 ac 2 a
and so
x = − b 2 a
b^2 − 4 ac 2 a
The solutions to this quadratic depend heavily on the value of b^2 − 4 ac. We give this a name (the discriminant) and a symbol (∆) and so discuss the discriminant
∆ = b^2 − 4 ac.
If the coefficients of the polynomial are integers and ∆ is a perfect square integer, we have rational roots. If the discriminant is positive, we have real roots. If the discriminant is zero, we have a single root. If the discriminant is negative, we have imaginary roots. We note for later that if the discriminant ∆ = b^2 − 4 ac is equal to zero then we have a single root and so our polynomial is a perfect square.
1.2 The general solution to the cubic equation
Every polynomial equation involves two steps to turn the polynomial into a slightly simpler polynomial.
- First divide by the leading term, creating a monic polynomial (in which the highest power of x has coefficient one.) This does not change the roots.
- Then, given xn^ + an− 1 xn−^1 + an− 2 xn−^2 + ...a 1 x + a 0 , substitute x = y − an− 1 n
to obtain an equation without the term of degree n − 1. (This is the depressed polynomial.) Since this step is reversible, solutions to the “depressed equation” give us solutions to the original equation.
Here are the first steps in Cardano’s method of solving the cubic. First, we do the two automatic steps:
- Divide by the leading term, creating a cubic polynomial x^3 + a 2 x^2 + a 1 x + a 0 with leading coefficient one.
- Then substitute x = y −
a 2 3 to obtain an equation without the term of degree two. This creates an equation of the form x^3 + P x − Q = 0.
Cardano would rewrite this equation in the form x^3 + P x = Q. He then noticed (!) the following algebra identity:
(a − b)^3 + 3ab(a − b) = a^3 − b^3. (1) Given x^3 + P x = Q, set ab =
P
and a^3 − b^3 = Q.
Solve this system by substitution and then assign x := a − b.
For example, we might replace a by 3 Pb (using the first equation) and then substitute into the second equation to obtain
(
P
3 b
)^3 − b^3 = Q.
Multiplying by b^3 , we have
(
P
)^3 − (b^3 )^2 = Qb^3
which is a quadratic equation in b^3. Rewrite this equation as
(b^3 )^2 + Qb^3 − (
P
)^3 = 0
If x 1 = α − β is a solution then so are
x 2 = αω^2 − βω =
2(e
3 π 4 ie 4 π 3 i^ − e π 4 ie 2 π 3 i) =
2(e
25 π 12 i^ − e 11 π 12 i) =
2(e
π 12 i^ + e −π 12 i)
and x 3 = αω − βω^2 =
2(e
34 π i e
23 π i − e
π 4 i e
43 π i ) =
2(e
1712 π i − e
1912 π i ) =
2(e
− 127 π i
712 π i ) By Euler’s formula, eθi^ + e−θi^ = 2 cos θ. So our answers are equivalent to
x 2 = 2
2 cos(
π 12
) and x 3 = 2
2 cos(
7 π 12
The value of cos( 12 π ) is not immediate, but we can find it from a trig identity. Since cos(2θ) = 2 cos^2 θ − 1
then
cos^2 θ =
1 + cos 2θ 2
Thus
cos π 12
and
x 2 =
In a similar manner we can find x 3. Our final answers are
x 2 =
3 and x 3 = −
Another example from Euler We solve Euler’s cubic: x^3 − 6 x = 9. (5) Since (a − b)^3 + 3ab(a − b) = a^3 − b^3 , we set
3 ab = −6 and a^3 − b^3 = 9.
(Let P := −6; Q := 9.) Then a = − 2 /b; (− 2 /b)^3 − b^3 = 9
so − 8 − b^6 = 9b^3 or b^6 + 9b^3 + 8 = 0. View this as a quadratic in b^3 so that
(b^3 )^2 + 9b^3 + 8 = 0 =⇒ (b^3 + 8)(b^3 + 1) = 0. (6) Therefore either b^3 = −8 or b^3 = − 1. Suppose b^3 = −8 and presumably b = − 2. Then a = 1 and x = a − b = 3.
If b = −1 then a = −2 and x = a − b = 3, so choosing the other factor does not give new information. We have found one solution, x = 3.
But what about other solutions? Set β = −2 as a solution to the cubic equation (6), so that β^3 = −8 and let α = 1 be the corresponding choice for a. Then b = βω = − 2 ω is also a solution to that cubic and in this case a = αω^2 = ω^2 is the corresponding choice for a. Then a second solution is
x = a − b = ω^2 + 2ω = (−
i) + (−1 +
3 i) =
3 i).
If instead we have a = αω and b = βω^2 then the final solution is
x = a − b = ω + 2ω^2 = (−
i) + (− 1 −
3 i) =
3 i).
So we have found all three solutions:
3 i). and
3 i).
1.3 Rational Solutions
We can avoid the lengthy computations, above, if we are lucky enough to find a rational solution to our polynomial. For example, in the last problem, if we had merely stumbled on the root x = 3, we could have divided the cubic polynomial x^3 − 6 x − 9 by x − 3 and rewritten it as
x^3 − 6 x − 9 = (x − 3)(x^2 + 3x + 3).
The quadratic equation, applied to x^2 + 3x + 3 would have given us the final two solutions without the extra work.
So how do we find these rational solutions when they occur?
If x = pq is a rational solution to the polynomial equation f (x) = 0 then qx − p is a factor of the polynomial f (x) and so we can use long division to write f (x) = (qx − p)g(x) where g(x) is a polynomial of smaller degree. We teach a version of this method in high school when students learn to solve quadratic equations by factoring. For example, one might solve the equation 3x^2 − 2 x − 8 = 0 by factoring the left-hand side into (3x + 4)(x − 2), obtain solutions x = − 43 and x = 2 and so avoid the quadratic formula.
We can try this method on polynomials of higher degree with integer coefficients. Descartes would point out, in the 1600s, that if f (x) = anxn^ + an− 1 xn−^1 + ...a 1 x + a 0 has a root x = p/q then q must divide an and p must divide a 0 (Notice how this works on the quadratic 3x^2 − 2 x − 8.) With then a short list of possible rational solutions, Descartes was willing to try these solutions and exhaust the possibilities for a rational root. The modern version of this is to pull out a graphing calculator, graph the polynomial equation y = f (x) and hope that the calculator identifies a nice rational (or even integer!) root. For example, with Euler’s cubic x^3 − 6 x − 9 , we discover that x = 3 is a root. When then divide the polynomial by x − 3 to obtain a quadratic polynomial and now we can go ahead and use the quadratic formula. This method is much faster than the general method, but it requires that we be “lucky” and stumble upon a root.
Exercises on cubic equations
- For each of the equations, below, do the appropriate substitution to turn the polynomial on the left-hand side into a “depressed” polynomial.
(a) 2x^2 − 5 x + 3 = 0 (b) x^3 − 6 x^2 + 2x + 3 = 0 (c) x^4 + 8x^3 + 3x − 12 = 0
- Complete your work in equation part (a), above, finding the solutions to the equation by completing the square.
- For parts (b) and (c) in problem 1, guess a rational solution and then use this to find all solutions to the polynomial.
- Find a rational solution for the following polynomials. Then use this solution to find all the roots.
(a) Viete and Cardano-Tartaglia examined this polynomial: x^3 + 63x − 316. (b) Viete and Cardano-Tartaglia also did this one: x^3 − 63 x − 162. (c) Euler’s cubic: x^3 − 6 x − 4.
