1 Introduction
In this essay I aim to introduce the quartic, and then discuss contributions by five distinct mathematicians, to be precise Ferrari, Viète, Harriot, Descartes and Euler. Examining each method step by step then further applying these procedures with my own running example.

2 What is a Quartic Polynomial?
Definition 2.1 (Quartic)
A quartic equation is a fourth-degree polynomial which has the following form
Ax^4+Bx^3+Cx^2+Dx+E=0(1)
Where A is nonzero [13]

Quartic functions are of even degree and contain three critical points. When A≥1 the function has a global minimum and increases to positive infinity. When A<0 the function, alternatively has a global maximum and decreases to negative infinity. [13]
Example 2.2
f(x)=x^4/4-(4x^3)/3+x^2+4x-2
The derivative is
f(x)^’=x^3-4x^2+2x+4
This quartic function has three critical points, at x=2,x=1-√3 and x=1+√3

Thus the critical values are
f(2)=2^4/4-(4(2)^3)/3+2^2+4(2)-2=4-32/3+4+8-2=10/3 f(1-√3)=(-1)/3-2√3 f(1+√3)=(-1)/3+2√3
This function has a global minimum at x=1-√3 and no global maxima

A distinctive type of quartic is known as a biquadratic, if the polynomial contains no odd degree terms. In other words a biquadratic is quadratic inx^2. (Note that if B=0 and D=0 for equation (1), we have a biquadratic equation [13].
The highest degree equations that can be expressed in terms of radicals are of fourth-degree, as proved by the Abel-Ruffini thereom (1824) [12].

Definition 2.3 (Radical)
A radix or radical symbol is used to write roots of equations
For example the square root, cube root and fourth roots of x are denoted √x,∛x,and ∜x respectively. In general for√(n&x), we call x the radicand and n the index. [12]
Throughout this essay I intend to find the roots of the following quartic

x^4-8x^3+25x^2-36x+21=0(2)
Clarifying the various intricate procedures needed to derive its solutions.

3 Lodovico Ferrari’s Method
Ferrari was the first mathematician to solve the quartic, and he did so in a very sophisticated manner. For supplementary information on Ferrari and each proceeding mathematician see Appendix I.
The first method adopted by Ferrari looks at an equation of the form (2) (following [1], Chapter 7.4 & [2], Chapter 6)
This can be rewritten by dividing the equation through by A so that the coefficient of the x^4 term is reduced to one. Further substituting x=X-B/4A gets rid of the x^3 term which can be seen below as follows. x^4+B/A x^3+C/A x^2+D/A x+E/A=0(X-B/4A)^4+B/A (X-B/4A)^3+C/A (X-B/4A)^2+D/A (X-B/4A)+E/A=0(3)

Expanding brackets and simplifying terms we have the following
D/A (X-B/4A)=DX/A-BD/(4A^2 ) C/A (X-B/4A)^2=C/A [(X-B/4A)(X-B/4A)] C/A [X^2-BX/4A-BX/4A+B^2/(16A^2 )] C/A [X^2-2BX/4A+B^2/(16A^2 )] C/A [X^2-BX/2A+B^2/(16A^2 )] (CX^2)/A-BCX/(2A^2 )+(B^2 C)/(16A^3 ) B/A (X-B/4A)^3=B/A [(X-B/4A)^2 (X-B/4A)] B/A [(X^2-BX/2A+B^2/(16A^2 ))(X-B/4A)] B/A [(X^3-(BX^2)/2A+(B^2 X)/(16A^2 )-(BX^2)/4A+(B^2 X)/(8A^2 )-B^3/(64A^3 ))] B/A [(X^3-(2BX^2)/4A-(BX^2)/4A+(B^2 X)/(16A^2 )+(〖2B〗^2 X)/(16A^2 )-B^3/(64A^3 ))] B/A [(X^3-(3BX^2)/4A+(3B^2 X)/(16A^2 )-B^3/(64A^3 ))] (BX^3)/A-(3B^2 X^2)/(4A^2 )+(3B^3 X)/(16A^3 )-B^4/(64A^4 ) (X-B/4A)^4=(X-B/4A)^2 (X-B/4A)^2 (X^2-BX/2A+B^2/(16A^2 ))(X^2-BX/2A+B^2/(16A^2 )) X^4-(BX^3)/2A+(B^2 X^2)/(16A^2 )-(BX^3)/2A+(B^2 X^2)/(4A^2 )-(B^3 X)/(32A^3 )+(B^2 X^2)/(16A^2 )-(B^3 X)/(32A^3 )+B^4/(256A^4 ) X^4-(2BX^3)/2A+(〖2B〗^2 X^2)/(16A^2 )+(B^2 X^2)/(4A^2 )-(〖2B〗^3 X)/(32A^3 )+B^4/(256A^4 ) X^4-(BX^3)/A+(B^2 X^2)/(8A^2 )+(〖2B〗^2 X^2)/(8A^2 )-(B^3 X)/(16A^3 )+B^4/(256A^4 ) X^4-(BX^3)/A+(〖3B〗^2 X^2)/(8A^2 )-(B^3 X)/(16A^3 )+B^4/(256A^4 )

Combining each of these terms we have that (3) equates to
X^4+((3B^2)/(8A^2 )-(3B^2)/(4A^2 )+C/A) X^2+((2B^3)/(16A^3 )-BC/(2A^2 )+D/A)X+(B^4/(256A^4 )-B^4/(64A^4 )+(B^2 C)/(16A^3 )-BD/(4A^2 )+E/A)=X^4+((8AC-3B^2)/(8A^2 )) X^2+((B^3-4ABC+8A^2 D)/(8A^3 ))X+((-3B^4+16AB^2 C-64A^2 BD+256A^3 E)/(256A^4 ))=0(4)Thus in general we have a reduced quartic of the form
X^4+aX^2+bX+c=0(5)
It follows
X^4+aX^2=-bX-c(6)
Now Ferrari added the term aX^2+a^2 to both sides of the equation, so we are able to complete the square as follows.
X^4+aX^2+aX^2+a^2=-bX-c+(aX^2+a^2 ) X^4+2a^2+a^2=(aX^2+a^2 )-bX-c(X^2+a)^2-a^2+a^2=aX^2+a^2-bX-c(X^2+a)^2=aX^2+a^2-bX-c
Additionally by the introduction of 2(X^2+a)y+y^2 we can rewrite this further like so
(X^2+a)^2+2(X^2+a)y+y^2=aX^2+a^2-bX-c+2(X^2+a)y+y^2 ((X^2+a)+y)^2=aX^2+a^2-bX-c+2X^2 y+2ay+y^2 (X^2+a+y)^2=(a+2y) X^2-bX+(a^2-c+2ay+y^2 )(7)
We want the right-hand side to be a perfect square yet this only occurs when the discriminant i.e. ‘B^2-4AC’ of a quadratic equation is zero.
Thus looking at the coefficients of the left-hand side of equation (7) we have the following
A=a+2yB=-bC=a^2-c+2ay+y^2
Plugging in these values we get the following expression for the discriminant
b^2=4(a+2y)(a^2-c+2ay+y^2 )⇒4(a+2y)(a^2-c+2ay+y^2 )-b^2=0
Now further collecting like terms we have the following, where y is the new unknown
(4a+8y)(a^2-c+2ay+y^2 )-b^2=04a^3-4ac+8a^2 y+4ay^2+8a^2 y-8cy+16ay^2+8y^3-b^2 8y^3+20ay^2+(16a^2-8c)y+(4a^3-4ac-b^2 )(8)
This is now in the form of a cubic equation, whereby equation (8) is called the resolvant cubic for the given quartic equation. Broadly speaking the resolvant cubic has three solutions and in general we can find y selecting any one of these solutions and taking square roots.
To complete the proof we can eventually use Cardano’s method for solving cubic equations.
With the aid of an example, I will apply the quartic method as outlined above to solve the quartic (2) mentioned earlier.
Example 3.1
To find the roots of x^4-8x^3+25x^2-36x+21=0 first note
A=1B=-8C=25D=-36E=21
Dividing through by A keeps the equation the same, so make the initial substitution and let
x=X-B/4A=X+2

Thus our equation becomes
(X+2)^4-8(X+2)^3+25(X+2)^2-36(X+2)+21=0
Using a result found earlier we can eliminate the x^3 term by plugging our values for A,B,C,D and Einto equation (4) as follows

X^4+((8(25)-3(-8)^2)/8) X^2+(((-8)^3-4(-8)(25)+8(-36))/8)X+((-3(-8)^4+16(-8)^2 (25)-64(-8)(-36)+256(21))/256) X^4+((200-192)/8) X^2+((-512+800-288)/8)X+((-12288+25600-18432+5376)/256) X^4+(8/8) X^2+(0/8)X+(256/256) X^4+X^2+1
Now that we no longer have the x^3 term we can rewrite this new equation in the form of (6) X^4+X^2=0X-1
Where a=1,b=0and c=1
Adding the term X^2+(1)^2 to both sides of the equation gives
X^4+X^2+X^2+1=0X-1+X^2+1X^4+2X^2+1=X^2 (X^2+1)^2-1+1=X^2 (X^2+1)^2=X^2
Now by introducing the term2(X^2+1)y+y^2, where y is a new unknown we have the following (X^2+1)^2+2(X^2+1)y+y^2=X^2+2(X^2+1)y+y^2 ((X^2+1)+y)^2=X^2+2〖yX〗^2+2y+y^2 (X^2+1+y)^2=(1+2y) X^2+(2y+y^2 )(9)
We require (X^2+1+y)^2 to be a perfect square. For this reason, substitute our values for a,b and c into equation (8).
8y^3+20(1) y^2+(16(1)^2-8(1))y+(4(1)^3-4(1)(1)-(0)^2 )8y^3+20y^2+(16-8)y+(4-4-0)8y^3+20y^2+8yy^3+5/2 y^2+y
We now have a cubic, where y is the new unknown. Thus we can find the roots accordingly
y^3+5/2 y^2+y=y(y^2+5/2 y+1)
Clearly y=0 is a solution and by inspection the other roots are y=(-1)/2 and y=-2
Thus any one of these solutions for y will result in a perfect square on the right-hand side of equation (9).
Choosing y=0 we have
(X^2+1+(0))^2=(1+2(0)) X^2+(2(0)+(0)^2 ) (X^2+1)^2=XX^2+1=±XX^2±X+1=0
Thus we have two quadratic equations and so we can solve these in the usual manner.
X^2+X+1=0X=(-(1)±√((1)^2-4(1)(1) ))/2(1) X=(-1±√(1-4))/2 X=(-1±i√3)/2
X^2-X+1=0X=(-(-1)±√((-1)^2-4(1)(1) ))/2(1) X=(1±√(1-4))/2 X=(1±i√3)/2
Therefore the solutions are X=(-1)/2+√3/2 i,-1/2-√3/2 i,1/2+√3/2 i and 1/2-√3/2 i
Since we made the initial substitutionx=X+2, the solutions of the original equation are 3/2+√3/2 i,3/2-√3/2 i,5/2+√3/2 i and 5/2-√3/2 i

4 Viète’s Method
Before discussing the subsequent method, first note that Viète was a very wordy mathematician. When it came to solving problems his representation of terms was as below, with a current notation to the left [4, Chapter 12 page 410].
A^2 – Aquadratum (or A quad)
B^3 – B cubus
C^4- C quadrato–quadratum (or Cquad-quad)
Likewise when terms were multiplied by one another he used the word in. Although he did use the German symbols for plus + and minus -, which are embraced today. He additionally used a curled symbol l in place of the now traditional square root sign and the word aequetus instead of and equals sign. [4, Chapter 12 page 410]

Example 4.1
(ABC-√B)/C^3 =D
Would be written as
(A∈B∈C-lB)/Ccubus aequetusD
Now in regards to Viète’s solution to the quartic, he further improved Ferrari’s approach. I will use current notation for simplicity, starting with a reduced quartic of the form (6) rearranged as below [following [1], Chapter 7.4, page 334]
X^4=-aX^2-bX-c
Now Viète added the term X^2 y^2+1/4 y^4 to both sides of the equation where y is a new unknown we have the following X^4+X^2 y^2+1/4 y^4=-aX^2-bX-c+X^2 y^2+1/4 y^4 (X^2+1/2 y^2 )^2=(y││2-a) X^2-bX+(1/4 y^4-c)
The right-hand side is a perfect square precisely when the discriminant of the left-hand side is zero.
With A=y^2-a,B=-b and C=1/4 y^4-c we have
B^2-4AC=0=(-b)^2-4(y^2-a)(1/4 y^4-c) b^2-4(1/4 y^6-cy^2-1/4 ay^4+ac) b^2-y^6+4cy^2+ay^4-4acb^2=y^6-4cy^2-ay^4+4ac
We now have a cubic iny^2, thus is solvable.

Another issue I will briefly converse are the main aspects of Viète’s formulas following [5].

Theorem 4.2 (Viète’s Formulas)
Viète further concluded that the roots of a quartic of the form (1)
Satisfy the following Viète formulas
x_1+x_2+x_3+x_4=(-B)/A x_1 x_2+x_1 x_3+x_1 x_4+x_2 x_3+x_2 x_4+x_3 x_4=C/A x_1 x_2 x_3+x_2 x_3 x_4+x_1 x_2 x_4+x_1 x_3 x_4=(-D)/A x_1 x_2 x_3 x_4=E/A
WithA≡1, and where x_1,x_2,x_3 andx_3 are the roots of the fourth-degree equation.
There are five special symmetrical polynomials contained in Viète’s formulas, based on a reduced quartic of the form (5). Symmetric polynomials have the property such that interchanging any variable will result in the same polynomial. Below we can see clearly that upon exchanging z_1^3 with z_4^3, we are left with the second polynomial unchanged.
z_1^2+z_2^2+z_3^2+z_4^2=-2az_1^3+z_2^3+z_3^3+z_4^3=-3bz_1^4+z_2^4+z_3^4+z_4^4=2a^2-4cz_1^5+z_2^5+z_3^5+z_4^5=5ab
We can further reduce these polynomials to exclude a,b and ccorrespondingly, to arrive at the below relations, along with their cyclic permutations.
z_1 z_2 (a+z_1^2+z_1 z_2+z_2^2 )-c=0z_1^2 z_2 (z_1+z_2 )-bz_1-c=0b+az_2+z_2^3=0

Example 4.3
Moving on, I will now look at our previous polynomial (2) and apply Vièta’s method for our reduced quartic rearranged as follows
X^4+X^2=0X-1X^4=-X^2-1
Where a=1,b=0 and c=1
Adding the term X^2 y^2+1/4 y^4 to both sides of the equation where y is a new unknown we have the following
X^4+X^2 y^2+1/4 y^4=-X^2-1+X^2 y^2+1/4 y^4 (X^2+1/2 y^2 )^2=(y││2-1) X^2+(1/4 y^4-1)(10)
We know that the right-hand side is a perfect square when B^2-4AC=0 thus
B^2-4AC=0=(0)^2-4(y^2-1)(1/4 y^4-1)-4(1/4 y^6-y^2-1/4 y^4+1)0=y^6-y^4-4y^2+4
This is a cubic iny^2 which is solvable and so it follows
-4=y^2 (y││3-y^2-4y)Thus by inspection the roots are y=-1,1,-√2,√2,-i√2 and i√2
By choosing any of these values for y the right-hand side of (10) will be a perfect square.
Hence substituting y=√2 we have (X^2+1/2 (√2)^2 )^2=((√2)││2-1) X^2+(1/4 (√2)^4-1) (X^2+1/2 (2))^2=(2-1) X^2+(1/4 (4)-1) (X^2+1)^2=X^2 X^2+1=±X
Thus we have two quadratic equations which we found earlier in Ferraris’ method, namely X^2+X+1=0 andX^2-X+1=0.
Finally solving these as before and noting the initial substitution we again arrive at the roots 3/2+√3/2 i,3/2-√3/2 i,5/2+√3/2 i and5/2-√3/2 i.

5 Harriot’s Method
Harriot’s system for writing mathematical problems was a huge breakthrough in mathematics and has a very present feel. Throughout his research, he used lower case letters, compared to Viète. On the other hand, he continued to express constant values with vowels, and known values with consonants. He also joint multiplied terms by pushing them together avoiding the multiplication sign altogether, as we do today. [4, Chapter 14.1 page 469]

Example 5.1
Comparing notation, an equation such as [again following 4, Chapter 14.1 page 469]
12=8x-13x^2+8x^3-x^4
Would be written as follows12=8a-13aa+8aaa–aaaa
Outlining Harriot’s method, to remove the cubic term he leta=2-e. However Ferrari would have made the substitutionx=X-B/4A=X-8/4(-1) =X+2.
The reduced equation via Harriot’s method is
-20e+11ee-eeee=0
Solving this cubic we have two real roots e=0 ande=-4, and a pair of complex rootsx=2±i. Thus after reducing to a cubic in a similar fasion to Ferrai we obtain
eee-11e=-20
Harriot found that one root wase=-4, and concluded that the addition of the remaining roots must be 4 and their product5.
The only values satisfying this are
e=2±√(-1)

Thus the original equation has the complex roots

a=2-(2±√(-1))±√(-1)
Adopting Harriot’s notation
Let X=a,where X^4=aaaa,X^3=aaa etc.
Harriot also used instead of,however we will stick with the traditional equal sign for simplicity. I will again explain his method but this time implementing his notation with an example from [10]
Example 5.2
We want to solve the reduced quartic equation
X^4-6X^2+136X-1155=0

On the first line place your reduced quartic
aaaa-6aa+136a=1155
————————————
Adding 4aa+1to both sides of the equation creates a perfect square on the left-hand side

(aa-1)(aa-1)=(2a-34)(-2a+34)

Next we write down the first possible solution aa-1=2a-34 and solve
aa-1=2a–3433=2a–aaaa-2a=-33

Adding 1 to both sides of the equation allows us to complete the square
aa-2a+1=+1–33
Result 1
a-1=√(-32)
Result 2
1-a=√(-32)
Simplified Result 1
a=1+√(-32)
Simplified Result 2
a=1-√(-32)
————————————
The second possible solution is aa-1=34-2a
aa-1=34-2aaa+2a=35
Again adding 1 to both sides of the equation allows us to complete the square
aa+2a+1=1+35
Result 3
a+1=√36
Simplified Result 3
a=√36-1=5
————————————
Simplified Result 4
-a-1=√36 a=-√36-1=-7
Thus the roots of the given quartic equation are X=5,-7,1+√(-32) and 1-√(-32)
After simplifying X=5,-7,1+4i√2 and 1-4i√2

Example 5.3
Now based on Harriot’s method, and using the form (5) for our equation we can find the roots as follows
aaaa+aa=1aaaa+aa+1⁄4=1+1⁄4 aa+1⁄2=5⁄4 aa=5⁄4-1⁄2 aa=3⁄4 a=√(3⁄4) a=-√(3⁄4) aa+1⁄2=(-5)⁄4 aa=(-5)⁄4-1⁄2 aa=(-7)⁄4 a=-√((-7)⁄4) a=√((-7)⁄4)
Thus the solutions are X=-√(3/4),√(3/4),-√((-7)/4),√((-7)/4)
Reverting back to the substitution x=X+2 it follows that in modern notation
(3+i√3)/2,(3-i√3)/2,(5+i√3)/2 and (5-i√3)/2

6 Descartes’ Method
The penultimate method discovered by Descartes’ expresses the quartic as a multiple of two quadratics, and then makes use of equating coefficients with those of the original polynomial. [Following [1], Chapter 8.2, page 373]
Taking an equation of the form (5), the shortfall of the x^3 term implies the X terms must have opposite signs and equivalent powers, namely the form
(X^2+pX+q)(X^2-pX+r)
Thus expanding these brackets we have
X^4-pX^3+rX^2+pX^3-p^2 X^2+prX+qX^2-pqX+qr
This simplifies to
X^4+(q+r-p^2 ) X^2+p(r-q)X+qr
Now comparing coefficients we have
X^4+aX^2+bX+c=X^4+(q+r-p^2 ) X^2+p(r-q)X+qr(11)
Leading to three equations
a=q+r-p^2 b=p(r-q)c=qr
For k≠0, the first two can be rearranged to give the simultaneous equations
a+p^2=q+r1 (-b)/p=q-r2
Adding 1 and 2 gives
a+p^2-b/p=2q
and solving 1 plus -1×2 gives
a+p^2-1((-b)/p)=q+r-1(q-r)a+p^2+b/p=2r
With (ap+p^3-b)/2p=q and(ap+p^3+b)/2p=r

We now have expressions for q andr. Therefore we can substitute them into the final equationc=qr.c=((ap+p^3-b)/2p)((ap+p^3+b)/2p)c×2p×2p=(ap+p^3-b)(ap+p^3+b)4cp^2=(ap+p^3-b)(ap+p^3+b)4cp^2=a^2 p^2+ap^4+abp+ap^4+p^6+p^3 b-abp-bp^3-b^2 0=p^6+2ap^4+(a^2-4c) p^2-b^2The result is a cubic inp^2, and hence is solvable.
Selecting any root such that p^2≠0 we will have two quadratic equations
X^2+pX+1/2 (a+p^2-b/p)=0X^2-pX+1/2 (a+p^2+b/p)=0(12)
These roots combined make up the four roots of the reduced quartic.
Example 6.1
Starting with our polynomial (2) in the reduced form
X^4+X^2+1=0
Since there is no x^3 term we can express the above as a pair of quadratics.
Comparing coefficients and using the result (11) with our values a=1,b=0 and c=1 we have
X^4+X^2+1=X^4+(q+r-p^2 ) X^2+p(r-q)X+qr
Leading to three equations
1=q+r-p^2 0=p(r-q)1=qr
Fork≠0, the first two can be rearranged to give the simultaneous equations
1+p^2=q+r10=r-q2
Adding 1 and 2 gives
1+p^2=2r
Thus solving 1 plus -1×2 we aquire
1+p^2=q+r-1(r-q)1+p^2=2q
With (1+p^2)/2=q and(1+p^2)/2=r where q=r
We now have expressions for q andr. Therefore we can substitute them into the final equation1=qr.1=((1+p^2)/2)((1+p^2)/2)1×2×2=(1+p^2 )(1+p^2 )4=1+2p^2+p^4 0=p^4+2p^2-30=p^6+2p^4-3p^2 (13)
The result is a cubic inp^2, and hence is solvable.
Clearly p=1 is a root since
(1)^6+2(1)^4-3(1)^2=1+2-3=0
By assessment, the other roots of the sextic are p=0,-1,i√3 and -i√3
Now substituting p=1 in (12) we arrive at the following pair of quadratic equations
0=X^2+(1)X+1/2 ((1)+(1)^2-((0))/((1) )) X^2+X+1/2 (2) X^2+X+1
0=X^2-(1)X+1/2 ((1)+(1)^2+((0))/((1) )) X^2-X+1/2 (2) X^2-X+1
Ultimately solving these as we did earlier, and remembering the original substitution it follows the roots of the initial quartic are 3/2+√3/2 i,3/2-√3/2 i,5/2+√3/2 i and5/2-√3/2 i.

7 Euler’s Method
The last method I will explore is one by Euler, which is primarily based on an extension of his solution of the cubic [following 3]
Initially let x=√r+√s+√t

Squaring this expression and rearranging we achieve
x^2=(√r+√s+√t)^2 x^2=r+√rs+√rt+√sr+s+√st+√rt+√st+tx^2=r+s+t+2(√rs+√rt+√st)
By subtracting (r+s+t) from both sides of the equation then further squaring a second time we acquire
x^2-(r+s+t)=2(√rs+√rt+√st) (x^2-(r+s+t))^2=(2(√rs+√rt+√st))^2 x^4-2(r+s+t) x^2+(r+s+t)^2=4(rs+rt+st)+8(√rsrt+√rsst+√rtst) x^4-2(r+s+t) x^2+(r+s+t)^2=4(rs+rt+st)+8√rst (√r+√s+√t)
We can now substitute x=√r+√s+√t and equate to zero to obtain the reduced quartic
x^4-2(r+s+t) x^2-8√rst x+(r+s+t)^2-4(rs+rt+st)=0
Hence starting with a quartic of the form (5) and by taking suitable r,s and t values, we know that a root of the original quartic will bex=√r+√s+√t. Thus equating coefficients it follows
a=-2(r+s+t)b=-8√rst c=(r+s+t)^2-4(rs+rt+st)
The system can be rewritten with the use of the following three variable symmetric functions
σ_1=r+s+tσ_2=rs+rt+stσ_3=rst
Thus it follows
σ_1=(-a)/2 c=σ_1^2-4σ_2 σ_2=1/4 (σ_1^2-c) 1/4 (a^2/4-c) (a^2-4)/16 b^2=64rstσ_3=b^2/64

Hence σ_1=(-a)/2,σ_1=(a^2-4)/16 and σ_3=b^2/64 (14)
We have now have an expression for our known coefficients a,band c in terms of the introduced r,s andt. Euler concluded that r,s and t make up the roots of a cubic with coefficients 〖-σ〗_1,σ_2and 〖-σ〗_3 such that
y^3+(a/2) y^2+((a^2-4)/16)y-b^2/64 (15)
Accordingly this cubic is solvable.

Example 7.1
Initially letX=√r+√s+√t , for our polynomial (5) with a=1,b=0 andc=1. Now using (14) we have the following values for our coefficients σ_1=(-1)/2 σ_2=(-3)/16 σ_3=0
Consequently we have a solvable cubic of the form (15) below with coefficients 〖–σ〗_1=1/2,σ_2=(-3)/16and 〖-σ〗_3=0
X^3+1/2 X^2-3/16 X=0(16)
Clearly X=0 is a root since
X(X^2+1/2 X-3/16)=0
Thus solving the quartic equation it follows
X=((-1)/2±√((1/2)^2-4(1)((-3)/16) ))/2(1) ( (-1)/2±√(1/4+12/16))/2 ( (-1)/2±√1)/2Therefor the solutions are X=1/4,-3/4,0 which are our values in no particular order for r,s andt.
Referring back to our earlier substitution for Xit follows X=√(1/4)+√((-3)/4)+√0 √1/√4+(i√3)/√4±1/2±(i√3)/2
Since x=X+2the four solutions as found earlier are 3/2+√3/2 i,3/2-√3/2 i,5/2+√3/2 i and 5/2-√3/2 i

8 Conclusion
After analysing all mathematicians’ methods for solving the quartic, my favourite method would have to be by Euler. Personally, this is by far the quickest and direct way to find quartic roots. Once you take the reduced quartic coefficients a,b and c and plug these values in to find σ_1,σ_2andσ_3, you subsequently form a cubic polynomial and solve accordingly.
One must notice, that using the substitution X=p^2/4 for the cubic found in Euler’s method will results as that of Descartes’ cubic inp^2.[3, Chapter 4.4, page 86]
Taking Euler’s cubic (16) and thus letting X=p^2/4 it can be shown thatX^3+1/2 X^2-3/16 X=0(p^2/4)^3+1/2 (p^2/4)^2-3/16 (p^2/4) p^6/64+p^4/32-(3p^2)/64 p^6+2p^4-3p^2
Which is the same result found earlier (13)
I admire the steps formed by Ferrari and he deserves the most credit for being the initial stepping stone to all further techniques.
As discussed throughout, we saw how the representation of equations has continuously changed. Many aspects of mathematics today wouldn’t be the same if it wasn’t for the consuming stages of research encountered by each mathematician.
Nowadays due to vast advances in technology, we can get solutions to polynomial equations via several sources online, including Wolfram Alpha!

Appendix I

Lodovico Ferrari (February 2, 1522 – October 5, 1565)
Lodovico Ferrari was a famous mathematician, known for solving fourth-degree equations. He was looked after by his father who was sadly killed, which steered him to live with his uncle Vincent. Ferrari had a mischievous cousin called Luke who decided to run away to Milan and take up a servant job for the famous Italian doctor and mathematician Girolamo Cardano. Not keeping his half of the deal, Luke up and left his new responsibilities and returned home. At the age of only fourteen, Ferrari was forced to go to Milan and complete the services his cousin had neglected.
Ferrari was a bright boy and for this reason Cardano alternatively employed him as a secretary. Cardano taught Ferrari mathematics throughout his teenage years and by the age of twenty Ferrari was teaching geometry in Milan at the Piatti Foundation.
In 1540, his noteworthy breakthrough in mathematics was the solution of the quartic equation. However, due to an oath with Niccolò Fontana Tartaglia, publishing was delayed until 1545 when the formula appeared in Cardano’s legendary algebra book Ars Magna. This resulted in uproar between Ferrari and Tartaglia, and thus a contest broke out. Ferrari triumphed and was offered numerous job opportunities, including an invitation to teach the Emperor’s son.
He ultimately became a wealthy man working as a tax inspector, which enabled him to take an early retirement. Nonetheless he returned to Bologna in 1565 but died soon after due to poisoning. Apparently he was betrayed by his sister Maddalena, who went on to inherit his immense fortune. As luck would have it, she didn’t get off lightly. Maddalena remarried two weeks after her brother’s death, yet her new partner unquestionably took off with all the assets, leaving her with nothing. [Ferrari’s biography follows ideas from 9]

François Viète (1540 – December 13, 1603)
Viète started out as a lawyer. After training at the University of Poitiers he returned to his home town to exercise his line of work. He was successful at what he practiced and advised the likes of Queen Mary of England, Eleanor of Austria and King Henry III of France. He only studied mathematics during his free time and it wasn’t until his later years that he worked on solving equations.
In 1593 a future companion, Adriaen van Roomen, posed a 45th-degree equation.
To be precise
45x-3795x^3+95634x^5-…+945x^41-45x^43+x^45=N[6]
Viète was able to provide 23 solutions, after highlighting that the equation was the expansion of sin45θ in powers ofsinθ. Back then, negative solutions were unfamiliar and subsequently not noted.
At present Viète is known by some as “the father of algebra” as so mentioned in [11] for his contributions to mathematics, for instance theories of equations and algebraic notation. He is also a legend for cracking a composite cipher for Henry IV, which was used throughout the war against Spain. [Viète’s biography following ideas from 6, chapter 6.8 pages 107-108]
Thomas Harriot (1560 – July 2, 1621)
Harriot was a mysterious English mathematician, who graduated from Oxford. Not much is known about his life or if he ever married. It was through his friend Nathaniel Torporley that he became accustomed to Viète. Harriot took a liking to Viète’s research and prior to publication, he went on to rewrite voluminous papers in a new, contemporary fashion. Harriot had an analogous approach to mathematical methods compared to Viète yet he never published any of his work. Soon after his death, due to nose cancer, executors attempted to gather his findings. Unfortunately the results found do not truly portray his remarkable mathematical encounters in depth [4, page 469].
Harriot’s remarks on cubic equations developed his further understanding for higher degree equations. He illustrated the following as stated in [4]
If a,b,c are roots of a cubic equation, then the cubic is
(x-a)(x-b)(x-c)=0
He is additionally known for his observations of moons, and the discovery of “greater than” and “less than” symbols. Despite the fact that these were later edited to the ones we currently know explicitly and. [6, pages 352-354]
Today Harriot’s legacy lives on and his home on the Thames has been renamed to reflect this. Even though his grave was destroyed during the great fire of London 1666, a replica monument can now be found situated within the Bank of England. [6, pages 352-354]

René Descartes (March 31, 1596 – February 11, 1650)
As a boy, due to delicate health reasons, Descartes was given distinct privileges. To name a few, as a student he was given his own room, granted approval to stay in bed late and read books off-limits to his peer group. This treatment did not benefit him during his later life. [6, pages 122-125]
To grant his father’s wishes, Descartes went on to study Law at the University of Poitiers. Upon completion, he volunteered in military school and began researching mechanics. Sometime later he enrolled in the Maximilian of Bavaria army, where he serviced for a year. [7]
In 1649 he moved to Stockholm to lecture Queen Christina in philosophy. She was keen to have lessons at 5am, which put a lot of strain and pressure on Descartes. Unfortunately, during one very cold winter, he caught pneumonia and drastically did not recover. [6, pages 122-125]

Leonhard Euler (April 15, 1707 – September 18, 1783)
Euler was a revolutionary Swiss mathematician. He was taught elementary mathematics from a young age by his father, Paul. Euler enrolled at the University of Basel at only 14 years old and by 1723 he had completed a master’s degree in philosophy.
Euler was privately taught mathematics by Johann Bernoulli, his father’s companion, who noticed Euler’s natural flair and willingness in the subject. Johan Bernoulli influenced Paul to let his son study mathematics instead of going down the ministry route.
Euler’s Notation
Symbol Year Classification
f(x) 1734 Functions
e 1736 est. The base of the natural logarithm
i 1777 Used for the square root of -1, “imaginary unit”
π 1736 est. Pi
∑ 1755 Sigma, for summation
∆yand ∆^2 – Finite differences
cos, sin, tan – Trigonometric functions
Euler went on to the Academy of Sciences in St. Petersburg, where he was professor of Mathematics and head of the Geography department. It was here that he stumbled upon Katharina Gsell. They later married and she gave him thirteen offspring.
Euler introduced voluminous new aspects in several areas of mathematics, including distinguished additions to Fermat’s Last Theorem, and mathematical symbols as seen in Figure 7.
One prominent example of his explorations would be Euler’s Formula
e^ix=cosx+isinx
He began to lose his sight in 1738, nonetheless he was a very keen determined man and this didn’t hold him back. Some even say that his astonishing memory was down to his lack of sight.
During the early 1740’s Euler was a highly accredited gentleman. On two occurrences he claimed joint first prize of the notable Grand Prize of Paris Academy. He resided in Berlin for over 20 years, until he eventually returned back home to St. Petersburg, where his sight significantly diminished.
[Euler’s abstract following ideas from 8 & 6, pages 200-202]
Figure 7 – Euler’s Notation

References

[1] Burton, D., M., The History of Mathematics – An Introduction, McGraw-Hill, 7th ed.(2010) (Chapter 7.4 pp 329-331)
[2] Hart, S., Problems in Mathematics 2014-15 Lecture 1: Polynomials, a brief history.
[3] Kalman, D., Uncommon Mathematical Excursions: Polynomia and Related Realms, The Mathematical Association of America (2009), ISBN 978-0-88385-341-2
[4] Katz, J.,V., A History of Mathematics, Pearson, 3rd ed.(2008)
[5] Math World, Quartic Equation, 17 Mar. 2015, http://mathworld.wolfram.com/QuarticEquation.html
[6] Stillwell, J., Mathematics and Its History (Undergraduate Texts in Mathematics), Springer; 3rd ed.(2010)
[7] The Mac History of Mathematics archive, Biographies, Descartes, 1 May. 2015, http://www-history.mcs.st-andrews.ac.uk/Biographies/Descartes.html
[8] The Mac History of Mathematics archive, Biographies, Euler, 1 May. 2015, http://www-history.mcs.st-andrews.ac.uk/Biographies/Euler.html
[9] The Mac History of Mathematics archive, Biographies, Ferrari, 1 May. 2015, http://www-history.mcs.st-andrews.ac.uk/Biographies/Ferrari.html
[10] The Mac History of Mathematics archive, Biographies, Harriot, 1 May. 2015, http://www-history.mcs.st-andrews.ac.uk/Biographies/Harriot.html
[11] The Mac History of Mathematics archive, Biographies, Viète, 1 May. 2015, http://www-history.mcs.st-andrews.ac.uk/Biographies/Viete.html
[12] Wikipedia, Nth root, 4 Apr. 2015, http://en.wikipedia.org/wiki/Nth_root
[13] Wikipedia, Quartic Function, 4 Apr. 2015, http://en.wikipedia.org/wiki/Quartic_function
[14] Wolfram Alpha, 6 May 2015, https://www.wolframalpha.com/input/?i=x^4%2F4-%284+x^3%29%2F3%2Bx^2%2B4+x-2