Question 4

4. Generate a cubic:

ax3+bx2+cx+d

where a and b are nonzero integers so that your cubic is the un-depressed version of the cubic in Chapter 6: x3+6x-20. Next, use the solution to the depressed cubic in Chapter 6 to find a solution to your cubic.

To do this, I have to start with the general form:

let x = y - b/3a

ay³ + by² + cy + d = 0
Ben then reminds me to obtain another way this equation is written from Journey through Genius
ay³ + 0y² + (c -  (b²/3a)y + (d+(2b³/27a²)-(bc/3a)) = 0
So then I compare this to the depressed cubic x³+6x-20=0 that I'm un-depressing
ay³ = x³
(c -  (b²/3a)y = 6x
(d+(2b³/27a²)-(bc/3a)) = -20
If, for just a moment, I ignore the difference in variables and assume that x and y are equal because I do have an equation to convert between them (x = y - b/3a) then:
ax³ = x³; so a = 1

(c -  (b²/3a)x = 6x; (c -  (b²/3a) = 6 - another way to write this is: c = 6 + b²/3a
(d+(2b³/27a²)-(bc/3a)) = -20; plug that value of "c" and "a" that we've figured out into this equation:
d+(2b³/27a²)-(b(6 + b²/3a)/3a)) = -20
d+(2b³/27(1)²)-(b(6 + b²/3(1))/3(1))) = -20
( d+(2b³/27)-(b(6 + b²)/3) = -20) 27 - multiply both sides by 27 to get rid of the fractions:
27d + 2b³ - 9b(6 + b²) = -540
27d + 2b³ - 9b³ - 54b = -540
27d - 7b³ - 54b = -540 - at this point, I don't have another equation to solve for d and b, but I can approximate their value by trial and error. I also know that b has to be a non-zero integer. But, other than this limitation, b can be any number as long as it satisfies the equation. I then decided to try b = 3 because 3 is a non-zero integer and I can easily find the value of 3³.
So:
27d - (7)3³ - 54(3) = -540
27d - 189 -  162 = -540
 27d - 351 = -540
27d = -189
d = -189/27 = -7
Then, I still have to solve for the actual value of c:
c = 6 + b²/3a
c = 6 + 3²/3(1)
c = 6 +9/3
c = 6 + 3
c = 9
Now, I that know all the values of a, b, c, and d, I can plug this back into the general form:
ay³ + by² + cy + d = 0, which can be rewritten as:
ax³ + bx² + cx + d = 0
x³ + 3x² + 9x - 7 = 0

To find the solution to my cubic I need to use this equation:
x = y - b/3a, which I can rewrite as:
x₁ = x₂ – b/3a; x₁ is the solution to my cubic and x₂ is the solution to x³+6x-20

Dunham had found in Journey through Genius that the solution to x³+6x-20 = 0 is that x = 2

Therefore, x₂ = 2, and I already know b and a from doing the problem earlier, b = 3 and a=1.

So:

 x₁ = 2 – (3/3(1))

 x₁ = 2 - (3/3)

x₁ = 2 - 1

x₁ = 1