Still looking at how to solve these...

Both are cubic polynomials?

43x^3 + 13x^2 + 18x - 456 = 0
DO I divide first by x-1?

2x(3x+2)^2=312

Honestly: I would graph the equations and find the roots first. I don't see them off hand.

graph y=43x^3 + 13x^2 + 18x - 456

and
y=2x(3x+2)^2 -312

Try to find the rational roots (if there are any).

If the equation

43x^3 + 13x^2 + 18x - 456 = 0

has rational roots x = p/q then p must be a divisor of 456 and q must be a divisor of 43. I didn't find anything , but I only tried for half a minute :)

If you find a rational root, say x = y, then you divide the polynomial by x-y and solve the resulting quadratic equation.

If there really aren't any rational roots you can solve the cubic equation exactly as follows.

First you substitute:

x = y - 13/3

and work out the polynomial as a function of y. This has the effect of eliminating the quadratic term. So, you get an equation of the form:

y^3 + p y + q = 0 (1)

You solve this equation by comparing this equation to the equation for
(A + B)^3:

(A + B)^3 = A^3 + 3 A^2B + 3 A B^2 + B^3

rearranging gives:

(A + B)^3 - 3AB(A + B)-(A^3 + B^3) = 0

What use is this? Well, suppose you can find A and B such that

3AB = -p (2)

and

A^3 + B^3 = -q (3)

then Y = A + B would satisfy the equation (1)

This should be possible because we need to solve two equations for two unknowns A and B. So, how do we go about solving for A and B? What you do is you take the third power of Eq. (2):

27 A^3 B^3 = -p^3

If you now put A^3 = s and B^3 = t then this means that

s*t = -p^3/27

while Eq. (3) implies that:

s + t = -q

Combining these two equations for
s and t gives you a quadratic equation for these quantities. By extracting the cube root you find A and B,. Add them together and you find y. Subtract the term 13/3 to find x.

It turns out that you need to make use of complex numbers to find all three roots even if they are all real.

To solve the first cubic polynomial equation, 43x^3 + 13x^2 + 18x - 456 = 0, one approach is to try to find any rational roots it might have. To do this, you can use the Rational Root Theorem.

The Rational Root Theorem states that if a polynomial equation has a rational root p/q, where p is a divisor of the constant term (456 in this case) and q is a divisor of the leading coefficient (43 in this case), then p/q is a possible rational root.

In this case, you would need to find the divisors of 456 and the divisors of 43, and check if any of the possible combinations yield a rational root. However, it seems that there are no rational roots based on initial inspection.

If you find a rational root, say x = y, then you can divide the polynomial by x-y and solve the resulting quadratic equation.

If there are no rational roots, you can solve the cubic equation using a substitution method. First, you substitute x = y - 13/3 to eliminate the quadratic term. This will give you an equation of the form y^3 + p y + q = 0. Let's call this equation (1).

To solve this equation, you can use a method called "depressed cubic equation." You compare equation (1) to the expanded form of (A + B)^3:

(A + B)^3 = A^3 + 3 A^2B + 3 AB^2 + B^3

By matching the terms, you can set up two equations:

3AB = -p (equation 2)
A^3 + B^3 = -q (equation 3)

Solving these two equations for A and B will allow you to find the value of Y = A + B, which satisfies equation (1). To solve for A and B, you can cube equation (2) to get 27A^3B^3 = -p^3. Let A^3 = s and B^3 = t, so we have st = -p^3/27. From equation (3), we have s + t = -q. Combining these two equations gives a quadratic equation for s and t.

Once you solve for s and t, you can extract the cube root to find A and B. Adding A and B together will give you the value of y. Finally, subtracting the term 13/3 will give you the value of x.

It's worth mentioning that sometimes solving cubic equations may require the use of complex numbers, even if all the roots are real.