-x^3 + 1.85x^2 - x +0.15 = 0
how can you solve for x?
An exact solution or a numerical solution?
Exact solutions can be obtained as follows:
x^3 - 1.85x^2 + x - 0.15 = 0
First get rid of the quadratic term by substituting:
x = (y + 1.85/3)
You then obtain an equation of the form:
y^3 + p y + q = 0
Write this as:
y^3 = -p y - q
Compare this to the formula for (a+b)^3 which can be written as:
(a+b)^3 = 3 a b (a+b) + a^3 + b^3
This looks very similar to the equation you want to solve with a + b playing the role of y. If you can somehow choose a and b such that:
3 a b = -p (1)
and
a^3 + b^3 = -q (2)
then the equation becomes exactly similar and the solution would thus be
y = a + b
But how do you choose a and b such that Eqations (1) and (2) are satisfied? If you take the third power of equation (1), you get:
27 a^3 b^3 = -p^3
if we put A = a^3 and B = b^3, then we have:
A B = -p^3/27
and (2) implies that:
A + B = -q
Solving for A and B is easy, if you use one equation to eliminate, say A and substitute that in the other, you get a quadratic equation. So, all you have to do is to solve that quadratic equation, calculate a and b from that.
if i already know x-1 is a factor, how can i solve it from here?
First of all multiply each term by -1 to start your equation with a positive term.
notice that the coefficients of the positive terms add up to 2 and the coefficients of the negative terms add up to -2, so x=1 is a solution (this was luck)
Divide the cubic by x-1 to get x^2 - .85x + .15 = 0
which factors to (x-.6)(x-.25)=0
So you have 3 rational solutions:
x=1, x=.6, and x=.25
If you know one of the factors, then divide the function by that factor.
To solve the equation `-x^3 + 1.85x^2 - x + 0.15 = 0` when you already know that `x-1` is a factor, you can follow these steps:
1. Multiply each term by -1 to start the equation with a positive term: `x^3 - 1.85x^2 + x - 0.15 = 0`
2. Divide the cubic by `x-1` to get a quadratic equation: `(x^3 - 1.85x^2 + x - 0.15) / (x-1) = x^2 - .85x + .15 = 0`.
3. Factor the quadratic equation to find the remaining solutions: `(x-.6)(x-.25) = 0`
This means that either `(x-.6) = 0` or `(x-.25) = 0`.
Solve each equation separately to find the remaining solutions:
If `(x-.6) = 0`, then `x = 0.6`.
If `(x-.25) = 0`, then `x = 0.25`.
So, the three rational solutions to the equation are: `x = 1`, `x = 0.6`, and `x = 0.25`.
By dividing the function by the known factor `x-1`, you simplified the equation from a cubic to a quadratic. Then, by factoring the quadratic, you found the remaining solutions.