math
posted by mary .
Still looking at how to solve these...
Both are cubic polynomials?
43x^3 + 13x^2 + 18x  456 = 0
DO I divide first by x1?
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 xy 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.
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