April 19, 2014

Homework Help: math

Posted by mary on Sunday, January 21, 2007 at 6:23pm.

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?


Honestly: I would graph the equations and find the roots first. I dont see them off hand.

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

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)


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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