Thursday
October 23, 2014

Homework Help: math

Posted by Steve on Tuesday, January 23, 2007 at 1:47pm.

Still having problems solving this...

2x(2x + 1)^2 = 312

I started with:
2x(2x+1)(2x+1)-312 = 0
2x(4x^2+4x+1)-312 = 0
8x^3+8x^2+2x -312 = 0

Now what?

If you don't have to solve for x, then then you have the final answer!

If you have to find x: then you have to use the group method.

(8x^3+8x^2)+ (2x-312)
8x^2(x + 1) + 2(x-156)

Since there doesn't seem to be any like terms after factor. The answer may have well been what you had above. Someone on here may second this.


I do have to solve for x...
Is this a cubic polynomial???
Still not sure.

8x^3+8x^2+2x -312 = 0

Now what?

divide by 2:

4x^3+ 4x^2+ x - 156 = 0

Try the Rational Roots theorem. If there is a root of the form p/q with p and q integers that don't have common factors, then p must be a divisor of 156 and q a divisor of 4. I don't see a candidate that works...

If you draw the graph of this function (it helps to calculate the derivative to find the local maximum and minimum), you see that this function has only one root at approximately x = 3. Using a high precision numerical algorithm I find that the root is at:

x = 3.06633690898....

This is certainly not a rational root, because the only possible rational roots would have to become an integer after multiplication by 4

It is possible to exactly calculate the roots of cubic equation. If we divide the equation by 4 we get:

x^3+ x^2 + 1/4x - 39 = 0


First you substitute:

x = y - 1/3

and work out the polynomial as a function of y. This has the effect of eliminating the quadratic term. We get:

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

with:

p = -1/12

q = -4213/108


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 1/3 to find x.

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