I don't have any links but I still think I can help you understand. Here's an example.
So say you have a function, ax^4 + bx^2 + c = 0; you can approach this the same way a typical quadratic.
ax^4 + bx^2 + c = 0
x^4 + (b/a)x^2 = -c/a
-i subtracted "c" from both sides and divided both sides by "a"
x^4 + (b/a)x^2 + (b/2a)^2 = -c/a + (b/2a)^2
(x^2 + b/2a)^2 = -c/a + (b/2a)^2
x^2 + b/2a = (+/-)[ (-4ac)/(4a^2) + (b^2)/(4a^2) ]^0.5
x^2 + b/2a = (+/-)[ [ b^2 - 4ac ]^0.5 ]/2a
x^2 = [ -b (+/-) [ b^2 - 4ac ]^0.5 ]/2a
--as you can see, this looks similar to the quadratic formula EXCEPT everything is equal to x^2. So to find x you would simply take the square root of the quadratic formula:
x = (+/-)[ [ -b (+/-) [ b^2 - 4ac ]^0.5 ]/2a ]^0.5
--obviously the only values that would work are the positive ones.
For ODD exponential polynomials, you cannot complete the square at FIRST because the technique only works for quadratic polynomials. So you have to approach odd exponential polynomials in a different manner. I prefer doing the "rational root" test and applying synthetic division. For example:
x^3 + 2x^2 - x - 2 = 0
Like I said, completing the square isn't going to work out too well, AT FIRST, so you do the "rational root" test to see if you can get it in a form that allows you to complete the square.
I'll test x = 1 to see if it works:
x = 1: 1 2 -1 -2
1 3 2
1 3 2 0
--remainder is zero, so x = 1 is a root, so you write that down
So now you can rewrite the polynomial as:
x^3 + 2x^2 - x - 2 = 0
(x - 1)(x^2 + 3x + 2) = 0
NOW, you can now complete the square for the second term in the paranthesis since you factored out x = 1:
x^2 + 3x + 2 = 0
x^2 + 3x + 9/4 = -2 + 9/4
(x + 3/2)^2 = 1/4
x + 3/2 = (+/-)[1/4]^0.5
x = -(1/2) + 3/2; x = (1/2) + 3/2
x = 1; x = 2
---these are the roots
So, x^3 + 2x^2 - x - 2 = 0 can is now factored and can rewritten as
(x - 1)(x + 1)(x - 2)
Hope that helped you out.
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