Solving Quartic Equations: A > 1

[navee]

How can we factor a question which is too the power of 4 or quartic. for example ax^4+bx^3+cx^2+dx+e, but a has a grreater value than 1.

 

[shaner-baner]

I'm not sure, but I think that fourth order equations are solvable with some weird equation like third order. In practice, you might start with the rational zeros test. What you do is: take all possible factors of 'e' and divide by all possible factors of 'a'. Then test these to see if they are zeros (if q is a zero, x-q is a factor). Using a procedure called synthetic division let's you use the factors that you find to reduce the order of the equation. When the equation has finally been reduced down to a quadratic, then you can use the quadratic formula. Alternately, there are 'numerical' methods that can find approximations of the zeros, (i.e. Newtons Method). There are a few other tricks. For example. If it looks like ax^4+bx^2+c, you can just use the quadratic formula to solve for x^2, and then take the square root of your answers.
In general factoring higher order polynomials is a tricky business. Hope I answered your question

 

[HallsofIvy]

Theoretically, any polynomial equation can be factored completely into linear factors if you use complex numbers, linear and quadratic factors if you stay in real numbers.

Yes, there exist a "quartic" formula. Here is a website with a (comparatively) simple explanation: http://web.usna.navy.mil/~wdj/book/node95.html

There is, however, no general method for factoring except by solving the equation and then using the solutions to get the factors: If a, b, c are solutions, the we can factor as p(x-a)(x-b)(x-c).