I know it has two x-intercepts but I'm not sure how to factor polynomials with degrees greater than 2..

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- Jun 14th 2012, 11:03 AMdaigoHow to factor 3*(x^4) + 4*(x^3) - 3
I know it has two x-intercepts but I'm not sure how to factor polynomials with degrees greater than 2..

- Jun 14th 2012, 11:33 AMrichard1234Re: How to factor 3*(x^4) + 4*(x^3) - 3
It doesn't factor completely over the integers. You can factor it as $\displaystyle x^3 (3x + 4) - 3$ but that's about all you can do.

- Jun 14th 2012, 11:52 AMdaigoRe: How to factor 3*(x^4) + 4*(x^3) - 3
There's no way I can find the exact values of the zeroes of this function algebraically?

- Jun 14th 2012, 12:01 PMrichard1234Re: How to factor 3*(x^4) + 4*(x^3) - 3
Both real roots are irrational. There is a quartic formula, but I wouldn't bother memorizing it. Newton's method is probably the best "algebraic" way of solving it.

Let $\displaystyle f(x) = 3x^4 + 4x^3 - 3$. For some initial "guess" $\displaystyle x_0$, let $\displaystyle x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} $ and, for $\displaystyle i \ge 2$, $\displaystyle x_i = x_{i-1} - \frac{f(x_{i-1})}{f'(x_{i-1})}$. As $\displaystyle i$ gets larger, you should approach a root of f(x).