When can we not use synthetic division to solve for a cubic polynomial? For example we can use synthetic division to solve $-t^3 -4t^2 +20t +48$. When I can't use synthetic division what are my other options? I know there is the difference of two cubes technique, factoring out an $x$ might work if possible(then factoring polynomial with $x^2$), or factoring by grouping.
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7 Personally, when I am required to solve a polynomial with degree greater than 2, I like to use the Rational Root Theorem to find potential roots, and then use synthetic division using a root when it is found in order to reduce it to a nicer polynomial.
Note there is an explicit formula for cubic polynomials, but it is messy.
Using your question as an example, we know that possible rational roots are (plus or minus) factors of $48$. Just from guessing, we know that $-2$ is a root. Therefore we can reduce to the quadratic $-x^2 - 2x + 24$, which can be easily solved using the quadratic equation.
- $\begingroup$ Right but is the rational root theorem + synthetic division always possible or does it not always work? $\endgroup$Bubbletea– Bubbletea2015-04-10 19:54:08 +00:00Commented Apr 10, 2015 at 19:54
- $\begingroup$ @Bubbletea If the polynomial has a rational root, then yes it is. Note that any polynomial with an odd degree must have at least one rational root. It can be difficult if the polynomial has a very high degree, because you simply reduce to another polynomial with a high degree... so generally numerical methods are best for solving standard polynomials. But for cubic and quintic, I think that synthetic division and rational root theorem is the best approach for a quick and easy solution. $\endgroup$Addison– Addison2015-04-10 19:57:45 +00:00Commented Apr 10, 2015 at 19:57
- $\begingroup$ Not all polynomials with odd degree have a rational root, viz. $x^3+x+1$ $\endgroup$dioid– dioid2015-04-10 20:31:41 +00:00Commented Apr 10, 2015 at 20:31
- $\begingroup$ You're right - I was confusing that with real roots. Regardless, if it has real roots and can be factored, then you can use this system. $\endgroup$Addison– Addison2015-04-10 20:33:20 +00:00Commented Apr 10, 2015 at 20:33
- $\begingroup$ @RobertLewis it is $-x^3$, not $x^3$. So we have $-(-2)^3 = 8$, not negative. $\endgroup$Addison– Addison2015-04-10 20:41:35 +00:00Commented Apr 10, 2015 at 20:41