Consider the two quartic equations below.
$\displaystyle \color{white}.\quad.$ $\displaystyle \begin{array}{crcl}
(1) & x^4+x^3-2x^2-x+1&=&0\\
(2) & x^4+5x^3+7x^2+2x&=&0
\end{array}$
Which of the two equations is easier to solve? And why?
Consider the two quartic equations below.
$\displaystyle \color{white}.\quad.$ $\displaystyle \begin{array}{crcl}
(1) & x^4+x^3-2x^2-x+1&=&0\\
(2) & x^4+5x^3+7x^2+2x&=&0
\end{array}$
Which of the two equations is easier to solve? And why?
At first glance the second one as you can take a factor of x out straight away however the cubic that is left; $\displaystyle x^3-5x^2+7x+2$ is quite difficult to find a factor of.
The first one however is much easier even though at first look it doesn't seem to be. This is because a factor of $\displaystyle (x-1)$ can be noticed relatively quickly leaving the cubic $\displaystyle x^3+2x^2-1$ in which the factor $\displaystyle (x+1)$ can be found leaving the quadratic $\displaystyle x^2+x-1$.
Well by inspection $\displaystyle x-1$ is a factor of the first, as is $\displaystyle x+1$ so:
$\displaystyle x^4+x^3-2x^2-x+1=(x-1)(x+1)(x^2+x-1)$
So the first is no more difficult to solve than a quadratic.
The second has $\displaystyle x$ as a factor leaving a cubic, which by applying the rational
root theorem we discover has a factor $\displaystyle x+2$, so:
$\displaystyle x^4+5x^3+7x^2+2x=x(x+2)(x^2+3x+1)$
So the second is also no more difficult to solve than a quadratic.
So they are equally easy to solve.
(I should add that the two quadratic factors do have real roots but they are not rational)
RonL
Thanks!
This was the result of trying to solve a problem posed by someone on another site. We were to solve the following equation:
$\displaystyle
\color{white}.\quad.
$ $\displaystyle (x+2)^2+(x+3)^3+(x+4)^4=2$
The idea of course is NOT to tackle the equation directly by expanding everything out into a mess, but to make a substitution to simplify things a bit. I suggested the substitution $\displaystyle y=x+3$, which would cause the constant terms to disappear, leading to $\displaystyle y^4+5y^3+7x^2+2y=0$ (which is Eqn (2) above).
But someone else suggested the substitution $\displaystyle y=x+4$ instead. This leads to Eqn (1): $\displaystyle y^4+y^3-2y^2-y+1=0$. At first glance, I was like Sean, thinking that my equation with constant term zero was easier to tackle. Now I know, thanks to CaptainBlack, that both subsitutions are equally good.
What I am not happy with is that the other person said my substitution was “stupid” when his was at best only as good as mine: http://www.mathisfunforum.com/viewtopic.php?id=9144
Immature f**l, because he has a solution he thinks it is the best.
Possibly he does not have the tools to easily spot rational roots other than +/-1.
Of course he may be refering to it being less work to expand a cube of the form
$\displaystyle (x+a)^3$ than to expand a fourth power like $\displaystyle (x+a)^4$
RonL
hmmmmmph .... mathisfun indeed ..... until someone loses an eye (no offence to Euler).
Some salient points:
1. The other person is a novice. You're a super duper member - with all the rep and cred that brings with it.
2. By their own signature, their English is not very good. Maybe they meant the word smart but got all muddled up. Maybe they thought it funny --> fun, like the forum name - in which case, they need to add 'my sense of humour is not very good, either' to their signature.
3. The other person posed the original question. Clearly your solution suggested a similar approach to them - unfortunately it also gave them delusions of grandeur.
4. The good Captain, as always, raises some good points:
5. At the end of the day, the mathematics does the talking. It says that neither solution is easier than the other. And that's that.
6. Get some perspective - Read the words to The Sunscreen Song (they're at the end).