Solve for all 6 complex roots of the equation $x^6+10x^5+70x^4+288x^3+880x^2+1600x+1792=0$.
The most common method that students would think of how to tackle this problem is by setting up three factors, toying around the possible candidates for the constants for each factor as the prime factorization of $1792=2^8(7)$.
The thing is, how to assign $2^8(7)$ and separate them to become a product of three positive integers could be very tricky, and it doesn't seem wise to conduct an exhaustive search to see what kind of combination of the constant values that their product leads up to $1792$...
You might be very fortunate and luck into the true case, but, in solving any mathematical problem, we don't want to rely on luck, what we desire is the ability to speak so fluently and professionally in math that we want to assure readers what we have done would be a good solution.
Chances are really bright too that you may become frustrated for obtaining the silly guess, like how I would show you in the following example:
Let
$x^6+10x^5+70x^4+288x^3+880x^2+1600x+1792$
$=(x^2+ax+2)(x^2+bx+4)(x^2+cx+2^5(7))$
$=(x^2+ax+2)(x^2+bx+4)(x^2+cx+224)$
I then expand the first two factors of the RHS and then compare the coefficients to obtain the values for $a,\,b,\,c$:
$x^6+10x^5+70x^4+288x^3+880x^2+1600x+1792$
$=(x^4+(a+b)x^3+(6+ab)x^2+(4a+2b)x+8)(x^2+cx+224)$
Compare the coefficients of $x^5,\,x^4,\,x^3,\,x^2,\,x$, we get:
$10=c+a+b$
$70=224+(a+b)c+6+ab$
$288=224(a+b)+(6+ab)c+4a+2b$
$880=224(6+ab)+(4a+2b)c+8$
$1600=224(4a+2b)+8c$
After some tedious working with the above system of equations, we found out that there is no real solutions exist.
Quch! What a giant waste of time! But if you don't have a plan, or strategy, you're definitely open to risk, risk of running out of patient and time to continue with this so-called trial and error methods, when the possibilities are seemingly "endless".
Being a mathematics proficient student, you need to know when to discontinue with inappropriate and unproductive strategy.
I encourage you to try the problem, before seeing my solution that I will be posting tomorrow.
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