Making decision by weighing the data

Solve for all 6 complex roots of the equation $x^6+10x^5+70x^4+288x^3+880x^2+1600x+1792=0$.


Previously we talked about how to assign $2^8(7)$ to be the constants for the three factors such that we have


$\small x^6+10x^5+70x^4+288x^3+880x^2+1600x+2^8(7)=(x^2+ax+?)(x^2+bx+?)(x^2+cx+?)$


We realized that we would get to correctly assign the number as the constants for the three factors if we have got lucky, what if we don't have the luck and have to exhaust all possibility?


As the math educator, I would ask you to stop solving for this problem for a short while and turn your focus to asking students the number of cases that are possible to rewrite $2^8(7)$ as the product of three positive integers.


Let your students to shift their mind from algebra to probability, this often generates a good course for students as in real life, we have to deal with the situation where we have to consider many aspects so to suggest good solutions than coming up with approach that would waste our precious time and effort hugely.


We could, list out all the possible cases and do it like follows:


For [MATH]\color{yellow}\bbox[5px,purple]{2^8(7)=1(7)(2^8)}[/MATH], we could split it down to such cases:

[MATH]\left.\begin{array}{I}\begin{align*}2^8(7)&=(2^1)(7)(2^7)\\&=(2^2)(7)(2^6)\\ &=(2^3)(7)(2^5)\\ &=\cancel{(2^5)(7)(2^3)}\end{align*}\end{array}\right\}\text{3 cases}[/MATH]

For [MATH]\color{yellow}\bbox[5px,green]{2^8(7)=1(7\cdot 2)(2^7)}[/MATH], we could split it down to such cases:

[MATH]\left.\begin{array}{I}\begin{align*}2^8(7)&=1(7\cdot 2)(2^7)\\&=(2^1)(7\cdot 2)(2^6)\\ &=(2^2)(7\cdot 2)(2^5)\\ &=(2^3)(7\cdot 2)(2^4)\\&=\cancel{(2^4)(7\cdot 2)(2^3)}\end{align*}\end{array}\right\}\text{4 cases}[/MATH]

For [MATH]\color{black}\bbox[5px,orange]{2^8(7)=1(7\cdot 2^2)(2^6)}[/MATH], we could split it down to such cases:

[MATH]\left.\begin{array}{I}\begin{align*}2^8(7)&=1(7\cdot 2^2)(2^6)\\&=(2^1)(7\cdot 2^2)(2^5)\\ &=(2^2)(7\cdot 2^2)(2^4)\\ &=(2^3)(7\cdot 2^2)(2^3)\\&=\cancel{(2^4)(7\cdot 2^2)(2^2)}\end{align*}\end{array}\right\}\text{4 cases}[/MATH]

For [MATH]\color{yellow}\bbox[5px,blue]{2^8(7)=1(7\cdot 2^3)(2^6)}[/MATH], we could split it down to such cases:

[MATH]\left.\begin{array}{I}\begin{align*}2^8(7)&=1(7\cdot 2^3)(2^5)\\&=(2^1)(7\cdot 2^3)(2^4)\\ &=(2^2)(7\cdot 2^3)(2^3)\\ &=(2^3)(7\cdot 2^3)(2^2)\\&=\cancel{(2^4)(7\cdot 2^3)(2^1)}\end{align*}\end{array}\right\}\text{4 cases}[/MATH]

For [MATH]\color{yellow}\bbox[5px,red]{2^8(7)=1(7\cdot 2^4)(2^4)}[/MATH], we could split it down to such cases:

[MATH]\left.\begin{array}{I}\begin{align*}2^8(7)&=1(7\cdot 2^4)(2^4)\\&=(2^1)(7\cdot 2^4)(2^3)\\ &=(2^2)(7\cdot 2^4)(2^2)\\ &=\cancel{(2^3)(7\cdot 2^4)(2^1)}\end{align*}\end{array}\right\}\text{3 cases}[/MATH]

For [MATH]\color{yellow}\bbox[5px,indigo]{2^8(7)=1(7\cdot 2^5)(2^3)}[/MATH], we could split it down to such cases:

[MATH]\left.\begin{array}{I}\begin{align*}2^8(7)&=1(7\cdot 2^5)(2^3)\\&=(2^1)(7\cdot 2^5)(2^2)\\ &=\cancel{(2^2)(7\cdot 2^5)(2^1)}\end{align*}\end{array}\right\}\text{2 cases}[/MATH]

For [MATH]\color{black}\bbox[5px,yellow]{2^8(7)=1(7\cdot 2^6)(2^2)}[/MATH], we could split it down to such cases:

[MATH]\left.\begin{array}{I}\begin{align*}2^8(7)&=1(7\cdot 2^6)(2^2)\\&=(2^1)(7\cdot 2^6)(2^1)\\ &=\cancel{(2^2)(7\cdot 2^5)(2^1)}\end{align*}\end{array}\right\}\text{2 cases}[/MATH]

For [MATH]\color{black}\bbox[5px,orange]{2^8(7)=1(7\cdot 2^7)(2)}[/MATH], we only have one case for that.


Similarly for [MATH]\color{white}\bbox[5px,purple]{2^8(7)=1(7\cdot 2^8)(1)}[/MATH], we only have one case for that.


So altogether, we would end up with a total of 26 possible cases to work.


Now, ask yourself by answer to yourself honestly, do you really want to work out all those 26 cases to look for the right answer to factor


$\small x^6+10x^5+70x^4+288x^3+880x^2+1600x+2^8(7)=(x^2+ax+?)(x^2+bx+?)(x^2+cx+?)$?


I'm fairly certain that you would beg for another alternative and put this one under the carpet by now. :D