Is there a nice factorization?

Let $x=2$ so the given equation becomes

$\large{{ x }^{ 8 }+{ \left( \sum _{ i=1 }^{ 8 }{ { a }_{ i } } \right) x^{ 7 } }+\left( \sum _{ 1\le i<j\le 8 }^{ }{ { a }_{ i }{ a }_{ j } } \right) { x }^{ 6 }+\left( \sum _{ 1\le i<j<k\le 8 }^{ }{ { a }_{ i }{ a }_{ j }{ a }_{ k } } \right) { x }^{ 5 }+...+\prod _{ i=1 }^{ 8 }{ { a }_{ i } } }=111546435$

Clearly this is equivalent to

$\large{\prod _{ i=1 }^{ 8 }{ \left( x+{ a }_{ i } \right) } =111546435}$

Since $111546435=3\times 5\times 7\times 11\times 13\times 17\times 19\times 23$ .

We assume that $a_{8} > a_{7} >....>a_{1}$ then $x+a_{1}=3 \longrightarrow a_{1}=1$ . This means we can keep changing the inequality above like $a_{8} > a_{7} >...>a_{1}>a_{2}$ this for $5, 7, 11, 13, 17, 23$ and will get $3,5,9,11,15,21$ respectively so the sum of all values are

$\large{1+3+5+9+11+15+21=82}$ .

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Things to note:

• The sign $\large{ \sum _{ i=1 }^{ 8 }{ { a }_{ i } }}$ denotes $a_{1}+a_{2}+...+a_{8}$

• The sign $\large{\sum _{ 1\le i<j\le 8 }^{ }{ { a }_{ i }{ a }_{ j } }}$ denotes the product of all $a_{i}$ for $i=1,2,...8$ taken two at a time, like in the expansion of $(x+1)(x+2)(x+3)$ the coefficient of $x$ denotes the product of 1, 2, 3 taken to at a time like $1\times 2 + 2\times 3 + 3\times 1$ and similarly the rest signs of he summation denotes the same thing.

• The sign $\large{\prod _{ i=1 }^{ 8 }{ { a }_{ i } }}$ the denotes the product of all $a_{i}$ .