Can We Solve A Septic, I Mean A Heptic 😄

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[MichaelRothwell]

Nice!
You took my route: x=-1 is a root by inspection, and the function is strictly increasing, so there are no other real roots.
You can see the function is strictly increasing by taking the derivative and finessing the zero at zero, or just algebraically by factorising f(b) - f(a)=(b-a)(something always positive) for b>a.

Note, however, that although there is no _general_ formula for a degree n polynomial for n≥5, you can solve a _particular_ polynomial in radicals if its Galois group is solvable.

[DavyCDiamondback]

Thanks, I spent an hour fiddling with trying to split the sextic equation into 4 quadratics and even assuming they would split up nicely into simple quadratics with interger coefficients, I called it quits

[Qermaq]

I'd love to see how one would go about finding the complex roots. Maybe not this one, or maybe.

[Fred-yqfs]

f(x) = x^7 + 5x^3 + 6 = 0
-1 is an obvious solution.
f' = 7x6+15x^2 >=0 (sum of squares)
so f is increasing over the reals
so it has one and only one real zero, which we found.
And we're done.

[youibklibrary]

Another solution
x⁷ + 5x³ + 6 = x⁷ – x³ + 6x³ + 6
= x³(x⁴– 1) + 6(x³ + 1)
= x³[(x²)²– 1²] + 6(x³ + 1³)
*** a² – b² = (a + b)(a – b), a³ + b³ = (a + b)(a² – ab + b²)
= x³(x² + 1)(x² – 1²) + 6(x + 1)(x² – x + 1² )
= x³(x² + 1)(x – 1)(x + 1) + 6(x + 1)(x² – x + 1² )
= (x + 1)[x³(x² + 1)(x – 1) + 6(x² – x + 1² )]
= (x + 1)[(x⁵ + x³)(x – 1) + 6x² – ６x + ６ ]
= (x + 1)(x⁶ – x⁵ + x⁴– x³ + 6x² – ６x + ６ )

[riccardofroz]

f(x)=x^7+5x^3+6=0
x^7-x^3+6x^3+6=0
x^3(x^4-1)+6(x^3+1)=0




x=-1

f'(x)=7x^6+15x^2
derivative is always positive so the function is strictily increasing;
Hence there is only one real solution x=-1.

[williamspostoronnim]

Уж очень бегло говорим, галопом по Европам!