Let's Solve A Special Polynomial Equation

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[@DonRedmond-jk6hj]

Since you do so many polynomial equations you perhaps should mention Descartes Rule of Signs, which in this case tells us that there are 0 or 2 positive roots and one negative root. Also the signs of the qutients are handy. Here you try 1 and get nonnegative coefficients for quotient and remainder, and so there are no roots greater than 1. Thus you do not have to try 2, 4, 8 or 16.

[@ShortsOfSyber]

good idea

[@ManjulaMathew-wb3zn]

Great job!
I like your saying “ trial and error” because my first attempt was to split 17 as 1 and 16 which didn’t work. But I still am happy to find the other option of 9 and 8 worked perfectly.

[@mikelivstone]

A little calculus may be informative before solving (if a bit of overkill).
Let y = x^3 -x^2 +2x +16. Then y' = 3x^2 - 2x + 2 > 0 for all real x. Therefore y is a strictly-increasing cubic function and must have exactly one real zero. Now, since y = 16 at x = 0, the real zero of y must occur where x is negative. So, there's one negative zero and a pair of complex conjugates.

[@mcwulf25]

17 is also a perfect square minus a perfect cube.

[@vladimirkaplun5774]

Rational roots of the polynomial: ±1,±2,±4,±8,±16. x=-2 works...

[@timmaths]

1st method: easy to see -2 is a root

[@BenDover69831]

i dont think that's a valid method

[@benjaminvatovez8823]

@@BenDover69831 It is valid: P(-2)=-8-4-4+16=0 then (x+2) divides P(x). Use Horner method then use the classical solving methods of quadratic equations.

[@BenDover69831]

@@benjaminvatovez8823 i mean it works but it's not valid because it's guess and check

[@ShortsOfSyber]

why not?

[@BenDover69831]

@@ShortsOfSyber cause it's just guessing (my opinion)