Question

hey any math (polynomial expert) please help.
my question is that for example we are given any degree polynomial and we are to solve a polynomial equal just for an example 2x^3 + 4x^2 + x^2 + 2 = any constant.
now we would solve it by taking the constant other side any making a general cubic polynomial form so we could find it zeroes.
But we are finding the values of x for which the polynomial equals zero not the constant on the r.h.s, why do we find zeroes we have to find the value(s) of x for which polynomial equals constant.

Answer 1

hey
in such questions we usually find zeroes. the question is not given in general cubic form to make it confusing so, first we shift the constant to make it in cubic equation . then the statement becomes equal to zero and by factorizing the equation we get the zeroes

Answer 2

We generally Convert a polynomial into its standard form as its solvable for its solutions.

So, E.g as you gave

$2 {x}^{3} + 4 {x}^{2} + x + 2 = k$

Now, as the polynomial is equal to a constant at any three values of x, so if we subtract the constant from the polynomial, then it must end up giving zero while cross checking the solution

$2 {x}^{3} + 4 {x}^{2} + x + (2 - k) = 0$

We can reduce this even more:

First we divide both sides: by 2

${x}^{3} + 2 {x}^{2} + \frac{x}{2} + \frac{2 - k}{2} = 0$

Then complete the cube

$(x + \frac{2}{3} ) {}^{3} - (x + \frac{2}{ 3} )( \frac{5}{6} ) + \frac{68 - 27k}{54}=0$

Now, we can apply the algebraic formula for the cubics.

$x_{1,2,3} = - \frac{2}{3} + \omega _{k} .\sqrt[3]{( \frac{27k - 68}{108} ) + \sqrt{( \frac{27k - 68}{108} ) {}^{2} - ( \frac{5}{18} ) {}^{3} } } + {\omega _{k}}^{2} .\sqrt[3]{( \frac{27k - 68}{108} ) - \sqrt{( \frac{27k - 68}{108} ) {}^{2} - ( \frac{5}{18} ) {}^{3} } }$

That's it for cubics! Solving for x in terms of constant will end up the same constant if we put back into the equation and if we subtract constant on both sides, we get zero.

I hope your doubts must be cleared now

:)