Question

How do you find the real solutions of the polynomial `x^3+x^2=(11x+10)/3`?

Answer 1

The three solutions are:

`x=-2; \ \ \ 1/2 -1/2sqrt(23/3); \ \ \ 1/2 +1/2sqrt(23/3)`

Explanation:

We have:

`x^3+x^2=(11x+10)/3`

Multiplying out the fraction we get

`3x^3 + 3x^2=11x+10`

`:. 3x^3 + 3x^2-11x-10 = 0`

Define, `f(x)` by:

`f(x) = 3x^3 + 3x^2-11x-10`

As there is no trivial method for factorising a cubic, and assuming that we are not looking for an approximate solution of `f(x)=0` via Numerical Methods we attempt to find a trivial solution by trial and error. We note that:

`f(-2)=0=>x=-2` is a solution of `f(x)=0`

Therefore by the factor theorem, then if `x=-2` is a solution of `f(x)=-0` then `(x+2)` is a factor of `f(x)`, thus we can write:

`f(x) = (x+2)Q(x)`

Where `Q(x)` is a quadratic we can find via algebraic long division, or inspection, to get:

`f(x) = (x+2)(3x^2-3x-5)`

And to find the other roots we use the quadratic formula to get:

`x=1/2 +-1/2sqrt(23/3)`

Hence the three solutions are:

`x=-2; \ \ \ 1/2 -1/2sqrt(23/3); \ \ \ 1/2 +1/2sqrt(23/3)`