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

Jul 20, 2017

The three solutions are:

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

#### Explanation:

We have:

${x}^{3} + {x}^{2} = \frac{11 x + 10}{3}$

Multiplying out the fraction we get

$3 {x}^{3} + 3 {x}^{2} = 11 x + 10$

$\therefore 3 {x}^{3} + 3 {x}^{2} - 11 x - 10 = 0$

Define, $f \left(x\right)$ by:

$f \left(x\right) = 3 {x}^{3} + 3 {x}^{2} - 11 x - 10$

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

$f \left(- 2\right) = 0 \implies x = - 2$ is a solution of $f \left(x\right) = 0$

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

$f \left(x\right) = \left(x + 2\right) Q \left(x\right)$

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

$f \left(x\right) = \left(x + 2\right) \left(3 {x}^{2} - 3 x - 5\right)$

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

$x = \frac{1}{2} \pm \frac{1}{2} \sqrt{\frac{23}{3}}$

Hence the three solutions are:

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