# How do you find the real solutions of the polynomial 2x^3=19x^2-49x+20?

Feb 20, 2017

The roots are $\frac{1}{2}$, $5$ and $4$.

#### Explanation:

Given:

$2 {x}^{3} = 19 {x}^{2} - 49 x + 20$

Subtract the right hand side from the left to get the standard form:

$2 {x}^{3} - 19 {x}^{2} + 49 x - 20 = 0$

By the rational root theorem, any rational zeros of this cubic are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 20$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational roots are:

$\pm \frac{1}{2} , \pm 1 , \pm 2 , \pm \frac{5}{2} , \pm 4 , \pm 5 , \pm 10 , \pm 20$

We find that $x = \frac{1}{2}$ is a zero...

$2 {\left(\textcolor{b l u e}{\frac{1}{2}}\right)}^{3} - 19 {\left(\textcolor{b l u e}{\frac{1}{2}}\right)}^{2} - 49 \left(\textcolor{b l u e}{\frac{1}{2}}\right) + 20 = \frac{1 - 19 - 98 + 80}{4} = 0$

Hence $\left(2 x - 1\right)$ is a factor:

$2 {x}^{3} - 19 {x}^{2} + 49 x - 20 = \left(2 x - 1\right) \left({x}^{2} - 9 x + 20\right)$

To factor the remaining quadratic, find a pair of factors of $20$ with sum $9$. The pair $5 , 4$ works and hence:

${x}^{2} - 9 x + 20 = \left(x - 5\right) \left(x - 4\right)$

So the other two roots are $x = 5$ and $x = 4$.