# How do you find the real solutions of the polynomial x^3+6=2x^2+5x?

Nov 16, 2017

Solution: $x = 1 , x = 3 , x = - 2$

#### Explanation:

${x}^{3} + 6 = 2 {x}^{2} + 5 x \mathmr{and} {x}^{3} + 6 - 2 {x}^{2} - 5 x = 0$ or

${x}^{3} - 2 {x}^{2} - 5 x + 6 = 0 \mathmr{and} {x}^{3} - {x}^{2} - {x}^{2} + x - 6 x + 6 = 0$ or

${x}^{2} \left(x - 1\right) - x \left(x - 1\right) - 6 \left(x - 1\right) = 0$ or

$\left(x - 1\right) \left({x}^{2} - x - 6\right) = 0 \mathmr{and} \left(x - 1\right) \left({x}^{2} - 3 x + 2 x - 6\right) = 0$ or

$\left(x - 1\right) \left\{x \left(x - 3\right) + 2 \left(x - 3\right)\right\} = 0$ or

$\left(x - 1\right) \left\{\left(x - 3\right) \left(x + 2\right)\right\} = 0$ or

$\left(x - 1\right) \left(x - 3\right) \left(x + 2\right) = 0$

$x - 1 = 0 \therefore x = 1 \mathmr{and} x - 3 = 0 \therefore x = 3$ or

$x + 2 = 0 \therefore x = - 2$

Solution: $x = 1 , x = 3 , x = - 2$ [Ans]