# How do you find the real solutions of the polynomial x^3-7x^2=7-x?

Jun 16, 2017

$x = 7$

#### Explanation:

${x}^{3} - 7 {x}^{2} = 7 - x$

$\implies {x}^{3} - 7 {x}^{2} + x - 7 = 0$

let$\text{ } f \left(x\right) = {x}^{3} - 7 {x}^{2} + x - 7$

for cubics there is a least one real root.

finding this root by the factor theorem. The root(s) must be a factor of $7$.

$f \left(7\right) = {7}^{3} - {7}^{2} \times 7 + 7 - 7$

$f \left(7\right) = \cancel{{7}^{3} - {7}^{3}} + \cancel{7 - 7} = 0$

$\therefore x = 7 \text{ is a root}$

to see if there are any other real root we factorise

${x}^{3} - 7 {x}^{2} + x - 7 = \left(x - 7\right) \left({x}^{2} + b x + c\right)$

comparing constants

$\implies c = 1$

comparing coefficients of$\text{ } {x}^{2}$

$L H S = - 7$

$R H S = - 7 + b \implies b = 0$

giving

${x}^{3} - 7 {x}^{2} + x - 7 = \left(x - 7\right) \left({x}^{2} + 1\right)$

other roots $\text{ } {x}^{2} + 1 = 0$

no real roots.

$\therefore$ the only real root is $x = 7$