# Question #dcc59

Feb 17, 2017

$b = - 2$

#### Explanation:

According with the conditions we have

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

Equating the coefficients

$\left\{\begin{matrix}1 + r = 0 \\ 1 + r = 0 \\ 1 + b - r = 0 \\ 1 - a = 0\end{matrix}\right.$

solving for $a , b , r$ we obtain

$a = 1 , b = - 2 , r = - 1$