Question

$x^3-15x^2+59x-45=0$ solve for $x$

Answer 1

put $x=1$

$⟹1-15+59-45=00=0$

So $x=1$ is root of equation

So by synthetic division

$1\left|\begin{array}{ccccccc}1& & -15& & 59& & -45\\ 0& & 1& & -14& & 45\\ 1& & -14& & 45& & 0\end{array}\right|$

$x^2-14x+45=0x^2-5x-9x+45=0x(x-5)-9(x-5)=0(x-5)(x-9)=0x=5,9$

So the value of $x=1,5,9$