Question 1163109
<br>
At the point where you have<br>
{{{x^3+4x^2+45 = x(x^2+4x+5)}}}<br>
you are finished with part (a); the remaining quadratic factor does not factor over the real numbers.<br>
However, every quadratic expression factors over the complex numbers; so you can do part (b).  Solve the equation<br>
{{{x^2+4x+5 = 0}}}<br>
using the quadratic formula.  That gives you the roots of the quadratic equation; therefore it shows you how to factor the quadratic over the complex numbers.<br>
{{{x = (-4 + sqrt(16-20))/2 = -2+i}}}<br>
and<br>
{{{x = (-4 - sqrt(16-20))/2 = -2-i}}}<br>
So<br>
{{{x^2+4x+5 = (x-(-2+i))(x-(-2-i))}}}<br>
And so, finally,<br>
{{{x^3+4x^2+5x = (x)(x-(-2+i))(x-(-2-i))}}}<br>