Question 703190
The possible rational roots for {{{P(x)=x^3-3x^2-9x-5}}} are 1, -1, 5, and -5.
{{{P(-1)=(-1)^3-3(-1)^2-9(-1)-5=-1-3+9-5=0}}},
so {{{-1}}} is a root of {{{P(x)}}} and
{{{P(x)}}} has {{{(x+1)}}} as a factor.
Using division (long or synthetic) you find that
{{{P(x)=(x^2-4x-5)(x+1)}}}
Factoring further, since {{{x^2-4x-5=(x-5)(x+1)}}} , you find that
{{{P(x)=(x-5)(x+1)(x+1)}}}
So, multiplying times {{{highlight((x-5))}}} , you would get the perfect square
{{{(x-5)P(x)=(x-5)^2(x+1)^2=((x-5)(x+1))^2=(x^2-4x-5)^2}}}
 
NOTE: I do not like to divide polynomials, so I factor by grouping, like this:
{{{P(x)=x^3-3x^2-9x-5=x^3+x^2-4x^2-4x-5x-5=x^2(x+1)-4x(x+1)-5(x+1)=(x^2-4x-5)(x+1)}}}
(It's also too difficult for me to get terms/coefficients to line up when typing stuff into this website).