Express $y= x^{3} - x^{2} - 5x - 3$ in its fully factorised form Don't know how to do this, please help. I have never done factorising cubic polynomials and don't know how to go about this 

The usual way to do this is to use the Factor Theorem.
A polynomial $\mathrm{f}(x)$ is divisible by $(x-a)$ if $\mathrm{f}(a)=0$.
In your case, you have $\mathrm{f}(x) = x^3 - x^2 - 5x - 3$, and you need to find an $a$ for which
$$\mathrm{f}(a) = a^3 - a^2 - 5a - 3 =0 $$
This can often be done by trial-and-error, or by using the table function on a scientific calculator. 
By checking whole number choices for $a$, with $-5 \le a \le 5$, we see that $a=-1$ work.
Since $\mathrm{f}(-1) = 0$, the Factor Theorem tells us that $(x-(-1)) = (x+1)$ divides $\mathrm{f}(x)$. Hence:
$$x^3 - x^2 - 5x - 3 \equiv (x+1)(x^2+px+q)$$
where $p$ and $q$ are numbers that you need to find. You can do one of two things:


*

*Expand the right-hand side and then compare coefficients,

*Use polynomial long division.


If we expand, we see that $(x+1)(x^2+px+q) \equiv x^3+(p+1)x^2+(p+q)x+q$.
Comparing coefficients gives $p+1=-1$, $p+q=-5$ and $q=-3$. Clearly $p=-2$ and $q=-3$. Hence
$$x^3 - x^2 - 5x - 3 \equiv (x+1)(x^2-2x-3)$$
Now we need only factorise the quadratic $x^2-2x-3$, which gives $(x-3)(x+1)$. Hence 
$$x^3 - x^2 - 5x - 3 \equiv (x+1)(x+1)(x-3)$$
