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 
 A: Depending on the level of familiarity that your assignments are assuming, you may be able to get some traction by trying some values.  For instance, assuming you know the quadratic formula, you know that if you can find one factor by trial-and-error, you can use the quadratic formula to factor the remaining two (if they are real).
Furthermore, homework assignments often use integer roots.  You might expect, therefore, that some possible roots would be $+1$, $-1$, $+3$, or $-3$ (since $3$ is divisible by all of those).  Indeed, if you try (say) $x = 3$, you get $3^3-3^2-5\cdot3-3 = 27-9-15-3 = 0$.  Because $x^3-x^2-5x-3 = 0$ when $x = 3$, you know that $x-3$ must be a factor of $x^3-x^2-5x-3$.  Use polynomial division to obtain the quotient when dividing $x^3-x^2-5x-3$ by $x-3$ to obtain a second-degree polynomial (a quadratic expression, in other words).  If you know how to factor those, then you are home free. 
A: $$x^3-x^2-5x-3=x^3+2x^2-3x^2-6x+x-3$$
$$x^3+2x^2+x-3x^2-6x-3=0$$
$$x(x^2+2x+1)-3(x^2+2x+1)=0$$
$$(x^2+2x+1)(x-3)=0$$
A: 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 
$$x^3 - x^2 - 5x - 3 \equiv (x+1)(x+1)(x-3)$$
