There are three prime numbers $a,b,c$ such that $abc=5159$. Can you find them? With a little time, yes, you probably could, but it is a lot easier if I tell you that $a=67$. Then $5159÷67=77$ and finding $b$ and $c$ is easy.

The situation with cubics is much the same. If I told you that there are three brackets such that

could you find those brackets? It's hard, but if I told you that one of those brackets (one of the factors) is $(x+3)$ it becomes much easier.

On the next page, we'll see exactly how to do this.

1. Given that $(x+3)$ is a factor of the cubic

   $$p(x)=x^3+4x^2+x-6$$

   write the cubic as a product of three linear factors.

2. Given that $(x+4)$ is a factor of the cubic

   $$f(x)=x^3+9x^2+6x-56$$

   factorise the cubic completely.

3. Given that $(x+1)$ is a factor of the cubic

   $$g(x)=x^3-43x-42$$

   solve the equation

   $$g(x)=0$$