The key here is to look at the constant at the end.

For part a, this is . Think about a quadratic. When factorizing , you'd look for factors of 8, and come up with:

So,

To factorize, though the co-efficients of x are important, it's the constant that determines the possible factors, which in this case are (+ or -) 1,2,4,8. The co-efficients just allow you to select which specific factors will work.

Now in part a, just apply this to a quartic:

Assume p(x)=0

Then

Now work out what the factors of 24 are, and you'll get

(Yuck: Including negatives, there're a lot to choose from. Perhaps someone at a more advanced level will know of an easier way)

To start, choose

Now think back to the example of the quadratic, which does not apply to your question but I am using as an example:

So

So, if x=1 gives a result of P(x)=0, then (x-1) is a factor of P(x).Now let's go back to your example.

So, if when we substitute x=1 into p(x) and the result is equal to 0, then (x-1) is a factor of p(x)

This gives a result which clearly does not equal zero, so is not a factor. You'd then try x=-1, then x=2, x=-2, etc...

As P(x) is a quartic, there will be a maximum of four factors. I'm afraid I don't know of a faster approach.

Once you find a factor, partially factorize the expression. This way, once you have found two factors, you'll be left with an easy quadratic.