Take the polynomial p(x)=x^4+x^3+2x^2+4x-8, use the factor theorem to write p(x) as two linear factors and an irreducible quadratic. An irreducible quadratic is a quadratic that can not be factorised.

We must use the factor theorem since that is what the question asks. So we take p(x) and think about the factors of 8 we then evaluate it at x=1 to start, then move onto x=-1,2,-2 and so on. We see that it has a root at x=1, so by the factor theorem we can tell that x-1 is a factor. We then divide the polynomial p(x) by x-1 (it makes sense to do this now because we're looking to express p(x) in the form of its factors) to get the cubic x^3+2x^2+4x+8. We can then use the same technique to find the second linear root, again considering the factors of 8. This time we see x+2 is a factor. We then divide the cubic by x+2 to get the quadratic x^2+4. We can tell this s an irreducible quadratic because all square numbers are positvie so this can never equal 0. And so p(x)=(x-1)(x+2)(x^2+4).