Factorising using synthetic division
The previous method works perfectly well but only finds the remainder. To find the quotient as well, use synthetic division as follows.
\(f(x) = 2{x^4} + 9{x^3} - 18{x^2} - 71x - 30\)
This working gives you:
\(2{x^4} + 9{x^3} - 18{x^2} - 71x - 30\)
\(= (x + 2)(2{x^3} + 5{x^2} - 28x - 15)\)
Now you need to factorise the second bracket.
There's no point in trying \((x - 1),\,(x + 1)\)and \((x + 2)\). You've already shown that these aren't factors. However \((x + 2)\) could be a repeated factor.
In this case, though, we can rule out \((x + 2)\) as well. Why? Because the constant is \(15\). It's not an even number, so it's not divisible by \(2\).
Instead, let's try values that are factors of \(15\), for example, \(3\) and \(5\).
The remainder is \(60\), so \((x + 3)\) is not a factor.
The remainder is \(0\), so \((x - 3)\) is a factor.
This now factorises the original expression to:
\(= (x + 2)(x - 3)(2{x^2} + 11x + 5)\)
The question asked you to factorise fully, so to give your final answer, factorise the quadratic.
\(2{x^4} + 9{x^3} - 18{x^2} - 71x - 30 = (x + 2)(x - 3)(x + 5)(2x + 1)\)
