Can we find the coefficients and factorise this quartic?

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Substituting \(x=-\frac{1}{2}\) we have \[\begin{align*} f\left(-\frac{1}{2}\right) &= 0\\ \implies\quad \frac{2}{16}-\frac{a}{8}+\frac{b}{4}+2-4 &= 0\\ \iff\quad 1 - a + 2b - 16 &= 0\\ \iff\quad -a+2b &= 15 \end{align*}\] Also we have that \[\begin{align*} f(2) &= 0\\ \implies\quad 32+8a+4b-8-4 &= 0\\ \iff\quad 2a+b &= -5 \end{align*}\]

We now have two simultaneous equations in \(a\) and \(b\). If we add twice the first to the second, to eliminate \(a\), we see that \(5b = 25\) and so \(b = 5\).

We can then use either equation to find that \(a = -5\).

Since \(f\left(-\frac{1}{2}\right)=0\) and \(f(2)=0\) we know that \(2x+1\) and \(x-2\) are two factors of \(f(x)\).

So \[f(x)=2x^4 - 5x^3 + 5x^2 - 4x - 4 =(2x+1)(x-2)g(x) = (2x^2 - 3x -2)g(x),\] where \(g(x)\) is a quadratic polynomial. By inspection, we can see that \(g(x)=x^2-x+2\).

Therefore we have \[f(x)=(2x+1)(x-2)(x^2-x+2).\]

To show that \(f(x)\) only has two real roots we need to show that \(g(x)=x^2-x+2\) has no real roots.

The discriminant of \(g(x)\) is \(1 - 4 \times 2 = -7\), and since this is negative, we see that \(g(x)\) has no real roots.

So, we can conclude that \(f(x)\) has exactly two real roots, which are \(x=-\dfrac{1}{2}\) and \(x=2\).

If we sketch the graph of \(f(x)\), we’ll clearly see the regions for which \(f(x)>0\). We know that there are only two real roots, and we observe that the \(y\)-intercept is \(-4\).

So the graph is

From this, we can see that the values of \(x\) for which \(f(x)>0\) are \(x<-\dfrac{1}{2}\) and \(x>2\).