The expression \[2x^4+ax^3+bx^2-4x-4\] where \(a\) and \(b\) are constants, is denoted by \(f(x)\). Given that \(f\left(-\dfrac{1}{2}\right)=0\) and \(f(2)=0\), find the values of \(a\) and \(b\).

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 \(a = -5\).

With these values for \(a\) and \(b\)

  1. express \(f(x)\) as the product of three algebraic factors, and hence show that the equation \(f(x)=0\) has only two real roots;

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 \(g(x)=x^2-x+2\).

To find the expression for \(g(x)\), let’s write \(g(x)=ax^2+bx+c\) and try to find \(a\), \(b\) and \(c\) such that \[ (2x^2 - 3x -2)(ax^2+bx+c)=2x^4 - 5x^3 + 5x^2 - 4x - 4. \] We can either multiply out the brackets on the left hand side, group the terms and find what \(a\), \(b\) and \(c\) must be, or we can try to `inspect’ the equation.

For example, the only terms that contribute to the \(x^4\) term will be \(2x^2\) from the first bracket and \(ax^2\) from the second.

This tells us that \(a=1\). Similarly, the constant term must come from the constant terms in both of the brackets, so we find that \(-2c=-4\) and so \(c=2\).

To find \(b\), we check the coefficient of \(x\), so that \(-2b-3c=-4\). We know already that \(c=2\), so we see that \(b=-1\). So \(g(x)=x^2-x+2\).

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 \(x=2\).

  1. find the set of values of \(x\) for which \(f(x)>0\).

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

Graph of the function. It tends to infinity as x tends to both plus and minus infinity, and dips below the x axis once, with roots at x = minus a half and x = 2.

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