Suppose that the equation \[x^4 + Ax^2 + B = (x^2 + ax + b)(x^2 - ax + b)\] holds for all values of \(x\).

- Find \(A\) and \(B\) in terms of \(a\) and \(b\).

- Use this information to find a factorisation of the expression \[x^4 - 20x^2 + 16\] as a product of two quadratics in \(x\).

We can write \[x^4 - 20x^2 + 16 = (x^2 + ax + b)(x^2 - ax + b)\] if we can find solutions to \[2b-a^2=-20, \qquad b^2 = 16.\]

These equations have solutions \[\begin{align*} b&=4, & a&=2\sqrt{7} \\ b&=-4,\quad & a&=2\sqrt{3}. \end{align*}\]We can use either one of these (we were only asked to find one factorisation). From the first solution, we get \[x^4 - 20x^2 + 16 = (x^2 + 2\sqrt{7}x + 4)(x^2 - 2\sqrt{7}x + 4),\] and from the second solution, we get \[x^4 - 20x^2 + 16 = (x^2 + 2\sqrt{3}x - 4)(x^2 - 2\sqrt{3}x - 4).\]

- Show that the four solutions of the equation \[x^4 - 20x^2 + 16 = 0\] can be written as \(\pm \sqrt{7} \pm \sqrt{3}\).

Take \[x^4 - 20x^2 + 16 = 0.\] From above, this is equivalent to \[(x^2 + 2\sqrt{7}x + 4)(x^2 - 2\sqrt{7}x + 4) = 0,\] so \[x^2 + 2\sqrt{7}x + 4 = 0\] or \[x^2 - 2\sqrt{7}x + 4 = 0.\] Now the first option leads to \[x = -\sqrt{7} \pm \sqrt{3}\] (using the quadratic formula or completing the square), while the second gives \[x = +\sqrt{7} \pm \sqrt{3}.\] Combining these we get \[x = \pm \sqrt{7} \pm \sqrt{3}.\]

Alternatively, if we begin with the factorisation \[(x^2 + 2\sqrt{3}x - 4)(x^2 - 2\sqrt{3}x - 4)=0,\] we also get the same solutions \[\pm \sqrt{7} \pm \sqrt{3}.\]