Review question

# Can we factorise $x^4 + Ax^2 + B$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7809

## Solution

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

1. Find $A$ and $B$ in terms of $a$ and $b$.
Expanding the right-hand side of the equation gives $x^4 + Ax^2 + B = x^4 + (2b-a^2)x^2 + b^2,$ so by equating coefficients we get \begin{align*} A &= 2b-a^2, \\ B &= b^2. \end{align*}
In fact, we have a difference of two squares here: \begin{align*} (x^2 + ax + b)(x^2 - ax + b) &=\bigl((x^2+b)+ax\bigr)\bigl((x^2+b)-ax\bigr)\\ &=(x^2+b)^2-(ax)^2\\ &=x^4+2bx^2+b^2-a^2x^2\\ &=x^4+(2b-a^2)x^2+b^2. \end{align*}
1. 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 $x^4 - 20x^2 + 16 = (x^2 + 2\sqrt{3}x - 4)(x^2 - 2\sqrt{3}x - 4).$

1. 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 $\pm \sqrt{7} \pm \sqrt{3}.$