Review question

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

Ref: R8219

## Solution

Show that $(x^2+2y^2)^2-4x^2y^2=x^4+4y^4,$

Let’s start from the left-hand side and multiply out the brackets. We find \begin{align*} (x^2+2y^2)^2-4x^2y^2 &=x^4+4x^2y^2+4y^4-4x^2y^2\\ &=x^4+4y^4, \end{align*}

which is what we wanted.

and hence factorise $x^4+4y^4$.

We know that $x^4+4y^4=(x^2+2y^2)^2-4x^2y^2.$ Notice that the right hand side is of the form $a^2-b^2$, where $a=x^2+2y^2$ and $b=2xy$.

We know that $a^2-b^2=(a+b)(a-b)$, and so $(x^2+2y^2)^2-(2xy)^2=(x^2+2y^2+2xy)(x^2+2y^2-2xy).$

Can we factorise the terms in the brackets any further? We can think of the first bracket as $x^2+2yx+2y^2,$ which is a quadratic in $x$ with coefficients that depend on $y$. If this factorised into $(x+ay)(x+by)$ for some $a$ and $b$, then there would be one or two real solutions to $x^2+2yx+2y^2=0$ for each real value of $y$. So we can check the discriminant,
$(2y)^2-4(2y^2)=4y^2-8y^2=-4y^2.$ But this is less than zero whenever $y$ is non-zero. Thus $x^2+2y^2+2xy=0$ doesn’t have any real solutions if $y\ne0$, and so we can’t factorise the first bracket any further.

For the second bracket, the discriminant is also $(-2y)^2-4(2y^2)=-4y^2,$ and so we can’t factorise the second bracket either. Therefore we’ve fully factorised $x^4+4y^4$.

Alternatively we can see $(x^2+2y^2+2xy)(x^2+2y^2-2xy) = ((x-y)^2+y^2)((x+y)^2+y^2)$, which won’t factorise over the real numbers.

It turns out therefore that $x^2+4y^2$ does not factorise, but $x^4+4y^4$ does. Why do they behave so differently?