• Fully factorise $x^4-1$

•   Define $u = x^2$. Then $u^2=x^4$.

•   $x^4-1 \equiv u^2-1 \equiv (u+1)(u-1) \equiv (x^2+1)(x^2-1) \equiv (x^2+1)(x+1)(x-1)$

1. Expand $(x+\frac{1}{2})(x+6)$. Is it possible factorise this expanded expression so that the factors only contain integer coefficients and constants?
2. Expand $(x+\frac{1}{2})(2x+6)$. Is it possible factorise this expanded expression so that the factors only contain integer coefficients and constants?

1.   $(x+\frac{1}{2})(x+6)\equiv x^2+\frac{13}{2}x+3$; no
2.   $(x+\frac{1}{2})(2x+6)\equiv 2x^2+7x+3 \equiv (2x+1)(x+3)$

Stan and Ollie are trying to factorise $x^2-5x+6$. Stan writes $(x-2)(x-3)$. Ollie writes $(2-x)(3-x)$. Are either of them correct? If so, who?

Hover for answers:

•   They’re both correct, but Stan’s factorisation is more conventional.