What method do I use to factorise $\displaystyle 16x^4y + 2xy$?
The answer is $\displaystyle 2xy(2x+1)(4x^2-2x+1)$
Thanks
One way to do it is to first take out the common factor of $\displaystyle 2xy$:
$\displaystyle 2xy(8x^3+1)$
You can then see (or test) values of $\displaystyle x$ that will make $\displaystyle 8x^3+1=0$
In this case:
$\displaystyle P\left (-\frac{1}{2}\right )=0$
$\displaystyle \therefore (2x+1)$ is a factor.
Then if you divide $\displaystyle 8x^3+1$ by $\displaystyle 2x+1$ you'll get the quadratic factor.
Lovemath is using the fact that $\displaystyle x^3+ y^3)= (x+ y)(x^2- xy+ y^2)$.
It is useful to know that, for all n, $\displaystyle x^n- y^n= (x- y)(x^{n-1}+ x^{n-2}y+$$\displaystyle \cdot\cdot\cdot+ xy^{n-2}+ y^{n-1})$
and, for all odd n, $\displaystyle x^n+ y^n= (x+ y)(x^{n-1}- x^{n-1}y+ \cdot\cdot\cdot- xy^{n-1}+ y^n)$.