1. ## Factorising algebraic expression

What method do I use to factorise $\displaystyle 16x^4y + 2xy$?

The answer is $\displaystyle 2xy(2x+1)(4x^2-2x+1)$

Thanks

2. 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.

3. $\displaystyle =2xy(8x^3+1)=2xy[(2x)^3+1]$
$\displaystyle =2xy(2x+1)(4x^2-2x+1)$

4. 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, $\displaystyle x^n+ y^n= (x+ y)(x^{n-1}- x^{n-1}y+ \cdot\cdot\cdot- xy^{n-1}+ y^n)$.