Factor completely

Im having trouble understanding how to factor these two problems completely. I know they are irregular but im always left with factors im not completely sure suffices what is being asked.

The following questions are:

$$m^2-n^2+4n-4,$$ and $$(2+y)^2-9z^6$$

I would appreciate the help so much! Thank you!

  • $\begingroup$ Please use MathJax. $\endgroup$ – José Carlos Santos Aug 16 '17 at 14:17
  • $\begingroup$ For both, difference of two squares (for first you need to factorise $n^2+4n-4$ before). $\endgroup$ – Shuri2060 Aug 16 '17 at 14:23



Notice that $n^2-4n+4 = \left(n-2\right)^2$ by completing the square, so you have:

$$m^2-n^2+4n-4 = m^2-\left(n-2\right)^2$$

Now use the formula $a^2-b^2=(a-b)(a+b)$; what are $a$ and $b$ in your case?


Use the same formula but with $a=2+y$ and $b=3z^3$ since $\left( 3z^3 \right)^2 = 9z^6$.


For factorisation, You will need this


$$a^2+b^2-2ab=(a-b)(a-b)=(a-b)^2$$ and

$$a^2-b^2=(a+b)(a-b) $$

for example $$x^2-6x+9=(x-3)^2$$

and $$16x^2-9y^2=(4x+3y)(4x-3y) .$$

Try now, you can and must do it.


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