Q.Use the difference of two squares method to factorise:
$\displaystyle (y + 4)^2 - (y - 2)^2$
the answer is $\displaystyle 12(y + 1)$ but i have absolutely no idea how to work it out.
thanks
You have said that $\displaystyle y+4-(y-2) = y+4 - y - 2$ which is incorrect. $\displaystyle (y+4)-(y-2) = y+4-y+2 = 6$ since subtracting a negative means you add it on.
Doing that gives you 6(2y+2) = 12y+12 = 12(y+1)
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$\displaystyle [(y+4)+(y-2)] \times [(y+4)-(y-2)]$
$\displaystyle (y+4+y-2) \times (y+4-y+2)$
$\displaystyle (2y+2) \times 6$
$\displaystyle 6(2y+2) = 6 \times 2(y+1) = 12(y+1)$