# Thread: [SOLVED] Difference of two Squares

1. ## [SOLVED] Difference of two Squares

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

2. Originally Posted by waven
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.
Do you know how to factor a difference of squares? Use

$\displaystyle a^2-b^2=(a+b)(a-b).$

Compare this with what you have. Then you should see that $\displaystyle a=y+4$ and $\displaystyle b=y-2.$ Go ahead and substitute, and then simplify.

3. Originally Posted by Reckoner
Do you know how to factor a difference of squares? Use

$\displaystyle a^2-b^2=(a+b)(a-b).$

Compare this with what you have. Then you should see that $\displaystyle a=y+4$ and $\displaystyle b=y-2.$ Go ahead and substitute, and then simplify.
ok i got
$\displaystyle =(y+4+y-2)(y+4-y-2)$
$\displaystyle =(2y+2)(2)$
$\displaystyle =2y+4$
$\displaystyle =2(y+2)$
thats definately wrong, could someone perhaps show me the full working please from beginning to end?

4. Originally Posted by waven
ok i got
$\displaystyle =(y+4+y-2)(y+4-y-2)$
$\displaystyle =2y+2+2$
$\displaystyle =2y+4$
$\displaystyle =2(y+2)$
thats definately wrong, could someone perhaps show me the full working please from beginning to end?
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)

----------

$\displaystyle [(y+4)+(y-2)] \times [(y+4)-(y-2)]$
$\displaystyle (y+4+y-2) \times (y+4-y+2)$
$\displaystyle (2y+2) \times 6$
$\displaystyle 6(2y+2) = 6 \times 2(y+1) = 12(y+1)$