# [SOLVED] Difference of two Squares

• Mar 24th 2009, 07:00 PM
waven
[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
• Mar 24th 2009, 07:13 PM
Reckoner
Quote:

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.
• Mar 24th 2009, 07:54 PM
waven
Quote:

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?
• Mar 24th 2009, 08:02 PM
e^(i*pi)
Quote:

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)

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