# Difference of the square - Tricky question for me...

• Sep 1st 2010, 07:00 PM
mindraked
Difference of the square - Tricky question for me...
Hi ! First post here !

I'm having some difficulty with a math problem in my math book and I'm starting to think the author made a mistake.

I'm trying to factorize $(a-b)^2 -4$ using the "difference of the squares" technique. According to my book, it should result in the following factors :

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

Now, if I multiply those two factors together, it gives me this :

$a^2 - ab + 2a - ab + b^2 - 2b - 2a + 2b - 4$

then

$a^2-2ab+b^2-4$

so, I'm tempted to conclude that $(a-b-2) (a-b+2)$ does not equal $(a-b)^2 -4$, and therefore that my book is wrong about it.

Am I right ? Did I make a mistake ?
• Sep 1st 2010, 07:15 PM
undefined
Quote:

Originally Posted by mindraked
Hi ! First post here !

I'm having some difficulty with a math problem in my math book and I'm starting to think the author made a mistake.

I'm trying to factorize $(a-b)^2 -4$ using the "difference of the squares" technique. According to my book, it should result in the following factors :

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

Now, if I multiply those two factors together, it gives me this :

$a^2 - ab + 2a - ab + b^2 - 2b - 2a + 2b - 4$

then

$a^2-2ab+b^2-4$

so, I'm tempted to conclude that $(a-b-2) (a-b+2)$ does not equal $(a-b)^2 -4$, and therefore that my book is wrong about it.

Am I right ? Did I make a mistake ?

$a^2-2ab+b^2-4 = (a-b)^2 -4$

because

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

Hopefully you didn't make the mistake of thinking

$(a-b)^2 = a^2-b^2\ \ \ \text{(Wrong)}$
• Sep 1st 2010, 07:16 PM
pickslides

$a^2-2ab+b^2-4= (a-b)(a-b)-4= (a-b)^2-4$
• Sep 2nd 2010, 09:43 AM
grgrsanjay
Quote:

Originally Posted by mindraked
Hi ! First post here ! welcome!!

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

Now, if I multiply those two factors together, it gives me this :

$a^2 - ab + 2a - ab + b^2 - 2b - 2a + 2b - 4$

then

$a^2-2ab+b^2-4$

so, I'm tempted to conclude that $(a-b-2) (a-b+2)$ does not equal $(a-b)^2 -4$, and therefore that my book is wrong about it. they both are equal and book is right

Am I right ? Did I make a mistake ? may be wrong in factorizing

it comes out to be true
(a-b-2)(a-b+2) =(a-b)^2 - 2^2 = $(a-b)^2 - 4$
• Sep 2nd 2010, 09:50 AM
HallsofIvy
Quote:

Originally Posted by mindraked
Hi ! First post here !

I'm having some difficulty with a math problem in my math book and I'm starting to think the author made a mistake.

I'm trying to factorize $(a-b)^2 -4$ using the "difference of the squares" technique. According to my book, it should result in the following factors :

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

Now, if I multiply those two factors together, it gives me this :

$a^2 - ab + 2a - ab + b^2 - 2b - 2a + 2b - 4$

then

$a^2-2ab+b^2-4$

And, of course, $a^2- 2ab+ b^2= (a- b)^2$ so that this is exactly $(a- b)^2- 4$.

Quote:

so, I'm tempted to conclude that $(a-b-2) (a-b+2)$ does not equal $(a-b)^2 -4$, and therefore that my book is wrong about it.

Am I right ? Did I make a mistake ?
• Sep 2nd 2010, 09:46 PM
mindraked
Quote:

Originally Posted by grgrsanjay
it comes out to be true
(a-b-2)(a-b+2) =(a-b)^2 - 2^2 = $(a-b)^2 - 4$

Thank you for this quick answer (and all the others), but, may I ask, how do you choose not to multiply each variable of the fist factor by each variable of the second factor and instead decide to couple the two $a-b$ into a $(a-b)^2$ ?

My understanding of all this is a bit fragile for the moment. Why would I chose to use two variables together or independently ?
• Sep 2nd 2010, 10:09 PM
undefined
Quote:

Originally Posted by mindraked
Thank you for this quick answer (and all the others), but, may I ask, how do you choose not to multiply each variable of the fist factor by each variable of the second factor and instead decide to couple the two $a-b$ into a $(a-b)^2$ ?

My understanding of all this is a bit fragile for the moment. Why would I chose to use two variables together or independently ?

Factoring, remember that x^2 - y^2 = (x-y)(x+y). Here you can let x = a-b and y = 2.