Why is (a-b)(a-b) equal to a^2-2ab+b^2?

Question

I know that:

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

For example, by substituting a=3 and b=1:


(3-1)(3-1)=4

or

3^2-2*3*1+1^2=9-6+1=4

I know they are equal but...

What I want to ask is that in the first place, how do people know we need to use the 3 in the first bracket to multiply the 3 in the second bracket and then use the 3 in the first bracket again to multiply the -1 in the second bracket and so on.

Thank you.

Answer 1

It is an implementation of the distributative law.

(a - b)c = ac - bc

c = a - b, so

(a -b)(a - b) = a(a - b) - b(a - b)

Then reapplying the distributative law on these two terms:

(a - b)(a - b) = [a(a) - a(b)] - [b(a) - b(b)]
= a^2 - ab - ab + b^2 = a^2 - 2ab + b^2

Answer 2

(a - b) (a + b)
= a(a - b) - b(a - b)
= a² - ab - ba + b²
= a² - ab - ab + b²
= a² - 2ab + b²

Also using a = 3 and b = 1 as you suggest:-
(a - b)(a - b) becomes:-
(3 - 1) (3 - 1)
= (2) (2)
= 4
OR
(3 - 1)(3 - 1)
= 9 - 3 - 3 + 1
= 10 - 6
= 4

Answer 3

we need to find what (a-b)(a-b) is for all values of a and b.

opening the bracket,

(a-b)(a-b) = a*a + (a)(-b) + (-b)(a) + (-b)(-b)

= a^2 -ab - ab - b^2

= a^2 - 2ab + b^2.


thus, for all values of a and b,

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

(this is an algebraic identity, that is, it is a statement that is true for all values of the letters in it).

Answer 4

FOIL?

Front, Outer, Inner, Last???

-or-

we can treat the first factor (a - b) to be just as one

so...

(a - b)(a) - (a - b)(b)

simplifying...

a² - ab - ab + b²

a² - 2ab + b²

Maybe they did it to have a shortcut of some sort...

Answer 5

0 is the answer

Answer 6

Because this is the rule. You must to follow.

Answer 7

(a - b) * ( c - d )

=

inner part of the brackets, b * c
plus
outter part of the brackers, a * -d
plus
left part of each brackets, a * c
plus
right part of each brackets, -b * - d

I dunno what you're talking about, but hope that helps.