Question

Express as a polynomial \[(a+b+c+d)^2\]

Answer 1

Use the algebraic identity \((a+b)^2=a^2+2ab+c^2\)

\(((a+b)+(c+d))^2\)
\((a+b)^2+2(a+b)(c+d)+(c+d)^2\)

Answer 2

Use the identity again, FOIL out the middle term;

\(a^2+2ab+b^2+c^2+2cd+d^2+2(ac+ad+bc+bd)\)

Answer 3

\(a^2+b^2+c^2+d^2+2ab+2cd+2ac+2ad+2bc+2bd\)

Answer 4

o.O... I'm trying to look beyond all the terms...

Answer 5

What do you mean? D:

Answer 6

Whew, okay I got thank same thing thanks zepp!

Answer 7

It just looks confusing... and hurts my eyes, lol...

Answer 8

there's a shortcut to this if you're interested :)
step 1: square first term = a^2
step 2: square second term = b^2
step 3: square third term = c^2
step 4: square fourth term = d^2
step 5: twice the product of first term and second term = 2ab
step 6: twice the product of first term and third term = 2ac
step 6: twice the product of first term and fourth term = 2ad
step 7: twice the product of second term and third term = 2bc
step 8: twice the product of second term and fourth term = 2bd
step 9: twice the product of third and fourth term = 2cd
step 10: combine everything = a^2 + b^2 + c^2 + d^2 + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd

this works for all square of polynomials

Answer 9

here is a tutorial where i proved this http://openstudy.com/updates/4f9dda4ae4b000ae9ed25338 :)

Answer 10

I am interested by so many steps...

Answer 11

it's just all squares and twice of products lol

Answer 12

I don't understand why you do step 5 to 9...

Answer 13

in such manner I guess...

Answer 14

OMG!!!... I see it, haha! thanks... that is beautiful @lgbasallote

Answer 15

haha glad to hear :)