x(y + z) = xy + xz (1)
If you substitute x = p + q, you get:
(p + q)(y+z) = (p+q)y + (p+q)z
Applying the rule (1) again:
(p+q)y = py + qy
(p+q)z = pz + qz
Therefore:
(p + q)(y+z) = py + pz + qy + qz
In general:
(a_1 + a_2 + a_3 + ... )*
(b_1 + b_2 + b_3 + ... )*
(c_1 + c_2 + c_3 + ... )*
......
Is equal to the sum of all the different products of the terms in the summation:
(a_1 + a_2 + a_3 + ... )*
(b_1 + b_2 + b_3 + ... )*
(c_1 + c_2 + c_3 + ... )*
......
=
a_1*b_1*c_1*... + a_2*b_1*c_1*....+ a_3*b_1*c_1*....+ ....+
a_1*b_2*c_1*....+ a_2*b_2*c_1*....+
a_3*b_2*c_1*....+.....+
a_1*b_3*c_1*..+..+ etc. etc. until you've summed all the possible products of the terms.
We can write (a+b)^2 as
(a+b)(a+b) = a*a + a*b + b*a +b*b =
a^2 + 2ab + b^2
If you replace b by -b you get:
(a-b)^2 = a^2 - 2ab + b^2
I don't understand how this is
if
(a ± b)² = a² ± 2ab + b²
i would think that is would be A squared plus or minues B squared
where does the 2ab come from??
1 answer
1 answer