Question

FOILing

Answer 1

The expression \[(x+8)^2\] is equivalent to:
\[2x+16\]
\[x^2+8x+64\]
\[x^2+64\]
\[x^2-64\]
\[x^2+16x+64\]

Answer 2

I wanna say C or
\[x^2+64\]
ThTs what I got after distributing the \[^2\]

Answer 3

Split the expression, since it is by power of 2 then the expression is multiplied by itself.

\(\bf{(x+8)(x+8)}\)

So distribute.

\(\bf{x^{2} +2x \times 8 + 8^{2}}\)

Answer 4

\[3x^2*8+64\]
I did it soooo wrong

Answer 5

make sure to pair and multiply right.

\(\bf{x^{2} + (2x \times8) + 8^{2}}\)

Answer 6

\[8x^2+16x\]

Answer 7

FOIL is just an acronym. The real rule to know regarding expanding binomials is the distributive property. You probably already know this one.

a(b + c) = ab + ac

In this case, let a = (x + 8) and b = x and c = 8

Then

(x + 8)^2 = (x + 8)(x + 8)
= (x + 8)x + (x + 8)8
= x^2 + 8x + 8x + 64
= x^2 + 16x + 64

Answer 8

:0

Answer 9

Hope I didn't confuse you further.

Answer 10

Only slightly. S’ lot to take in at once.

Answer 11

If you simply apply the distributive property, and then simplify afterwards, you'll have it forever. No need for that silly FOIL again.

Answer 12

So the first part of the binomial is always 8?

Answer 13

A* not 8

Answer 14

Yes

Answer 15

The first binomial is always a