(4x 3 + 7y 3z 4) 2
I got 16x^6+70x^3y^3z^4+49y^6z^8
somewhere i screwed up..does anyone see it?

Question

(4x 3 + 7y 3z 4) 2
I got 16x^6+70x^3y^3z^4+49y^6z^8
somewhere i screwed up..does anyone see it?

anonymous · Answer

Just to make sure I know what you started with; is this the expression?
$$(4x^3+7y+3z+4)^2$$

anonymous · Answer

Or is that really all multiplication after the +?

anonymous · Answer

ohh no wait i typed it wrong sorry... (4x^3+7y^3z^4)^2

anonymous · Answer

Ah. Ok, so expand it out
$$(4x^3+7y^3z^4)^2 \rightarrow (4x^3+7y^3z^4)(4x^3+7y^3z^4)$$
Then do FOIL distribution (first-outer-inner-last)
Is that what you tried first?

anonymous · Answer

I think your only error was the adding of the two middle terms.

anonymous · Answer

yessir! FOIL is what i did on all 12 of these problems,

anonymous · Answer

In the middle you'll have two of these
$$(4x^3) \cdot (7y^3z^4)$$

anonymous · Answer

Looks like you did 5×7 instead of 4×7..

anonymous · Answer

okay, thank you!

anonymous · Answer

waaaaaaaaaait...........where?

anonymous · Answer

(4x^3)⋅(7y^3z^4) = 28x^3y^3z^4.
I think you got 35x^3y^3z^4, and since there are two of those pieces, it doubled to 70 instead of 56.