1. Create your own third degree polynomial that when divided by x + 2 has a remainder of –4. 2. Create your own division of polynomials problem. Demonstrate how this problem would be solved using both long division and synthetic division.

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1. Create your own third degree polynomial that when divided by x + 2 has a remainder of –4. 2. Create your own division of polynomials problem. Demonstrate how this problem would be solved using both long division and synthetic division.

Mathematics
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what is the degree if you multiply x^2*(x+2) ?
\[\huge x^2(x+2) = \]
just multiply it out...

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not quite... \[\large x^2(x+2)=x^3+2x \]... agreed?
yes
so that's a third degree polynomial.. and what is the remainder if we take \(\large x^3+2x \) divided by \(\large x+2 \) ???
remember we just multiplied it out to get that third degree polynomial...
-4
no...
confused...................
if I ask you to multipl 2 times 3, what's the product?
**multiply...
6
now... what's the remainder if I ask you 6 divided by 3 ?
0
that's what i was asking in when we were working with the polynomial x^3 + 2x... the remainder when \(\large x^3+2x \) divided by \(\large x+2 \) is zero....
still with me?
yeah
good... but we want a remainder of -4.... so what do you think we need to do?
not sure
|dw:1340429487175:dw|
we need to stick something in that box so that our remainder is -4....
take a wild guess...
ok... how 'bout this....
???
still there sara?
yeah what goes in the box?
ok... sorry.. i made some typos up there... but this should be more clear...
|dw:1340430131260:dw|
|dw:1340430256983:dw|
see the question mark? what should it be in order for our remainder to be -4?
-4
yes... i'm sorry i shuda done this way first.....
|dw:1340430491392:dw|
okay gottcha thats the equation thats is a third degree polynomial that when divided by x + 2 has a remainder of –4. :)
yes...:) again.... sorry....
its okay
thanks for sticking with me....:)

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