sara1234
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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dpaInc
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what is the degree if you multiply x^2*(x+2) ?
dpaInc
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\[\huge x^2(x+2) = \]
dpaInc
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just multiply it out...
dpaInc
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not quite...
\[\large x^2(x+2)=x^3+2x \]... agreed?
sara1234
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yes
dpaInc
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so that's a third degree polynomial.. and what is the remainder if we take
\(\large x^3+2x \) divided by \(\large x+2 \) ???
dpaInc
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remember we just multiplied it out to get that third degree polynomial...
sara1234
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-4
dpaInc
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no...
sara1234
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confused...................
dpaInc
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if I ask you to multipl 2 times 3, what's the product?
dpaInc
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**multiply...
sara1234
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6
dpaInc
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now... what's the remainder if I ask you 6 divided by 3 ?
sara1234
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0
dpaInc
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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....
dpaInc
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still with me?
sara1234
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yeah
dpaInc
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good... but we want a remainder of -4....
so what do you think we need to do?
sara1234
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not sure
dpaInc
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|dw:1340429487175:dw|
dpaInc
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we need to stick something in that box so that our remainder is -4....
dpaInc
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take a wild guess...
dpaInc
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ok... how 'bout this....
sara1234
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???
dpaInc
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still there sara?
sara1234
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yeah what goes in the box?
dpaInc
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ok... sorry.. i made some typos up there... but this should be more clear...
dpaInc
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|dw:1340430131260:dw|
dpaInc
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|dw:1340430256983:dw|
dpaInc
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see the question mark? what should it be in order for our remainder to be -4?
sara1234
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-4
dpaInc
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yes... i'm sorry i shuda done this way first.....
dpaInc
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|dw:1340430491392:dw|
sara1234
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okay gottcha thats the equation thats is a third degree polynomial that when divided by x + 2 has a remainder of –4. :)
dpaInc
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yes...:)
again.... sorry....
sara1234
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its okay
dpaInc
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thanks for sticking with me....:)