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Brooke2013
Group Title
Use synthetic division to find P(2) for P(x) = x^4 + 3x^3  6x^2  10x + 8
Choices :)
A.) 2
B.) 28
C.) 4
D.) 16
 one year ago
 one year ago
Brooke2013 Group Title
Use synthetic division to find P(2) for P(x) = x^4 + 3x^3  6x^2  10x + 8 Choices :) A.) 2 B.) 28 C.) 4 D.) 16
 one year ago
 one year ago

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jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
You don't need synthetic division really. You can just plug in x = 2 and evaluate.
 one year ago

Brooke2013 Group TitleBest ResponseYou've already chosen the best response.0
what does it have to equal to ?
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
P(x) = x^4 + 3x^3  6x^2  10x + 8 P(2) = (2)^4 + 3(2)^3  6(2)^2  10(2) + 8 P(2) = ???
 one year ago

Brooke2013 Group TitleBest ResponseYou've already chosen the best response.0
I got 76 ? that's not a choice
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
P(2) = (2)^4 + 3(2)^3  6(2)^2  10(2) + 8 P(2) = 16 + 3(8)  6(4)  10(2) + 8 P(2) = 16 + 24  24  20 + 8 P(2) = 4
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
not sure where you went wrong, but I'm guessing you lost a sign somewhere
 one year ago

Brooke2013 Group TitleBest ResponseYou've already chosen the best response.0
so you don't do the exponents ?
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
what do you mean by "do exponents"?
 one year ago

Brooke2013 Group TitleBest ResponseYou've already chosen the best response.0
Like if you do 2^4 is 16 and 3(2) ^3 is 216 and 6(2) ^2 is 144 and 10(2) is 20 that's how I got 76 I added 16 + 216  144  20 + 8
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
you do 3(2)^3 = 3(8) = 24 NOT 3(2)^3 = 6^3 = 216 because exponents come before multiplication in PEMDAS
 one year ago

Brooke2013 Group TitleBest ResponseYou've already chosen the best response.0
O, ok thanks so much for explaining :) it makes more since now
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
yw
 one year ago
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