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anonymous
 5 years ago
How do you use synthetic division to find P(2) and P(1)? P(x) = x^5  4x^3 + 2x
I have this so far, but I can't remember how to find P(1) 2  1 0 4 0 2 0 ........2 4 0 0 4 _____________ ....1 2 0 0 2  4 So, P(2) = 4
anonymous
 5 years ago
How do you use synthetic division to find P(2) and P(1)? P(x) = x^5  4x^3 + 2x I have this so far, but I can't remember how to find P(1) 2  1 0 4 0 2 0 ........2 4 0 0 4 _____________ ....1 2 0 0 2  4 So, P(2) = 4

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0list the coefficients and put 2 in the box: 1 0 4 0 2 0 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the remainder will be P(2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you know the procedure for synthetic division?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01 0 4 0 2 0 2 ________________________

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01 0 4 0 2 0 2 2 4 0 0 2 _________________________________ 1 2 0 0 2 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0last line is a mistake. it should be 2*2=4 and 0 + (4) = 4 sorry

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh wow..thank you so much...i just couldn't figure that out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would you like me to write the next one?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0same coefficients but this time you put 1 in the box

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes please..that would help

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm actually studying for my final

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok list the coefficients again 1 0 4 0 2 0 1 ___________________________________

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0bring down the one from the top line 1 0 4 0 2 0 1 ___________________________________ 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01*1=1 goes on the second line 1 0 4 0 2 0 1 1 ___________________________________ 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0add 0+1=2 1 0 4 0 2 0 1 1 ___________________________________ 1 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01*=1 1 0 4 0 2 0 1 1 1 ___________________________________ 1 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.04+1=.3 1 0 4 0 2 0 1 1 1 ___________________________________ 1 1 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01*3=3 1 0 4 0 2 0 1 1 1 3 ___________________________________ 1 1 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.00+3=3 1 0 4 0 2 0 1 1 1 3 ___________________________________ 1 1 3 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01*3=3 1 0 4 0 2 0 1 1 1 3 3 ___________________________________ 1 1 3 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02+3=1 1 0 4 0 2 0 1 1 1 3 3 ___________________________________ 1 1 3 3 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01*1=1 1 0 4 0 2 0 1 1 1 3 3 1 ___________________________________ 1 1 3 3 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.00+1=1 1 0 4 0 2 0 1 1 1 3 3 1 ___________________________________ 1 1 3 3 1 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0on the other hand if i had to evaluate P(1) i would just plug it in because it is easier: \[P(1)=(1)^5 4(1)^3 + 2(1)=1+42=1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes this is correct by either calculation
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