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
list the coefficients and put -2 in the box: 1 0 -4 0 2 0 -2
the remainder will be P(2)
do you know the procedure for synthetic division?
1 0 -4 0 2 0 -2 ________________________
oops
1 0 -4 0 2 0 -2 -2 4 0 0 -2 _________________________________ 1 -2 0 0 2 -2
last line is a mistake. it should be -2*2=-4 and 0 + (-4) = -4 sorry
so P(2)=-4
Oh wow..thank you so much...i just couldn't figure that out
would you like me to write the next one?
same coefficients but this time you put 1 in the box
Yes please..that would help
I'm actually studying for my final
ok list the coefficients again 1 0 -4 0 2 0 1 ___________________________________
bring down the one from the top line 1 0 -4 0 2 0 1 ___________________________________ 1
1*1=1 goes on the second line 1 0 -4 0 2 0 1 1 ___________________________________ 1
add 0+1=2 1 0 -4 0 2 0 1 1 ___________________________________ 1 1
1*=1 1 0 -4 0 2 0 1 1 1 ___________________________________ 1 1
-4+1=-.3 1 0 -4 0 2 0 1 1 1 ___________________________________ 1 1 -3
1*-3=-3 1 0 -4 0 2 0 1 1 1 -3 ___________________________________ 1 1 -3
0+-3=-3 1 0 -4 0 2 0 1 1 1 -3 ___________________________________ 1 1 -3 -3
1*-3=-3 1 0 -4 0 2 0 1 1 1 -3 -3 ___________________________________ 1 1 -3 -3
2+-3=-1 1 0 -4 0 2 0 1 1 1 -3 -3 ___________________________________ 1 1 -3 -3 -1
1*-1=-1 1 0 -4 0 2 0 1 1 1 -3 -3 -1 ___________________________________ 1 1 -3 -3 -1
0+-1=-1 1 0 -4 0 2 0 1 1 1 -3 -3 -1 ___________________________________ 1 1 -3 -3 -1 -1
done
on the other hand if i had to evaluate P(-1) i would just plug it in because it is easier: \[P(-1)=(-1)^5 -4(-1)^3 + 2(-1)=-1+4-2=-1\]
okay, so P(-1) = -1
yes this is correct by either calculation
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