Use synthetic substitution to evaluate P(x) = x3 + 3x2 - 6 for x = 5. P(5) =

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Use synthetic substitution to evaluate P(x) = x3 + 3x2 - 6 for x = 5. P(5) =

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airricuhh why don't you try this one? if you get stuck I'll help you
so you start at 5 3 3 6 and then i don't know what comes under that..
goes"

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Other answers:

Let me teach you,
\[5^3+3(5^2)-6=194\]
got it?
no i don't saljudieh,
we have x=5, it means we are dividing by x-5 to get the value of P(5) write the coefficients of P(x)= x^3+3x^2-6 coefficients are 1 3 -6
got till here?
where did you get the 1 coefficient from? the x right?
yeah x^3 = 1 3x^2= 3 -6 as it is
okay, so where do you get the numbers that go under neath,
now we are using synthetic division first step write this way _______________ 5 | 1 3 -6
wait ash, i messed up the problem was Use synthetic substitution to evaluate P(x) = x3 + 3x2 - 6 for x = -1. P(-1) =
now we have _______________ 5 | 1 3 -6 bring the first term 1 as it is down , _______________ 5 | 1 3 -6 1 next multiply , the term brought down by 5 and ____________________________ 5 | 1 3 -6 5(5*1) 1
:/ im sorry, so it would be _______________ -1 | 1 3 -6
no we have to evaluate at x=5 so it's _______________ 5 | 1 3 -6
i know but i typo'd the problem its P(-1) :/
so it would be 3 5 which is 15 so then would i divide 15 by -6 and get the answer?
hey arricuhh u wanna learn? watch thishttp: http://www.youtube.com/watch?v=-aTMzg4syy8
well thanks.
did it help?
so is the answer -8 ?
nope, it did help. but i still got the wrong answer.
wait is it -5?
|dw:1328504343410:dw| ok look at this:
\[x^3+3x^2(no x so you PUT ZERO) -6\]you were messing a term of zero instead of the x term.. coz you had
saljudieh07 i messed up tho and the question is Use synthetic substitution to evaluate P(x) = x3 + 3x2 - 6 for x = -1. P(-1) = i accidently put 5. im going to try it again because i forgot the 0.
yeh whe you have the zero it will work
gdluck i g2g for now byee
okay thank you! (:
is that really you on the picture?
yea,

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