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mskyeg
Group Title
Use synthetic division to find P(3) for P(x) = x^4 – 6x^3 – 4x^2 – 6x – 2.
 2 years ago
 2 years ago
mskyeg Group Title
Use synthetic division to find P(3) for P(x) = x^4 – 6x^3 – 4x^2 – 6x – 2.
 2 years ago
 2 years ago

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richyw Group TitleBest ResponseYou've already chosen the best response.1
\[P(x)=x4−6x2−4x2−6x−2\] The polynomial remainder theorem says that If a polynomial \(P(x)\) is divided by \((x−r)\), then the remainder is a constant given by \(P(r)\).
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
so in this case you have \(r=3\), and you want to divide \(P(x)\) by \((x3)\)
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
now bear with me as I have no idea how to typeset long division
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
Alright and I have no intention on how to learn! I'll do it by hand real quick
 2 years ago

mskyeg Group TitleBest ResponseYou've already chosen the best response.0
ya i dont know how to do it either:)
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
haha alright I didn't want to waste time by writing out each step by hand so I just filmed it. I'll upload the photo, but if it confuses you just ask me for the video! (because it is more of a process that you have to see been done to get). I made a mistake in the last step of the video though. It's fixed in the picture. Just let me upload them real quick
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
So basically what you do is multiply 3 by the row 3 column 1, then put that answer in row 2 column 2. Then in row 3 column 2 you put the sum of row 1 column 2 and row 2 column 2. Then repeat the process.
 2 years ago

mskyeg Group TitleBest ResponseYou've already chosen the best response.0
oooh thats pretty simple thanx for explaining it:)
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
no worries. And video embedding would actually be a really sweet Idea now that I think of it. I wonder if it already works.
 2 years ago
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