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Given $p(x)=x^5+(1+2i)x^4-(1+3i)x^2+8+44i$ check with the Horner-scheme if $(-2-i)$ is a root of $p(x)$.

First I have to guess a root, then proceed with the Horner-method and if i factorized it, i can say if $(-2-i)$ is a root or not, but how can i guess the first root, are there any tricks ?

OBDA
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2 Answers2

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$p(x) = 8 + 44i + x^2(-(1+3i) + x^2(1+2i + x))$

Now, let $x=-2-i$, thus, $x^2 = 4-1-4i = 3+4i$

$\begin{align}p(x) &= 8 +44i + (3+4i)(-(1+3i) + (3+4i)(-1+i)))\\ &=8+44i + (3+4i)(-(1+3i) -3+3i-4i-4))\\ &=8+44i +(3+4i)(-1-3i-3+3i-4i-4)\\ &=8+44i + (3+4i)(-8 - 4i)\\ &=8+44i -24 -12i -32i +16\\ &=0 \end{align} $

Thus $-2-i$ is a root.

Guy
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  • Entirely giving away what clearly looks like a homework problem is perhaps not the most optimal thing to do the OP's education. – gt6989b Mar 12 '14 at 12:14
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    @gt6989b I have seen homework questions with proper attempts by this user before, so I am assuming that he tried on this one. I doubt he's using this site like an answer machine like some of the "1 rep wonders". It is just that, initially, I posted my attempts when asking for help with homework, but now I expect the community to understand that I am not a new user, and have probably already done my best in case I post homework, without needing to prove that by posting my attempt. I am just vesting the same trust in the OP. – Guy Mar 12 '14 at 12:19
  • Fair enough, I agree. – gt6989b Mar 12 '14 at 13:06
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You don't need to factor the entire thing. $r$ is a root of $p(x)$ if and onyl if $p(r)=0$. So use Horner's scheme to evaluate $p(-2-i)$ and see if you get $0$ or not.

gt6989b
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