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remainder theorem - p(-4): P(x)=x^3+3x-8 ???
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hmmm
?
\(\bf p(-4)\iff x=-1\implies (x+4)=0 \\ \quad \\ p(x)=x^3+3x-8 \iff p(x)=1x^3{\color{brown}{ +0x^2}}+3x^1-8\qquad thus \\ \quad \\ \begin{array}{llllll} -4&|&1&0&3&-8\\ &|&\square ? \\\hline\\ &&1 \end{array}\) in your synthetic division, to get the remainder, and thus use it in p(x) I assume you missed the next exponent of "x", 2, from 3 to 1, should be 3, 2,1 anyhow and you also used 8 for the constant, but looking closer, is -8
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