use long division to determine if 2x-1 is a factor of p(x)=6x^3+7x^2-x-2

Rui Baldwin 2021-09-29 Answered
use long division to determine if 2x-1 is a factor of \(\displaystyle{p}{\left({x}\right)}={6}{x}^{{3}}+{7}{x}^{{2}}-{x}-{2}\)

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Expert Answer

averes8
Answered 2021-09-30 Author has 5719 answers

\(\displaystyle{6}{x}^{{3}}+{7}{x}^{{2}}-{x}-{2}\) \ 2x-1
\(\displaystyle-{6}{x}^{{3}}+{3}{x}^{{2}}\) \ \(\displaystyle{3}{x}^{{2}}+{5}{x}+{2}\)
\(\displaystyle{10}{x}^{{2}}-{x}-{2}\)
\(\displaystyle-{10}{x}^{{2}}+{5}{x}\)
\(4x-2\)
\(-4x+2\)
\((2x-1)\) in a .... of p(x)
\(\displaystyle{p}{\left({x}\right)}={\left({2}{x}-{1}\right)}{\left({3}{x}^{{2}}+{5}{x}+{2}\right)}\)

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