The polynomials:
\(\displaystyle{P}{\left({x}\right)}={8}{x}^{{4}}+{6}{x}^{{2}}-{3}{x}+{1},{D}{\left({x}\right)}={2}{x}^{{2}}-{x}+{2}\)
To determine:
The polynomials \(Q(x)\) and \(R(x)\) such that \(\displaystyle{P}{\left({x}\right)}={D}{\left({x}\right)}{Q}{\left({x}\right)}+{R}{\left({x}\right)}\)
Solution:
The long division of polynomial \(\displaystyle{P}{\left({x}\right)}={8}{x}^{{4}}+{6}{x}^{{2}}-{3}{x}+{1}\) by \(\displaystyle{D}{\left({x}\right)}={2}{x}^{{2}}-{x}+{2}\) is given by:
According to division algorithm, we have,
\(\displaystyle{8}{x}^{{4}}+{6}{x}^{{2}}-{3}{x}+{1}={\left({2}{x}^{{2}}-{x}+{2}\right)}{\left({4}{x}^{{2}}+{2}{x}\right)}+{\left(-{7}{x}+{1}\right)}\)
So, \(\displaystyle{Q}{\left({x}\right)}={4}{x}^{{2}}+{2}{x}\) and \(\displaystyle{R}{\left({x}\right)}=-{7}{x}+{1}\)
Conclusion:
Hence, \(\displaystyle{Q}{\left({x}\right)}={4}{x}^{{2}}+{2}{x}\) and \(\displaystyle{R}{\left({x}\right)}=-{7}{x}+{1}\)
Let P(x)=8x^4+6x^2-3x+1 and D(x)=2x^2-x+2. Find polynomials Q(x) and R(x) such that P(x)=D(x)\cdot Q(x)+R(x)
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