# 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)

Let $$\displaystyle{P}{\left({x}\right)}={8}{x}^{{4}}+{6}{x}^{{2}}-{3}{x}+{1}$$ and $$\displaystyle{D}{\left({x}\right)}={2}{x}^{{2}}-{x}+{2}$$. Find polynomials $$Q(x)$$ and $$R(x)$$ such that $$\displaystyle{P}{\left({x}\right)}={D}{\left({x}\right)}\cdot{Q}{\left({x}\right)}+{R}{\left({x}\right)}$$.

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wheezym

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 $$\displaystyle{R}{\left({x}\right)}=-{7}{x}+{1}$$