# 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)

Tahmid Knox 2021-09-03 Answered

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

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