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)

Question

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

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Expert Answer

wheezym

Answered 2021-09-04 Author has 103 answers

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