Answer:
![(4 \cdot x^4 - 5\cdot x^3 + 2 \cdot x^2 - x + 5)/(x^2 + x + 1) = 2 \cdot x^2 - 9 \cdot x + 7 \ Remainder \ (x - 2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/3d2wvgpb9a3h9tfchp88fds4620hwj4c6t.png)
Explanation:
Part I
The problem can be expressed as follows;
The dividend is 4·x⁴ - 5·x³ + 2·x² - x + 5
The divisor is x² + x + 1
![(4 \cdot x^4 - 5\cdot x^3 + 2 \cdot x^2 - x + 5)/(x^2 + x + 1)](https://img.qammunity.org/2022/formulas/mathematics/high-school/772bvqea4dqzq5ycvjli67zazh0crunwwm.png)
Part II
The number of times x² goes into the larest term, 4·x⁴ = 4·x² times
2·x² - 9·x + 7
![(4 \cdot x^4 - 5\cdot x^3 + 2 \cdot x^2 - x + 5)/(x^2 + x + 1)](https://img.qammunity.org/2022/formulas/mathematics/high-school/772bvqea4dqzq5ycvjli67zazh0crunwwm.png)
4·x⁴ + 4·x³ + 4·x²
-9·x³ - 2·x² - x + 5
-9·x³ - 9·x² - 9·x
7·x² + 8·x + 5
7·x² + 7·x + 7
x - 2
Therefore, we have;
![(4 \cdot x^4 - 5\cdot x^3 + 2 \cdot x^2 - x + 5)/(x^2 + x + 1) = 2 \cdot x^2 - 9 \cdot x + 7 \ Remainder \ (x - 2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/3d2wvgpb9a3h9tfchp88fds4620hwj4c6t.png)