Question

HELPP

Answer 1

Answer:

$\dfrac{4 \cdot x^4 - 5\cdot x^3 + 2 \cdot x^2 - x + 5}{x^2 + x + 1} = 2 \cdot x^2 - 9 \cdot x + 7 \ Remainder \ (x - 2)$

Step-by-step 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

$\dfrac{4 \cdot x^4 - 5\cdot x^3 + 2 \cdot x^2 - x + 5}{x^2 + x + 1}$

Part II

The number of times x² goes into the larest term, 4·x⁴ = 4·x² times

2·x² - 9·x + 7

$\dfrac{4 \cdot x^4 - 5\cdot x^3 + 2 \cdot x^2 - x + 5}{x^2 + x + 1}$

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;

$\dfrac{4 \cdot x^4 - 5\cdot x^3 + 2 \cdot x^2 - x + 5}{x^2 + x + 1} = 2 \cdot x^2 - 9 \cdot x + 7 \ Remainder \ (x - 2)$