Question

2x⁷+2x⁶+2x⁵+2x⁴+2x³+2x²+2x ÷ x+1​

Answer 1

Answer:

x¹⁰ - x⁹ + x⁸ - 2x⁷ + 2x⁶ - 2x⁵ + 2x⁴ - 2x³ + x² - x + 1 = 0

x¹⁰ - x⁹ + x⁸ - x⁷ - x⁷ + x⁶ + x⁶ - x⁵ - x⁵ + x⁴ + x⁴ - x³ - x³ + x² - x + 1 = 0

x⁹(x - 1) + x⁷(x - 1) - x⁶(x - 1) + x⁵(x - 1) - x⁴(x - 1) + x³(x - 1) - x²(x - 1) - (x - 1) = 0

(x - 1)(x⁹ + x⁷ - x⁶ + x⁵ - x⁴ + x³ - x² - 1) = 0

(x - 1)(x⁹ - x⁸ + x⁸ - x⁷ + 2x⁷ - 2x⁶ + x⁶ - x⁵ + 2x⁵ - 2x⁴ + x⁴ - x³ + 2x³ - 2x² + x² - x + x - 1) = 0

(x - 1)(x⁸(x - 1) + x⁷(x - 1) + 2x⁶(x - 1) + x⁵(x - 1) + 2x⁴(x - 1) + x³(x - 1) + 2x²(x - 1) + x(x - 1) + (x - 1)) = 0

(x - 1)²(x⁸ + x⁷ + 2x⁶ + x⁵ + 2x⁴ + x³ + 2x² + x + 1) = 0

(x - 1)²(x⁸ + x⁷ + x⁶ + x⁶ + x⁵ + x⁴ + x⁴ + x³ + x² + x² + x + 1) = 0

(x - 1)²(x⁶(x² + x + 1) + x⁴(x² + x + 1) + x²(x² + x + 1) + (x² + x + 1)) = 0

(x - 1)²(x² + x + 1)(x⁶ + x⁴ + x² + 1) = 0

(x - 1)²(x² + x + 1)(x⁴(x² + 1) + (x² + 1)) = 0

(x - 1)²(x² + x + 1)(x² + 1)(x⁴ + 1) = 0 {all working shown so you can see I haven't just plugged it into an app to get the factorization}

The only real root is the double root at 1.

The other roots are:

From x² + x + 1 = 0 we get ⁻½ ± i½√3 (complete the square)

From x² + 1 = 0 we get 0 ± i

From x⁴ + 1 = 0 we get x² - i = 0 and x² + i = 0 which yield ±√½ ± i√½ and ±√½ ∓ i√½

Mark me as brainlist