Question

9. Divide 4n-4n²+n5-n⁴ by n³-4​

Answer 1

Step-by-step explanation:

N is odd integers .

so, Let n = 2k + 1 , k ∈ ℕ

now, n⁴ + 4n² + 11

= (2k +1)² + 4(2k + 1)² + 11

= (4k² + 4k + 1)² + 4(4k² + 4k + 1) + 11

= (16k⁴ + 16k² + 1 + 32k³ + 8k + 8k²) + 16k² + 16k + 4 + 11

= 16k⁴ + 32k³ + 40k² + 24k + 16

= 16[k⁴ + 2k³ + 1] + 40k² + 24k

= 16[k⁴ + 2k² + 1] + (80k² + 48k)/2

= 16[k⁴ + 2k² + 1] + 16k(5k + 3)/2

here, k(5k + 3)/2 is also a integer for all natural number of k

so, assume k(5k + 3)/2 = m

now, n⁴ + 4n² + 11

= 16[k⁴ + 2k² + 1] + 16m

= 16[k⁴ + 2k² + 1 + m]

hence, it is clear that 16 divides n⁴ + 4n² + 11 when n is odd integers