送分題目...簡易多項式的運算 ( 1 )

1) a^2 + 5 - 21^2 - (-3) - (3a^2-1)
= a^2 + 5 - 441 + 3 - 3a^2 + 1
= a^2 + 3a^2 + 5 + 3 + 1 - 441
= 4a^2 - 433

2) a^2b + 3 - 7a^3 - 8 +2a
= -7a^3 + a^2b + 2a + 3 - 8
= -7a^3 + a^2b + 2a - 5

3) b^3a^2 + 5a^4 - 4a^2b^3 + (a^2)^2
= b^3a^2 + 5a^4 - 4a^2b^3 + a^4
= 5a^4 + a^4 + b^3a^2 - 4a^2b^3
= 6a^4 - 3a^2b^31

4) 2(a+1)^2 + 3(a^2) + 3(a^2-1)
= 2(a^2+1) + 3a^2 + 3a^2 - 3
= 2a^2 + 2 + 3a^2 + 3a^2 -3
= 2a^2 + 3a^2 + 3a^2 + 2 - 3
= 8a^2 - 1

5) (3a+1)(a-2)^2
= (3a+1)(a^2-4)
= 3a^3 - 12a + a^2 - 4
= 3a^3 + a^2 - 12a - 4

6) (2a^2+1)(a^2-ab-3)
= 2a^4 + a^2 - 2a^3b - ab - 6a^2 -3
= 2a^4 - 2a^3b + a^2 - 6a^2 - ab - 3
= 2a^4 - 2a^3b - 5a^2 - ab - 3

* n^x是n的x次方

1)a(2次)+5-21(2次)-(-3)-(3a2次-1)
21(2次) or 21a(2次)?