Question

把下列各式分解因式

1. (a²+4)²-16a²
2. a^4-8a²b²+16b^4
(x+1)(x+2)+1/4

Answer 1

1. (a²+4)²-16a²
解：原式= (a²+4)²-（4a）²
= [(a²+4)+4a][(a²+4)-4a]
=(a+2)²(a-2）²
2. a^4-8a²b²+16b^4
原式= （a²）²-2×a²×4b²+（4b²）²
= （a²-8a²b²+4b²）²
(x+1)(x+2)+1/4
解：原式=x²+3x+9/4
=x²+2×（3／2）x+（3/2）²
=（ x²+（3/2））²

Answer 2

(a+2)²(a-2)²
(a+2)(a-2)(b+2)(b-2)
(x+3/2)²

Answer 3

1. (a2+4)2-16a2
=（a^2+4+4a^2)(a^2+4-4a^2)
=(4+5a^2)(4-3a^2)
=(4+5a^2)(2+√3a)(2-√3a)
2. a^4-8a2b2+16b^4
(a^2-4b^2)^2
=(a+2b)^2(a-2b)^2
(x+1)(x+2)+1/4
x^2+3x+9/4
(x+3/2)^2