Question

因式分解：求(x^2+2x+1)(x^2+2x-3)-6的最大值或最小值

Answer 1

(x^2+2x+1)(x^2+2x-3)-6
=(x²+2x)²-2(x²+2x)-3-6
=(x²+2x)²-2(x²+2x)+1-10
=(x²+2x-1)²-10
=[(x+1)²-2]²-10
因为[(x+1)²-2]²≥0，所以[(x+1)²-2]²-10≥-10
所以(x^2+2x+1)(x^2+2x-3)-6的最大值不存在，最小值为-10

Answer 2

(x^2+2x+1)(x^2+2x-3)-6
=(x^2+2x+1)[(x^2+2x+1)-4]-6
=(x^2+2x+1)^2-4(x^2+2x+1)+4-10
=[(x^2+2x+1)-2]^2-10
=[(x+1)^2-2]^2-10,
所以当(x+1)^2=2时有最小值-10.

Answer 3

=(x²+2x)²-2(x²+2x)-3-6
=(x²+2x-1)²-10
当x²+2x-1=0时有最小值-10
即x=-1±√2时有最小值-10

Answer 4

(x^2+2x+1)(x^2+2x-3)-6
=[(x^2+2x-1)+2]*[(x^2+2x-1)-2]-6
=(x^2+2x-1)^2-10
设t=x^2+2x-1 x∈R ∴t∈[-2,+∞)
y=t^2-10 ∴y∈[-10,+∞)
∴(x^2+2x+1)(x^2+2x-3)-6有最小值-10，无最大值