Herbie 寫到:3x<sup>4</sup>+42x<sup>3</sup>+204x<sup>2</sup>+390x+225
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要有做法喔
3x^4+42x^3+204x^2+390x+225=3(x^4+14x^3+68x^2+130x+75)
Let f(x)=x^4+14x^3+68x^2+130x+75
f(-1)=(-1)^4+14(-1)^3+68(-1)^2+130(-1)+75=0
f(-3)=(-3)^4+14(-3)^3+68(-3)^2+130(-3)+75=0
f(-5)=(-5)^4+14(-5)^3+68(-5)^2+130(-5)+75=0
Thus, (x+1), (x+3) and (x+5) are factors of f(x).
Let f(x)=(x+1)(x+3)(x+5)(x+a)
Comparing the constant term,
75=1*3*5*a
a=5
Thus, f(x)=(x+1)(x+3)(x+5)^2
ie. 3x^4+42x^3+204x^2+390x+225=3(x+1)(x+3)(x+5)^2