✨ 最佳解答 ✨
f(1)=g(1)=0 => (x-1) | f(x);
f(2)=g(2) 沒啥用;
f(3)=g(3)=2f(x) => (x-3) | f(x) => f(x) = k((x-1)(x-3));
f(4)=-12 => 3*1*k=3k=-12 => k=-4;
f(5)=4*2*(-4)=-32
✨ 最佳解答 ✨
f(1)=g(1)=0 => (x-1) | f(x);
f(2)=g(2) 沒啥用;
f(3)=g(3)=2f(x) => (x-3) | f(x) => f(x) = k((x-1)(x-3));
f(4)=-12 => 3*1*k=3k=-12 => k=-4;
f(5)=4*2*(-4)=-32
看了這個問題的人
也有瀏覽這些問題喔😉
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