x4−2x3+5x2+2x=x(x3−2x2+5x+2)x^4 - 2x^3 + 5x^2 + 2x = x(x^3 - 2x^2 + 5x + 2)x4−2x3+5x2+2x=x(x3−2x2+5x+2) x=1+5ix = 1 + \sqrt{5}ix=1+5i を代入して (1+5i){(1+5i)3−2(1+5i)2+5(1+5i)+2}(1 + \sqrt{5}i)\{(1 + \sqrt{5}i)^3 - 2(1 + \sqrt{5}i)^2 + 5(1 + \sqrt{5}i) + 2\}(1+5i){(1+5i)3−2(1+5i)2+5(1+5i)+2} =(1+5i){1+35i−15−55i−2(1+25i−5)+5+55i+2}= (1 + \sqrt{5}i)\{1 + 3\sqrt{5}i - 15 - 5\sqrt{5}i - 2(1 + 2\sqrt{5}i - 5) + 5 + 5\sqrt{5}i + 2\}=(1+5i){1+35i−15−55i−2(1+25i−5)+5+55i+2} =(1+5i)(1−5i)= (1 + \sqrt{5}i)(1 - \sqrt{5}i)=(1+5i)(1−5i) 1+5=61 + 5 = 61+5=6