해법
(8x3+125)=(2x+a)(bx2−10x+c)
해법
x=4(b−5)−c−ab+5a+12+a2b2+25a2−10a2b+120a+10ac−24ab−2abc+500b+c2−24c−2356,x=4(b−5)−c−ab+5a+12−a2b2+25a2−10a2b+120a+10ac−24ab−2abc+500b+c2−24c−2356;b=5
솔루션 단계
(8x⋅3+125)=(2x+a)(bx⋅2−10x+c)
8x⋅3+125 확장 :24x+125
(2x+a)(bx⋅2−10x+c) 확장 :4bx2−20x2+2cx+2abx−10ax+ac
24x+125=4bx2−20x2+2cx+2abx−10ax+ac
측면 전환4bx2−20x2+2cx+2abx−10ax+ac=24x+125
125를 왼쪽으로 이동
4bx2−20x2+2cx+2abx−10ax+ac−125=24x
24x를 왼쪽으로 이동
4bx2−20x2+2cx+2abx−10ax+ac−125−24x=0
표준 양식으로 작성 ax2+bx+c=0(4b−20)x2+(2c+2ab−10a−24)x+ac−125=0
쿼드 공식으로 해결
x1,2=2(4b−20)−(2c+2ab−10a−24)±(2c+2ab−10a−24)2−4(4b−20)(ac−125)
(2c+2ab−10a−24)2−4(4b−20)(ac−125)단순화하세요:2a2b2+25a2−10a2b+10ac+120a−24ab−2abc+500b+c2−24c−2356
x1,2=2(4b−20)−(2c+2ab−10a−24)±2a2b2+25a2−10a2b+10ac+120a−24ab−2abc+500b+c2−24c−2356;b=5
솔루션 분리x1=2(4b−20)−(2c+2ab−10a−24)+2a2b2+25a2−10a2b+10ac+120a−24ab−2abc+500b+c2−24c−2356,x2=2(4b−20)−(2c+2ab−10a−24)−2a2b2+25a2−10a2b+10ac+120a−24ab−2abc+500b+c2−24c−2356
x=2(4b−20)−(2c+2ab−10a−24)+2a2b2+25a2−10a2b+10ac+120a−24ab−2abc+500b+c2−24c−2356:4(b−5)−c−ab+5a+12+a2b2+25a2−10a2b+120a+10ac−24ab−2abc+500b+c2−24c−2356
x=2(4b−20)−(2c+2ab−10a−24)−2a2b2+25a2−10a2b+10ac+120a−24ab−2abc+500b+c2−24c−2356:4(b−5)−c−ab+5a+12−a2b2+25a2−10a2b+120a+10ac−24ab−2abc+500b+c2−24c−2356
2차 방정식의 해는 다음과 같다:x=4(b−5)−c−ab+5a+12+a2b2+25a2−10a2b+120a+10ac−24ab−2abc+500b+c2−24c−2356,x=4(b−5)−c−ab+5a+12−a2b2+25a2−10a2b+120a+10ac−24ab−2abc+500b+c2−24c−2356;b=5