百度试题 结果1 题目(b) Factorise fully 6x^3-27x^2+30x=3x(2x2-9x+10) [M1]=3x(2x-5)(x-2) 相关知识点: 试题来源: 解析 =3x(2x²-9x+10) [M1] = 3x(2x-5)(x-2) 反馈 收藏
解析 (x+3)^6[(x+3)+(2x-1)] (x+3)^6(3x+2) 结果一 题目 Factorise fully(x+3)^7+(x+3)^6(2x-1)Do not attempt to expand brackets. 答案 (x+3)^6[(x+3)+(2x-1)] (x+3)^6(3x+2)相关推荐 1Factorise fully(x+3)^7+(x+3)^6(2x-1)Do not attempt to expand bracke...
百度试题 结果1 题目Factorise fully(x+3)^7+(x+3)^6(2x-1)Do not attempt to expand brackets. 相关知识点: 试题来源: 解析 (x+3)^6[(x+3)+(2x-1)] (x+3)^6(3x+2) 反馈 收藏
25 Given that f(x)=2x^3-17x^2-58x+33(a) show that f(11)=0(2)(b)hence fully factorise f(x).(4) 相关知识点: 试题来源: 解析 (a)f(11)= 2(11)^3- 17(11)² -58(11) + 33(subst.)Note: 2662- 2057- 638 + 33 earns M markORattempt at dividing 2x^3- 17x² - ...