If 5 + 3 = 8 and -9 + 1 = -8, does that imply that X = 1 is a solution in all cases? That is, if A+C=-(B+D), is x = 1 always a solution? In other words as long as A + C = +-(B + D), then X = 1 is a solution.
You should explain more carefully why you "change the sign" when you list the solution set. The reason is that the product of these factors is equal to zero. That means one or all of these factors are zero. Therefore you set each factor equal to zero and when you solve for the variable the constant term is subtracted from both sides and that is why the sign changes.
Wonderful trick! An amazing teacher, thank you for sharing this trick!
If 5 + 3 = 8 and -9 + 1 = -8, does that imply that X = 1 is a solution in all cases? That is, if A+C=-(B+D), is x = 1 always a solution? In other words as long as A + C = +-(B + D), then X = 1 is a solution.
Thank you so much
What if that 13 over 7 is 5 over 3 and 1over7 is 2 over 3 how will u find ur factors
good video
You should explain more carefully why you "change the sign" when you list the solution set. The reason is that the product of these factors is equal to zero. That means one or all of these factors are zero. Therefore you set each factor equal to zero and when you solve for the variable the constant term is subtracted from both sides and that is why the sign changes.
#CubicEquation #factoring #factor