It is given that x + 3ax + 2a = (x + 2)(x + b) - b - 1, where a and b are constants.

[2B08]

I don't know how to do this question.

It is given that x + 3ax + 2a = (x + 2)(x + b) - b - 1, where a and b are constants. By comparing the
coefficients of the like terms, find the values of a and b.

 

[stapel]

I don't know how to do this question.

It is given that x + 3ax + 2a = (x + 2)(x + b) - b - 1, where a and b are constants. By comparing the
coefficients of the like terms, find the values of a and b.

To get started, try multiplying things out and simplifying on the right-hand side, so that you can then compare the coefficients. Where does this lead?

 

[Dr.Peterson]

I don't know how to do this question.

It is given that x + 3ax + 2a = (x + 2)(x + b) - b - 1, where a and b are constants. By comparing the
coefficients of the like terms, find the values of a and b.

I think you probably meant x^2 + 3ax + 2a = (x + 2)(x + b) - b - 1, since both sides should be quadratic.

This does give a nice answer when you do what they say to do: expand, and set corresponding coefficient expressions equal.

 

[Steven G]

Hint: If for example 3x² -4x +9 = ax²+bx+c, then a=3, b=-4 and c=9