Question

distributive law to factor 3 + 27b +6c. So far I have 3(1 + 81b + 18c). I do know that 3 goes into both 18 and 81 but I'm not exactly sure what the next steps are.

Answer 1

I missed the last three days of class and am trying to catch up

Answer 2

Ok here's what you need to do.
The distributive law states that \[a(b+c) = ab + ac\]
Going backwards, you can factorise if you know the common factor.
In this case, as you have rightly pointed, 3 is a factor of all of those terms.
3 + 27b +6c, so that goes outside of the brackets.
3( ).
What goes inside? Well you need to think about 3 times what inside the brackets gives you each term. The first term is 3. We know 3 x 1 = 3 so we put a 1 inside the brackets.
The second term is 27b. 3 times what gives 27b? That would be 9b by division.
So put +9b into the brackets.
The third term is 6c. 3 times what gives 6c? That would be 2c.
So put that in.
We have 3 + 27b +6c = 3(1 + 9b +2c).
To sum up, find the highest common factor of all terms.
Put that outside the brackets. Then find out what that number must be multiplied by to give you each terms and put those inside the brackets.
To check if your answer is right, expand it! If the expansion is not the same as the original unfactorised expression you have made a mistake.

Answer 3

Thank You So Much!!!!