(2 questions)

Factor each expressions

1) 2c + 2d - cd - d^2

2) abd - abe + acd -ace

First rearrange:
2c - cd + 2d - d^2 =
c(2-d)+d(2-d)=
(2-d)(c+d).
The second one is done using the same kind of procedure. Post your work and tell us what you don't understand if you get stuck.

To factor the expression 2c + 2d - cd - d^2, we can use the distributive property to rearrange it:

2c + 2d - cd - d^2 = (2c - cd) + (2d - d^2)

Now, we can factor out the common term from each pair of terms:

= c(2 - d) + d(2 - d)

Finally, we can further simplify by factoring out the common factor (2 - d):

= (2 - d)(c + d)

So, the factored form of the expression 2c + 2d - cd - d^2 is (2 - d)(c + d).

Now let's move on to the second expression:

abd - abe + acd - ace

We can rearrange it using the distributive property:

abd - abe + acd - ace = ab(d - e) + ac(d - e)

Next, we can factor out the common factor (d - e):

= (d - e)(ab + ac)

Therefore, the factored form of the expression abd - abe + acd - ace is (d - e)(ab + ac).

I hope this explanation helps! If you have any further questions, feel free to ask.