(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.