We start with

and wish to write it as

First, pull 2 out from the first two terms:

Let’s look at what is in parenthesis. In the final form this needs to be a perfect square. Right now we have

and we can obtain -10x by adding -5x and -5x. That is, we can build the following perfect square:

The “problem” with what we just did is that we added to what was given. Let’s put the expression together. We have

and when we multiply that out it does not give us what we started with. It gives us

So you see our expression is not right. It should have a -53 but instead has a -3. So to correct it we need to subtract another 50.
We do this as follows:

which gives us the final expression we seek:

If you multiply this out you will get the exact expression we were given. This means that:
a = 2
d = -5
e = -103
We are asked for the sum of a, d and e which is 2 + (-5) + (-103) = -106
In an arithmetic equation, there is no variable in the 'meat' of the equation(example: 5-5=0). In an algebraic equation, there is a variable in the meat of the equation(example: 5-x=0).
Hope this helps and please give brainliest!
Answer:

Step-by-step explanation:
Given

Required
The equivalent expression
We have:

Expand

Further expand

Apply laws of indices

Answer:
(9,-3)
Step-by-step explanation: