The easier way is synthetic division.
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5)2..-13...0...75..2..0.-50
..___10.-15.-75_0.10..50
...2...-3.-15...0...2.10....0 <=remainder of 0
If you put x = 5 into the numerator you will get 0 indicating that x-5 is a factor of that equation.
Start by writing (x - 5) and the other factor
(x - 5) (.........)
Now you need to get 2x^6 so you need to multiply by 2x^5
(x - 5) (2x^5 ..........) which leaves a -10x^5
Since you need -13x^5 multiply by -3x^4
(x - 5) (2x^5 - 3x^4 ......) which leaves 15x^4
And so on.....
To begin, its not any equation. Pl revise its meaning!!!
To divide Nr by x-5 rearrange the terms so that x-5 is a factor.
Thus (2x^6 -10x^5) -(3x^5 - 15x^4) - (15x^4 -75x^4) and so on...
Each bracket has x-5 as a factor.
Try the rest.