it is between 1 and 3 but i would go for 3, if choose all that would choose 1,2,3
Hello,
x^2-y^2=(x+y)(x-y)
x^3-y^3=(x-y)(x²+xy+y²)
Let's use Horner's division
.........|a^3|a^2.|a^1..........|a^0
.........|1....|5....|6..............|8....
a=p...|......|p....|5p+p^2....|6p+5p^2+p^3
----------------------------------------------------------
.........|1....|5+p|6+5p+p^2|8+6p+5p^2+p^3
The remainder is 8+6p+5p^2+p^3 or 8+6q+5q^2+q^3
Thus:
8+6p+5p^2+p^3 = 8+6q+5q^2+q^3
==>p^3-q^3+5p^2-5q^2+6p-6p=0
==>(p-q)(p²+pq+q²)+5(p-q)(p+q)+6(p-q)=0
==>(p-q)[p²+pq+q²+5p+5q+6]=0 or p≠q
==>p²+pq+q²+5p+5q+6=0
And here, Mehek are there sufficients explanations?
It’s C, y + 5 = 2(x - 4).
In y = mx + b form:
y = 2x - 13
