f.4 maths

6x^3 -13x^2 +px -5 = (2x - 1)(3x^2 -qx +1) + 3x+r
= 6x^3 - 2qx^2 + 2x - 3x^2 + qx - 1 + 3x + r
= 6x^3 - (2q + 3)x^2 + (q + 5)x + (r - 1)
Comparing coefficient :
2q + 3 = 13....(1)
q + 5 = p....(2)
r - 1 = - 5....(3)
By (3) : r = - 4
By (1) : q = 5 , sub to (2) :
p = 10

Remember the basic principle: When Y is divided by X, if the quotient is Q and the remainder is R, then Y = XQ + R.
For this question,
6x^3 - 13x^2 + px - 5 = (3x^2 - qx + 1)(2x - 1) + (3x + r).
Comparing the constant term,
- 5 = - 1 + r, so r = - 4.
Comparing the x term,
px = qx + 2x + 3x
p = q + 5 .......(1)
Comparing the x^2 term,
- 13x^2 = - 3x^2 - 2qx^2
- 13 = - 3 - 2q, so q = 5.
From (1) p = 5 + 5 = 10.