F4 math problem~!!

Given that f(x) = 10x³ + mx² - x + 6 is divisible by 5x - 3 and the quotient is
2x² +nx - 2.
(a) Find the values of m and n.
As f(x) is divisible by 5x-3 and the quotient is 2x² +nx - 2, so
f(x) = (5x-3)(2x²+nx-2) ......... (*)
f(x) = 5x(2x²+nx-2) - 3(2x²+nx-2)
f(x) = 10x³+5nx²-10x - 6x²-3nx+6
f(x) = 10x³ + (5n - 6)x² + (-10 - 3n)x + 6
So 10x³ + mx² - x + 6 = 10x³ + (5n - 6)x² + (-10 - 3n)x + 6
Comparing the coefficients of terms of x, we have
m = 5n - 6 .......... (1)
-1 = -10 - 3n ......... (2)
From (2),
-1 = -10-3n
1 = 10+3n
3n = -9
n = -3
Substitute n=-3 into (1)
m = 5(-3) - 6
m = -15-6
m = -21
So m = -21, n = -3.
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(b) Hence factorize f(x).
By (a), substitute m=-21, n=-3 into (*), we have
f(x) = (5x-3)(2x²+(-3)x-2)
f(x) = (5x-3)(2x²-3x-2)
f(x) = (5x-3)(x-2)(2x+1)
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(c) Solve f(x) = 0.
From (b),
f(x) = 0
(5x-3)(x-2)(2x+1) = 0
5x-3 = 0 or x-2 = 0 or 2x+1 = 0
x = 3/5 or x = 2 or x = -1/2