If x3+mx2+nx+6 has (x-2) as a factor & leaves a remainder. . .

Question : If x³+mx²+nx+6 has (x-2) as a factor & leaves a remainder 3 when divided by (x-3), find the values of m and n.

Doubt by Bhumika

Solution : 

Let p(x) = x³+mx²+nx+6

(x-2) is a factor of p(x)

x-2=0
x=2
p(2)=0
(2)³+m(2)²+n(2)+6=0
8+4m+2n+6=0
4m+2n=-14
2(2m+n)=-14
2m+n=-14/2
2m+n=-7 — (1)

Also

When p(x) is divided by (x-3) then remainder is 3.

x-3=0
x=3
p(3)=3
(3)³+m(3)²+n(3)+6=3
27+9m+3n+6=3
9m+3n+33=3
9m+3n=3-33
9m+3n=-30
3(3m+n)=-30
3m+n=-30/3
3m+n=-10 — (2)

Subtracting equation (2) from (1)

2m+n-(3m+n)=-7-(-10 )
2m+n-3m-n=-7+10
-m+0=3

-m=3
m=-3

Putting m=-3 in equation (1)

2(-3)+n=-7 
-6+n=-7
n=-7+6
n=-1

∴ m=-3 & n=-1