Question

if x3+mx2+nx+6 has x-2 as a factor and leaves a remainder 3 when divided by x-3 ,find values of m and n

plz tell

Answer 1

Letp(x)=x^3+mx^2+nx+6
We have x-2 as factor of p(x).Then according to factor therom

p(2)=0

So,
p(2)=2^3+m(2)^2+2n+6
p(x)=8+4m+2n+6
p(x)=14+4m+2n


14+4m+2n=0
4m+2n= -14 [On dividing botj side by 2]
2m+n= -7 ......................(i)

Now,on dividing p(x) by x-3;we get
remainder=p(3)

p(3)=3^3+m3^2+3n+6
p(3)=27+9m+3n+6
p(3)=33+9m+3n


33+9m+3n=3
9m+3n=3-33
9m+3n= -30 [On dividing both side by 3]
3m+n= -10 .....................(ii)


On subtracting eq.i from ii we get

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


On substituting m=-3 in eq.(ii) we get;
3(-3)+n= -10
-9+n= -10
n= -10 +9
n= -1 Ans.