Question

If the polynomial x^3+mx^2+nx+6 has (x-2) as a factor and leaves remainder 3 when divided by (x-3) then find the value of m and n

Answer 1

Let(x-2)=0
x=2
f(x)=x^3+mx^2+nx+6
f(2)=8+4m+2n+6
=14+4m+2n
since(x-2) is a factor,
therefore remainder=0
so, 14+4m+2n=0
2(7+2m+n)=0
2m+n=-7. Name it as eq.1
Now, let(x-3)=0
x=3
f(x)=x^3+mx^2+nx+6
f(3)=27+9m+3n+6
=33+9m+3n
Acc.to ques:
remainder=3
so,33+9m+3n=3
3(11+3m+n)=3(1)
11+3m+n=1
3m+n=-10.Name it as eq.2
On substracting eq.1 and 2, we get:
-m=3
m=-3. Put this value of m in any eq.1 or 2, to get n=-1

Answer 2

Let(x-2)=0

x=2

f(x)=x^3+mx^2+nx+6

f(2)=8+4m+2n+6

=14+4m+2n

since(x-2) is a factor,

therefore remainder=0

so, 14+4m+2n=0

2(7+2m+n)=0

2m+n=-7. Name it as eq.1

Now, let(x-3)=0

x=3

f(x)=x^3+mx^2+nx+6

f(3)=27+9m+3n+6

=33+9m+3n

Acc.to ques:

remainder=3

so,33+9m+3n=3

3(11+3m+n)=3(1)

11+3m+n=1

3m+n=-10.Name it as eq.2

On substracting eq.1 and 2, we get:

-m=3

m=-3. Put this value of m in any eq.1 or 2, to get n=-1