Question

If (x-2) is a factor of the expression xcube + mxsquare + nx + 6 and when this expression is divided by(x-3) , it leaves the remainder 3. Then , find the value of m and n ???????

Answer 1

$\large{\bf{\underline{\blue{ANSWER}}}}$

x-2=0

x=2.

Put the value of x in given p(x)

$2^3+m(2)^2+n(2)+6=0$

$8+4m+2n+6=0$

$14+4m+2n=0$

$2(7+2m+n)=0$......1

Now,

x-3=0

X=3.

Now again,put the value of x.

$(3)^3+m(3)^2+n(3)+6=3$

$27+9m+3n+6=3$

$33+9m+3n=3$

$3(11+3m+n)=1$......2

Now, Subtracting eq.1 and eq.2

$7+2m+n-(11+3m+n)=0-1$

$7+2m+n-11-3m-n=-1$

$-4-m=-1$

$-m=-1+4$

$m=-3$

Now,put the value of m in eq 1

$7+2(-3)+n=0$

$7-6+n=0$

$1+n=0$

$n=-1$

Hence, The value of m is -3 and n is -1

Thank You!

Answer 2

Answer:

Step-by-step explanation:

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