Question

if x^3+mx^2+nx+6 has (x-2) as a factor and leave remainder 3 , when divided by (x-3) find the values of m and n

Answer 1

Since (x-2) is a factor, putting the value of 'x' as '2' in the given expression should give us a zero (0).
When we do so we get:-
》2^3 + m*(2)^2 + n*2 + 6 = 0 (Equating the expression to zero.)
》8 + 4m + 2n + 6 = 0
》4m + 2n + 14 = 0
》2m + n + 7 = 0 (Divided each term of the previous equation by 2. Also let this be Equation 1.)

Now when the expression is divided by (x-3) gives us a remainder of 3 hence we will subtract the expression by 3 to make the expression divisible by (x-3).
New Expression: x^3 + mx^2 + nx + 3
Now that it is divisible, we shall repeat the same step as above but now putting the value of 'x' as '3' and equate the expression to zero (0).
When we do we get:-
》3^3 + m*(3)^2 + n*3 + 3 = 0
》27 + 9m + 3n + 3 = 0
》9m + 3n + 30 = 0
》3m + n = 10 (Divided each term of the previous equation by 3. Let this be Equation 2.)

From Eqn 1 we get:》 n = -7 -2m
From Eqn 2 we get:》n = -10 - 3m

Since LHS of both equations is same then the RHS must also be the same. Hence:-
》-7 -2m = -10 - 3m
》3m - 2m = -10 + 7
》m = -3

Put the value of 'm' as '-3' in Equation 1 or 2 and you will find the value of n.
For e.g. in Eqn 1.
》 2m + n + 7 = 0
》 2*(-3) + n + 7 = 0
》-6 + n + 7 = 0
》 n + 1 = 0
》 n = -1.

Hence the values are:
☆》m = -3
☆》n = -1