1.

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

Answer»

Answer:

The values are:☆》m = -3☆》n = -1

Step-by-step explanation:

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 + 3Now 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 -2mFrom Eqn 2 we get:》n = -10 - 3mSince LHS of both equations is same then the RHS must also be the same. Hence:-》-7 -2m = -10 - 3m》3m - 2m = -10 + 7》m = -3Put 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

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