The polynomial 6x^3+mx^2+nx-5 has a factor of x+1. When divided by x-1, the remainder is -4. What are m and n?

6 years ago

Perform a synthetic division of polynomial by -1 and since x+1 is a factor, remainder should be zero. That would give you m - n =11

Perform a synthetic division of polynomial by +1 and remainder should be -4. That would give you m + n =-5

Solve the system of 2 equations and 2 unknowns (m,n) which would give you m = 3 and n = -8

We have two unknowns, m and n, so we need to find two equations to solve the problem.

Equation 1 is (6x^{3}+mx^{2}+nx-5)/(x+1) has a remainder of 0

Equation 2 is (6x^{3}+mx^{2}+nx-5)/(x-1) has a remainder of -4

--

Performing division on Equation 1 gives us that the remainder is -11 - n + m, which is equal to 0.

So now Equation 1 is m-n = 11

Performing division on Equation 2 give us the remainder of n + m + 1 which is equal to -4.

So then Equation 2 become n + m = -5

Combining Equations 1 and 2 togethers gives the final answer of m = 3 and n = -8.

6 years ago

## Answered By Moe B

Perform a synthetic division of polynomial by -1 and since x+1 is a factor, remainder should be zero. That would give you m - n =11

Perform a synthetic division of polynomial by +1 and remainder should be -4. That would give you m + n =-5

Solve the system of 2 equations and 2 unknowns (m,n) which would give you m = 3 and n = -8

6 years ago

## Answered By Timothy B

We have two unknowns, m and n, so we need to find two equations to solve the problem.

Equation 1 is (6x

^{3}+mx^{2}+nx-5)/(x+1) has a remainder of 0Equation 2 is (6x

^{3}+mx^{2}+nx-5)/(x-1) has a remainder of -4--

Performing division on Equation 1 gives us that the remainder is -11 - n + m, which is equal to 0.

So now Equation 1 is m-n = 11

--

Performing division on Equation 2 give us the remainder of n + m + 1 which is equal to -4.

So then Equation 2 become n + m = -5

--

Combining Equations 1 and 2 togethers gives the final answer of m = 3 and n = -8.