To find the values of [tex]\( m \)[/tex] and [tex]\( n \)[/tex] such that the polynomial [tex]\( 2x^3 + mx^2 + nx - 14 \)[/tex] is exactly divisible by [tex]\( x^2 + x - 2 \)[/tex], we need to ensure that the polynomial [tex]\( 2x^3 + mx^2 + nx - 14 \)[/tex] has [tex]\( x^2 + x - 2 \)[/tex] as a factor. Here are the steps:
1. Factor the divisor:
[tex]\[
x^2 + x - 2 = (x - 1)(x + 2)
\][/tex]
This means that the roots of [tex]\( x^2 + x - 2 \)[/tex] are [tex]\( x = 1 \)[/tex] and [tex]\( x = -2 \)[/tex].
2. Setting up conditions for the polynomial:
For [tex]\( 2x^3 + mx^2 + nx - 14 \)[/tex] to be divisible by [tex]\( x^2 + x - 2 \)[/tex], it must be zero when [tex]\( x = 1 \)[/tex] and [tex]\( x = -2 \)[/tex].
3. Substitute [tex]\( x = 1 \)[/tex] into the polynomial:
[tex]\[
P(1) = 2(1)^3 + m(1)^2 + n(1) - 14 = 2 + m + n - 14
\][/tex]
Simplify to:
[tex]\[
2 + m + n - 14 = 0 \quad \Rightarrow \quad m + n - 12 = 0
\][/tex]
So the first equation is:
[tex]\[
m + n = 12
\][/tex]
4. Substitute [tex]\( x = -2 \)[/tex] into the polynomial:
[tex]\[
P(-2) = 2(-2)^3 + m(-2)^2 + n(-2) - 14 = 2(-8) + m(4) + n(-2) - 14
\][/tex]
Simplify to:
[tex]\[
-16 + 4m - 2n - 14 = 0 \quad \Rightarrow \quad 4m - 2n - 30 = 0
\][/tex]
So the second equation is:
[tex]\[
4m - 2n = 30
\][/tex]
5. Solve the system of equations:
[tex]\[
\begin{cases}
m + n = 12 \\
4m - 2n = 30
\end{cases}
\][/tex]
First, solve the second equation for [tex]\( n \)[/tex]:
[tex]\[
4m - 2n = 30 \quad \Rightarrow \quad 2m - n = 15 \quad \Rightarrow \quad n = 2m - 15
\][/tex]
Substitute [tex]\( n = 2m - 15 \)[/tex] into the first equation:
[tex]\[
m + (2m - 15) = 12
\][/tex]
Simplify:
[tex]\[
3m - 15 = 12 \quad \Rightarrow \quad 3m = 27 \quad \Rightarrow \quad m = 9
\][/tex]
Substitute [tex]\( m = 9 \)[/tex] back into [tex]\( n = 2m - 15 \)[/tex]:
[tex]\[
n = 2(9) - 15 = 18 - 15 = 3
\][/tex]
6. Conclude the result:
The values are [tex]\( m = 9 \)[/tex] and [tex]\( n = 3 \)[/tex]. Therefore, [tex]\( m + n = 9 + 3 = 12 \)[/tex].
Thus, the value of [tex]\( m + n \)[/tex] is [tex]\(\boxed{12}\)[/tex].
