If x2+x-2 is a factor of f(x)=2x3+9x2+mx−n, find the values of m and n.

We factorize the quadratic expression:

$$x^2 + x - 2 = 0$$ $$(x+2)(x-1) = 0$$

The roots are $x = -2$ and $x = 1$.

Since $x=-2$ and $x=1$ are roots, we must have $f(-2) = 0$ and $f(1) = 0$.
The polynomial is $f(x) = 2x^3 + 9x^2 + mx - n$.

For $x=1$:

$$f(1) = 2(1)^3 + 9(1)^2 + m(1) - n = 0$$ $$2 + 9 + m - n = 0$$ $$11 + m - n = 0$$ $$m - n = -11 \quad \text{(Equation 1)}$$

For $x=-2$:

$$f(-2) = 2(-2)^3 + 9(-2)^2 + m(-2) - n = 0$$ $$2(-8) + 9(4) - 2m - n = 0$$ $$-16 + 36 - 2m - n = 0$$ $$20 - 2m - n = 0$$ $$-2m - n = -20$$ $$2m + n = 20 \quad \text{(Equation 2)}$$

We have the system:

1. $m - n = -11$
2. $2m + n = 20$

Add Equation 1 and Equation 2:

$$(m - n) + (2m + n) = -11 + 20$$ $$3m = 9$$ $$m = \frac{9}{3}$$ $$m = 3$$

Substitute the value of $m=3$ into Equation 1:

$$3 - n = -11$$ $$-n = -11 - 3$$ $$-n = -14$$ $$n = 14$$

The values are $m=3$ and $n=14$.

The final answer is $m=3, n=14$.