Question

1. the number n(t) (in hundreds), of mosquitoes in a camping area after t weeks can be modelled using the equation n(t)=2t^4 - 5t^3 - 16t^2 + 45t. according to this model, when will the population of mosquitoes be greater than 1800?

Explanation:

Step1: Set up the inequality

Since $n(t)$ is in hundreds, we want $n(t)>18$. So, $2t^{4}-5t^{3}-16t^{2}+45t > 18$. Rearrange to get $2t^{4}-5t^{3}-16t^{2}+45t - 18>0$.

Step2: Try to find rational roots

By the Rational - Root Theorem, the possible rational roots of the polynomial $2t^{4}-5t^{3}-16t^{2}+45t - 18$ are factors of $\frac{18}{2}=9$. The possible values are $\pm1,\pm2,\pm3,\pm 6,\pm9$. Test $t = 1$: $2(1)^{4}-5(1)^{3}-16(1)^{2}+45(1)-18=2 - 5-16 + 45-18=8
eq0$. Test $t = 2$: $2(2)^{4}-5(2)^{3}-16(2)^{2}+45(2)-18=32-40 - 64 + 90-18=0$. So, $(t - 2)$ is a factor.

Step3: Perform polynomial long - division

Divide $2t^{4}-5t^{3}-16t^{2}+45t - 18$ by $(t - 2)$ using polynomial long - division or synthetic division. Using synthetic division:

2	-5	-16	45	-18
2	-1	-18	9	0

The quotient is $2t^{3}-t^{2}-18t + 9$.

Step4: Factor the quotient

Group the terms of $2t^{3}-t^{2}-18t + 9$: $t^{2}(2t - 1)-9(2t - 1)=(2t - 1)(t^{2}-9)=(2t - 1)(t - 3)(t + 3)$.
So, $2t^{4}-5t^{3}-16t^{2}+45t - 18=(t - 2)(2t - 1)(t - 3)(t + 3)$.

Step5: Find the intervals

Set each factor equal to zero: $t-2 = 0\Rightarrow t = 2$, $2t - 1=0\Rightarrow t=\frac{1}{2}$, $t - 3=0\Rightarrow t = 3$, $t+3=0\Rightarrow t=-3$.
The intervals are $(-\infty,-3),(-3,\frac{1}{2}),(\frac{1}{2},2),(2,3),(3,\infty)$.
We can use test points from each interval in the inequality $(t - 2)(2t - 1)(t - 3)(t + 3)>0$.
For $t=-4$: $(-4 - 2)(2(-4)-1)(-4 - 3)(-4 + 3)=(-6)(-9)(-7)(-1)=378>0$.
For $t = 0$: $(0 - 2)(2(0)-1)(0 - 3)(0 + 3)=(-2)(-1)(-3)(3)=-18<0$.
For $t=\frac{3}{2}$: $(\frac{3}{2}-2)(2(\frac{3}{2})-1)(\frac{3}{2}-3)(\frac{3}{2}+3)=(-\frac{1}{2})(2)(-\frac{3}{2})(\frac{9}{2})=\frac{27}{4}>0$.
For $t=\frac{5}{2}$: $(\frac{5}{2}-2)(2(\frac{5}{2})-1)(\frac{5}{2}-3)(\frac{5}{2}+3)=(\frac{1}{2})(4)(-\frac{1}{2})(\frac{11}{2})=-\frac{11}{2}<0$.
For $t = 4$: $(4 - 2)(2(4)-1)(4 - 3)(4 + 3)=(2)(7)(1)(7)=98>0$.
Since $t$ represents the number of weeks, $t\geq0$.

Answer:

$t>3$ weeks.