Respuesta :
Given a modelled account balance for the period of d days as shown below:
[tex]\begin{gathered} f(d)=d^3-2d^2-8d+3 \\ \text{where,} \\ f(d)\text{ is the account balance} \\ d\text{ is the number of days} \end{gathered}[/tex]Given that the account balance is $38, we would calculate the number of days by substituting for f(d) = 38 in the modelled equation as shown below:
[tex]\begin{gathered} 38=d^3-2d^2-8d+3 \\ d^3-2d^2-8d+3-38=0 \\ d^3-2d^2-8d-35=0 \end{gathered}[/tex]Since all coefficients of the variable d from degree 3 to 1 are integers, we would apply apply the Rational Zeros Theorem.
The trailing coefficient (coefficient of the constant term) is −35.
Find its factors (with plus and minus): ±1,±5,±7,±35. These are the possible values for dthat would give the zeros of the equation
Lets input x= 5
[tex]\begin{gathered} 5^3-2(5)^2-8(5)-35=0 \\ 125-2(25)-40-35=0 \\ 125-50-75=0 \\ 125-125=0 \\ 0=0 \end{gathered}[/tex]Since, x= 5 is a zero, then x-5 is a factor.
[tex]\begin{gathered} d^3-2d^2-8d-35=(d-5)(d^2+3d+7)=0 \\ (d-5)(d^2+3d+7)=0 \\ d-5=0,d^2+3d+7=0 \\ d=0, \end{gathered}[/tex][tex]\begin{gathered} \text{simplifying } \\ d^2+3d+7\text{ would give} \\ d=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ a=1,b=3,c=7 \end{gathered}[/tex][tex]\begin{gathered} d=\frac{-3\pm\sqrt[]{3^2-4\times1\times7}}{2\times1} \\ d=\frac{-3\pm\sqrt[]{9-28}}{2} \\ d=\frac{-3\pm\sqrt[]{-17}}{2} \end{gathered}[/tex]It can be observed that the roots of the equation would give one real root and two complex roots
Therefore,
[tex]d=5,d=\frac{-3\pm\sqrt[]{-17}}{2}[/tex]Since number of days cannot a complex number, hence, the number of days that would give a balance of $38 is 5 days
