Respuesta :
Answer:
The given equation can be written as [tex]5t^3-4t^2+7t+16=0[/tex]
The factors of the equation is
[tex]5t^3-4t^2+7t+16=(t+1)(\frac{9+i\sqrt{239}}{10})(\frac{9-i\sqrt{239}}{10})[/tex]
Step-by-step explanation:
Given equation is [tex]t+\frac{4}{t}+\frac{3}{t}-4=-\frac{16}{t^2}-4t[/tex]
Now to solve the given equation:
[tex]t+\frac{4}{t}+\frac{3}{t}-4=-\frac{16}{t^2}-4t[/tex]
[tex]t+\frac{7}{t}-4=-\frac{16}{t^2}-4t[/tex]
[tex]\frac{t^2+7-4t}{t}=\frac{-16-4t^3}{t^2}[/tex]
Multiply the above equation into [tex]t^2[/tex] on both sides
[tex]\frac{t^2+7-4t}{t}\times t^2=\frac{-16-4t^3}{t^2}\times t^2[/tex]
[tex]t(t^2+7-4t)=-16-4t^3[/tex]
[tex]t^3+7t-4t^2=-16-4t^3[/tex]
[tex]t^3+7t-4t^2+16+4t^3=0[/tex]
Adding the like terms
[tex]5t^3+7t-4t^2+16=0[/tex]
Rewritting the above equation
[tex]5t^3-4t^2+7t+16=0[/tex]
We can solve this equation by synthetic division method
-1_| 5 -4 7 16
0 -5 9 -16
_______________
5 -9 16 0
Therefore t+1 is a factor of this cubic equation
We have the quadratic equation [tex]5t^2-9t+16=0[/tex]
For quadratic equation [tex]ax^2+bx+c=0[/tex] the solution is
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
where a and b are coefficients of [tex]x^2[/tex] and x respectively
Here a=5 ,b=-9 and c=16
[tex]t=\frac{-(-9)\pm\sqrt{(-9)^2-4(5)(16)}}{2(5)}[/tex]
[tex]t=\frac{9\pm\sqrt{81-320}}{10}[/tex]
[tex]t=\frac{9\pm\sqrt{-239}}{10}[/tex]
[tex]t=\frac{9\pm\sqrt{239i^2}}{10}[/tex] where [tex]i^2=-1[/tex]
[tex]t=\frac{9\pmi\sqrt{239}}{10}[/tex]
Therefore [tex]t=\frac{9+i\sqrt{239}}{10}[/tex] and [tex]t=\frac{9-i\sqrt{239}}{10}[/tex]
Therefore the factors are t+1,[tex]\frac{9+i\sqrt{239}}{10}[/tex] and [tex]\frac{9-i\sqrt{239}}{10}[/tex]
[tex]5t^3-4t^2+7t+16=(t+1)(\frac{9+i\sqrt{239}}{10})(\frac{9-i\sqrt{239}}{10})[/tex]
