<u>Answer:</u>
f(x) will not have any roots.
<u>Solution:
</u>
The roots of f(x) means the solution of f(x) = 0 which are the points where the function f(x) crosses the x axis. As the given equation, the highest power is 3, hence the equation will have total 3 roots.
Let us assume the three roots are a, b, c
Hence, ![x^{3}+2 x^{2}+3 x+4=(x-a)(x-b)(x-c)=0](https://tex.z-dn.net/?f=x%5E%7B3%7D%2B2%20x%5E%7B2%7D%2B3%20x%2B4%3D%28x-a%29%28x-b%29%28x-c%29%3D0)
Multiplying the brackets we get ![a\times b\times c =4](https://tex.z-dn.net/?f=a%5Ctimes%20b%5Ctimes%20c%20%3D4)
So a, b, c must be the factors of 4
The possibilities of factors of 4 are +1, -1, +2, -2, +4, -4
Substituting the values we get,
![f(1)=1^{3}+2 \times 1^{2}+3 \times 1+4=10](https://tex.z-dn.net/?f=f%281%29%3D1%5E%7B3%7D%2B2%20%5Ctimes%201%5E%7B2%7D%2B3%20%5Ctimes%201%2B4%3D10)
![f(-1)=\left(-1^{3}\right)+2 \times(-1)^{2}+3 \times(-1)+4=-1+2-3+4=2](https://tex.z-dn.net/?f=f%28-1%29%3D%5Cleft%28-1%5E%7B3%7D%5Cright%29%2B2%20%5Ctimes%28-1%29%5E%7B2%7D%2B3%20%5Ctimes%28-1%29%2B4%3D-1%2B2-3%2B4%3D2)
![f(2)=2^{3}+2 \times 2^{2}+3 \times 2+4=26](https://tex.z-dn.net/?f=f%282%29%3D2%5E%7B3%7D%2B2%20%5Ctimes%202%5E%7B2%7D%2B3%20%5Ctimes%202%2B4%3D26)
![f(-2)=\left(-2^{3}\right)+2 \times(-2)^{2}+3 \times(-2)+4=-2](https://tex.z-dn.net/?f=f%28-2%29%3D%5Cleft%28-2%5E%7B3%7D%5Cright%29%2B2%20%5Ctimes%28-2%29%5E%7B2%7D%2B3%20%5Ctimes%28-2%29%2B4%3D-2)
![f(4)=4^{3}+2 \times 4^{2}+3 \times 4+4=112](https://tex.z-dn.net/?f=f%284%29%3D4%5E%7B3%7D%2B2%20%5Ctimes%204%5E%7B2%7D%2B3%20%5Ctimes%204%2B4%3D112)
![f(-4)=-4^{3}+2 \times(-4)^{2}+3 \times(-4)+4=-40](https://tex.z-dn.net/?f=f%28-4%29%3D-4%5E%7B3%7D%2B2%20%5Ctimes%28-4%29%5E%7B2%7D%2B3%20%5Ctimes%28-4%29%2B4%3D-40)
So, there are no values that satisfy the equation.
Hence f(x) will not have any roots.