The cubic equation has roots x = 7, x = -3 + 3i, and x = -3 - 3i. The quartic equation has roots x ≈ -2.17 + 0.46i, x ≈ -2.17 - 0.46i, x ≈ -0.83 + 1.45i, and x ≈ -0.83 - 1.45i.
The first problem is a cubic equation, f(x) = x^3 - x^2 + 25x - 252.
To find its real and imaginary roots, we can use the Rational Root Theorem to test for possible rational roots.
By testing the factors of the constant term, we find that x = 7 is a root. So, we can divide f(x) by (x - 7) to solve for the remaining roots.
After dividing, we get x^2 + 6x + 36, which is a quadratic equation.
Solving this quadratic equation using the quadratic formula gives us two complex roots: x = -3 + 3i and x = -3 - 3i.
Therefore, the solutions to the cubic equation are x = 7, x = -3 + 3i, and x = -3 - 3i.
The second problem is a quartic equation, f(x) = x^4 + 6x^3 + 14x^2 + 54x + 45.
To find its real and imaginary roots, we can use the Rational Root Theorem again. However, in this case, there are no rational roots.
To find the complex roots, we can use methods like factoring, synthetic division, or numerical approximations.
By using numerical methods, we find the complex roots to be x ≈ -2.17 + 0.46i, x ≈ -2.17 - 0.46i, x ≈ -0.83 + 1.45i, and x ≈ -0.83 - 1.45i.
Therefore, the solutions to the quartic equation are x ≈ -2.17 + 0.46i, x ≈ -2.17 - 0.46i, x ≈ -0.83 + 1.45i, and x ≈ -0.83 - 1.45i.
The probable question may be:
1. Find all the real and imaginary roots for the problem.
f(x)=x^3-x^2+25x-25
2. Find all the real and imaginary roots for the problem.
f(x)=x^4+6x^3+14x^2+54x+45