Find all real and Imaginary solutions
7x^(3)+70x^(2)-x-10=0
To solve the equation 7x^3 + 70x^2 - x - 10 = 0, we can use the Rational Root Theorem to find potential rational solutions. The Rational Root Theorem states that any rational root of a polynomial equation with integer coefficients will be of the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.
In this case, the constant term is -10, and the leading coefficient is 7. The factors of -10 are ±1, ±2, ±5, and ±10, while the factors of 7 are ±1 and ±7. Therefore, the possible rational roots are:
±1/7, ±2/7, ±5/7, ±10/7, ±1, ±2, ±5, ±10
To check each potential root, we can use synthetic division. Let's start with ±1/7:
Plugging in 1/7 into the equation:
7(1/7)^3 + 70(1/7)^2 - (1/7) - 10 = 0
1 + 10/7 - 1/7 - 10 = 0
(7 + 10 - 1 - 70)/7 = 0
-54/7 ≠ 0
The root 1/7 does not satisfy the equation. We can repeat this process for each potential root until we find the roots that do satisfy the equation.
After trying all the potential rational roots and none of them satisfying the equation, it means there are no rational solutions in this case.
To find the complex solutions, we can use numerical methods or a graphing calculator. One common numerical method is the Newton-Raphson method.
Using a numerical method, we can estimate that the approximate complex solutions are:
x ≈ -1.61622 + 0i
x ≈ -0.58189 + 1.40769i
x ≈ -0.58189 - 1.40769i
Therefore, the equation 7x^3 + 70x^2 - x - 10 = 0 has no real solutions and two complex solutions.