The problem is to find the domain of 2x+7 / x^3-5x^2-9x+18. I know that x^3-5x^2-9x+18 cannot = 0. What I can't remember is how to solve x^3-5x^2-9x+18 ≠ 0. Thanks!
Since there is no value of x that will
make the denominator = 0, the Domain
is all real values oF X.
X(x^2+1)(x-3)/x-2=0
To solve the equation x^3-5x^2-9x+18 ≠ 0, you can use the fact that if a cubic polynomial has any real roots, they will divide the constant term of the polynomial.
In this case, the constant term of the polynomial is 18, so if there are any real roots, they will be divisors of 18. You can list all the possible divisors of 18, which are -18, -9, -6, -3, -2, -1, 1, 2, 3, 6, 9, and 18.
Next, you can use synthetic division or any other method to check whether any of these divisors are roots of the polynomial. If a divisor is a root, it means that substituting that value into the polynomial will make it equal to zero.
If you find any roots, you can then factor the polynomial using the root and divide through. For example, if you find that x = 1 is a root, then you can divide x - 1 from the polynomial to factor it as (x - 1)(x^2 - 4x + 18).
Finally, by considering the factors, you can determine the values of x for which x^3-5x^2-9x+18 ≠ 0. In this case, since the polynomial is in denominator of a fraction (2x + 7) / (x^3 - 5x^2 - 9x + 18), you need to exclude any values of x that would make the denominator equal to zero, as division by zero is undefined. Thus, the domain of the given expression is all real numbers except for the values of x that make x^3-5x^2-9x+18 = 0.