Find the roots of x$^3$ - 19x - 30=0

Analyze the question: We need the three values of x that satisfy x³ - 19x - 30 = 0. Since it is a cubic, we use the Factor Theorem and test the options.

Concept: If f(a) = 0, then (x - a) is a factor of f(x). For a cubic with no x² term, the sum of the roots must be 0.

Solution:

Let f(x) = x³ - 19x - 30.

Test x = 5: f(5) = 125 - 95 - 30 = 0 ✔ so (x - 5) is a factor.

Test x = -2: f(-2) = -8 + 38 - 30 = 0 ✔ so (x + 2) is a factor.

Test x = -3: f(-3) = -27 + 57 - 30 = 0 ✔ so (x + 3) is a factor.

So x³ - 19x - 30 = (x - 5)(x + 2)(x + 3).

Check by expanding: (x + 2)(x + 3) = x² + 5x + 6 (x - 5)(x² + 5x + 6) = x³ + 5x² + 6x - 5x² - 25x - 30 = x³ - 19x - 30 ✔

Roots: x = 5, -2, -3

Checking the other options:

• A: -5, 3, 2 → f(3) = 27 - 57 - 30 = -60, so it fails.
• B: 5, 3, -2 → f(3) = -60, so it fails.
• D: -5, -3, 2 → f(2) = 8 - 38 - 30 = -60, so it fails.

Correct answer: C. 5, -3 and -2

Tip: For a cubic missing the x² term, the roots must add up to zero. Option C gives 5 - 3 - 2 = 0 ✔. Also, the product of the roots equals -(constant term) = 30 and 5 × (-3) × (-2) = 30 ✔. These quick checks let you eliminate wrong options in seconds.