1. **Problem Statement:** Determine if the polynomial $12x^3 + 6x^2 - 1$ is factorable. 2. **Formula and Rules:** To factor a polynomial, first look for a greatest common factor (GCF). If none exists, check if it can be factored by grouping, special formulas (difference of squares, sum/difference of cubes), or by using the quadratic formula for degree 2 polynomials. 3. **Step 1: Find the GCF** The terms are $12x^3$, $6x^2$, and $-1$. The coefficients are 12, 6, and 1. The GCF of 12, 6, and 1 is 1. The variable $x$ appears in the first two terms but not in the last, so no variable factor common to all terms. 4. **Step 2: Try factoring by grouping or special formulas** Since there are three terms, check if it is a cubic polynomial that can be factored as a product of a linear and a quadratic polynomial: Assume $12x^3 + 6x^2 - 1 = (ax + b)(cx^2 + dx + e)$. 5. **Step 3: Use the Rational Root Theorem to test possible roots** Possible rational roots are factors of the constant term ($rac{±1}{1}, ±1$) divided by factors of the leading coefficient (12): $±1, ±\frac{1}{2}, ±\frac{1}{3}, ±\frac{1}{4}, ±\frac{1}{6}, ±\frac{1}{12}$. 6. **Step 4: Test $x=\frac{1}{2}$:** $$12\left(\frac{1}{2}\right)^3 + 6\left(\frac{1}{2}\right)^2 - 1 = 12\times\frac{1}{8} + 6\times\frac{1}{4} - 1 = 1.5 + 1.5 - 1 = 2 \neq 0$$ 7. **Step 5: Test $x=-\frac{1}{2}$:** $$12\left(-\frac{1}{2}\right)^3 + 6\left(-\frac{1}{2}\right)^2 - 1 = 12\times\left(-\frac{1}{8}\right) + 6\times\frac{1}{4} - 1 = -1.5 + 1.5 - 1 = -1 \neq 0$$ 8. **Step 6: Test $x=1$:** $$12(1)^3 + 6(1)^2 - 1 = 12 + 6 - 1 = 17 \neq 0$$ 9. **Step 7: Test $x=-1$:** $$12(-1)^3 + 6(-1)^2 - 1 = -12 + 6 - 1 = -7 \neq 0$$ 10. Since none of the rational roots work, the polynomial has no rational roots and is not factorable over the rationals. **Final answer:** The polynomial $12x^3 + 6x^2 - 1$ is not factorable using rational numbers.