what is the completely factored form of $f(x)=6x^{3}-13x^{2}-4x + 15$?\n$(x + 1)(6x^{2}-19x + 15)$\n$(x +…

Explanation:

Step1: Use the Rational Root Theorem

The Rational Root Theorem states that if a polynomial (a_nx^n+\cdots + a_1x + a_0) has a rational root (p/q), then (p) is a factor of (a_0) and (q) is a factor of (a_n). For (f(x)=6x^{3}-13x^{2}-4x + 15), (a_n = 6) and (a_0=15). The possible rational roots are (\pm1,\pm3,\pm 5,\pm15,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{5}{2},\pm\frac{15}{2},\pm\frac{1}{3},\pm\frac{5}{3},\pm\frac{1}{6},\pm\frac{5}{6}). Try (x=-1): (f(-1)=6(-1)^{3}-13(-1)^{2}-4(-1)+15=-6 - 13 + 4+15=0). So ((x + 1)) is a factor.

Step2: Perform polynomial long - division

Divide (6x^{3}-13x^{2}-4x + 15) by ((x + 1)). Using polynomial long - division: (6x^{3}-13x^{2}-4x + 15=(x + 1)(6x^{2}-19x + 15)).

Step3: Factor the quadratic

Factor (6x^{2}-19x + 15). We need to find two numbers that multiply to (6\times15 = 90) and add up to (-19). The numbers are (-10) and (-9). Then (6x^{2}-19x + 15=6x^{2}-10x-9x + 15=2x(3x - 5)-3(3x - 5)=(2x - 3)(3x - 5)). So (f(x)=(x + 1)(2x - 3)(3x - 5)).

Answer:

D. ((x + 1)(2x - 3)(3x - 5))