which is a factor of $f(x)=60x^{4}+86x^{3}-46x^{2}-43x + 8$? use the rational root theorem to help you find…

Explanation:

Step1: Recall Rational Root Theorem

The Rational Root Theorem states that if a polynomial (f(x)=a_nx^n + a_{n - 1}x^{n-1}+\cdots+a_1x + a_0) has a rational root (p/q), then (p) is a factor of the constant term (a_0) and (q) is a factor of the leading - coefficient (a_n). For (f(x)=60x^4 + 86x^3-46x^2-43x + 8), (a_n = 60) and (a_0=8). The possible rational roots are of the form (p/q), where (p=\pm1,\pm2,\pm4,\pm8) and (q=\pm1,\pm2,\pm3,\pm4,\pm5,\pm6,\pm10,\pm12,\pm15,\pm20,\pm30,\pm60).

Step2: Use the factor - root relationship

If (x - r) is a factor of (f(x)), then (r) is a root of (f(x)), i.e., (f(r)=0). For a linear factor (ax - b), the root is (x=\frac{b}{a}).

Check (x - 6):

If (x-6) is a factor, then (x = 6) is a root. (f(6)=60\times6^4+86\times6^3-46\times6^2-43\times6 + 8=60\times1296+86\times216-46\times36-258 + 8\neq0).

Check (5x - 8):

If (5x - 8) is a factor, then (x=\frac{8}{5}) is a root. (f(\frac{8}{5})=60\times(\frac{8}{5})^4+86\times(\frac{8}{5})^3-46\times(\frac{8}{5})^2-43\times\frac{8}{5}+8\neq0).

Check (6x - 1):

If (6x - 1) is a factor, then (x=\frac{1}{6}) is a root. [ \begin{align*} f(\frac{1}{6})&=60\times(\frac{1}{6})^4+86\times(\frac{1}{6})^3-46\times(\frac{1}{6})^2-43\times\frac{1}{6}+8\ &=60\times\frac{1}{1296}+86\times\frac{1}{216}-46\times\frac{1}{36}-\frac{43}{6}+8\ &=\frac{60}{1296}+\frac{86}{216}-\frac{46}{36}-\frac{43}{6}+8\ &=\frac{5}{108}+\frac{43}{108}-\frac{138}{108}-\frac{774}{108}+\frac{864}{108}\ &=\frac{5 + 43-138-774 + 864}{108}\ &=\frac{0}{108}=0 \end{align*} ]

Check (8x + 5):

If (8x+5) is a factor, then (x=-\frac{5}{8}) is a root. (f(-\frac{5}{8})=60\times(-\frac{5}{8})^4+86\times(-\frac{5}{8})^3-46\times(-\frac{5}{8})^2-43\times(-\frac{5}{8})+8\neq0).

Answer:

C. (6x - 1)