Which of the numbers below are some potential roots of p(x) 36x2 -7x-60 according to the rational root theorem?

[tex]\pm 1,\pm 2,\pm 3,\pm 4,\pm 5,\pm 6,\pm 10,\pm 12,\pm 15,\pm 20,\pm 30,\pm 60,\pm \dfrac{1}{2},\pm \dfrac{3}{2},\pm \dfrac{5}{2},\pm \dfrac{15}{2},\pm \dfrac{1}{3},[/tex]

[tex]\pm \dfrac{2}{3},\pm \dfrac{4}{3},\dfrac{5}{3},\pm \dfrac{10}{3},\pm \dfrac{20}{3},\pm \dfrac{1}{3},\pm \dfrac{2}{3},\pm \dfrac{4}{3},\pm \dfrac{5}{3},\pm \dfrac{10}{3},\pm \dfrac{20}{3},\pm \dfrac{1}{4},\pm \dfrac{3}{4},\pm \dfrac{5}{4},[/tex]

[tex]\pm \dfrac{15}{4},\pm \dfrac{1}{6},\pm \dfrac{5}{6},\pm \dfrac{1}{9},\pm \dfrac{2}{9},\pm \dfrac{4}{9},\pm \dfrac{5}{9},\pm \dfrac{10}{9},\pm \dfrac{20}{9},\pm \dfrac{1}{12},\pm \dfrac{5}{12},\pm \dfrac{1}{18},\pm \dfrac{5}{18},\pm \dfrac{1}{36}[/tex]

Step-by-step explanation:

The given polynomial is

[tex]p(x)=36x^2-7x-60[/tex]

We need to find the potential roots of given polynomial according to the rational root theorem.

According to the rational root theorem, the potential roots are

[tex]x=\pm \dfrac{p}{q}[/tex]

where, p is a factor of constant term and q is the factor of leading term.

In the given polynomial constant term is -60 and leading term 36.

Factors of 60 are ±1,±2,±3,±4,±5,±6,±10,±12,±15,±20,±30,±60.

Factors of 36 are ±1,±2,±3,±4,±6,±9,±12,±18,±36.

Now, the potential roots are

[tex]\pm 1,\pm 2,\pm 3,\pm 4,\pm 5,\pm 6,\pm 10,\pm 12,\pm 15,\pm 20,\pm 30,\pm 60,\pm \dfrac{1}{2},\pm \dfrac{3}{2},\pm \dfrac{5}{2},\pm \dfrac{15}{2},\pm \dfrac{1}{3},[/tex]

[tex]\pm \dfrac{2}{3},\pm \dfrac{4}{3},\dfrac{5}{3},\pm \dfrac{10}{3},\pm \dfrac{20}{3},\pm \dfrac{1}{3},\pm \dfrac{2}{3},\pm \dfrac{4}{3},\pm \dfrac{5}{3},\pm \dfrac{10}{3},\pm \dfrac{20}{3},\pm \dfrac{1}{4},\pm \dfrac{3}{4},\pm \dfrac{5}{4},[/tex]

[tex]\pm \dfrac{15}{4},\pm \dfrac{1}{6},\pm \dfrac{5}{6},\pm \dfrac{1}{9},\pm \dfrac{2}{9},\pm \dfrac{4}{9},\pm \dfrac{5}{9},\pm \dfrac{10}{9},\pm \dfrac{20}{9},\pm \dfrac{1}{12},\pm \dfrac{5}{12},\pm \dfrac{1}{18},\pm \dfrac{5}{18},\pm \dfrac{1}{36}[/tex]