Question

Use the rational zeros theorem to list all possible rational zeros of the following.

Answer 1

`\sf (p)/(q)=(\pm1)/(\pm1),(\pm1)/(\pm3),(\pm7)/(\pm1),(\pm7)/(\pm3)=\pm 1, (\pm1)/(\pm3),\pm 7, (\pm7)/(\pm3)`

Explanation:

Given polynomial:

`f(x)=-3x^3-5x^2+x-7`

Rational Root Theorem

If P(x) is a polynomial with integer coefficients and if p/q is a root of P(x), then p is a factor of the constant term of P(x) and q is a factor of the leading coefficient of P(x).

Possible p-values

Factors of the constant term: ±1, ±7

Possible q-values

Factors of the leading coefficient: ±1, ±3

Therefore, all the possible values of p/q:

`\sf (p)/(q)=(\pm1)/(\pm1),(\pm1)/(\pm3),(\pm7)/(\pm1),(\pm7)/(\pm3)=\pm 1, (\pm1)/(\pm3),\pm 7, (\pm7)/(\pm3)`

Substitute each possible rational root into the function:

`x=-1 \implies f(-1)=-3(-1)^3-5(-1)^2+(-1)-7=-10`

`x=1 \implies f(1)=-3(1)^3-5(1)^2+(1)-7=-14`

`x=-7 \implies f(-7)=-3(-7)^3-5(-7)^2+(-7)-7=770`

`x=7 \implies f(7)=-3(7)^3-5(7)^2+(7)-7=-1274`

`x=-(1)/(3) \implies f\left(-(1)/(3)\right)=-3\left(-(1)/(3)\right)^3-5\left(-(1)/(3)\right)^2+\left(-(1)/(3)\right)-7=-(70)/(9)`

`x=(1)/(3) \implies f\left((1)/(3)\right)=-3\left((1)/(3)\right)^3-5\left((1)/(3)\right)^2+\left((1)/(3)\right)-7=-(22)/(3)`

`x=-(7)/(3) \implies f\left(-(7)/(3)\right)=-3\left(-(7)/(3)\right)^3-5\left(-(7)/(3)\right)^2+\left(-(7)/(3)\right)-7=(14)/(9)`

`x=(7)/(3) \implies f\left((7)/(3)\right)=-3\left((7)/(3)\right)^3-5\left((7)/(3)\right)^2+\left((7)/(3)\right)-7=-70`

As f(p/q) ≠ 0, none of the possible rational roots are actual roots of the given polynomial.