Question

the equation below has real roots that are rational. use the rational zero theorem to list all possible rational roots. then graph the polynomial function in the given viewing rectangle to determine which possible rational roots are actual roots of the equation.

21x³ - 79x² + 51x - 9 = 0; -4,6,1 by -300,300,100

list all possible rational roots.

(type an integer or a simplified fraction. use a comma to separate answers as needed. type each answer only once.)

Explanation:

Step1: Identify \(p\) and \(q\)

For the polynomial \(21x^{3}-79x^{2}+51x - 9\), by the Rational Zero Theorem, if the polynomial is \(a_{n}x^{n}+a_{n - 1}x^{n-1}+\cdots+a_{1}x + a_{0}\), here \(n = 3\), \(a_{n}=21\), \(a_{0}=-9\). Let \(p\) be the factors of the constant term \(a_{0}\) and \(q\) be the factors of the leading - coefficient \(a_{n}\).
The factors of \(p=-9\) are \(p=\pm1,\pm3,\pm9\). The factors of \(q = 21\) are \(q=\pm1,\pm3,\pm7,\pm21\).

Step2: Calculate \(\frac{p}{q}\)

The possible rational roots are \(\frac{p}{q}=\pm1,\pm\frac{1}{3},\pm\frac{1}{7},\pm\frac{1}{21},\pm3,\pm\frac{3}{7},\pm9,\pm\frac{9}{7}\)

Answer:

\(\pm1,\pm\frac{1}{3},\pm\frac{1}{7},\pm\frac{1}{21},\pm3,\pm\frac{3}{7},\pm9,\pm\frac{9}{7}\)