the equation below has real roots that are rational. use the rational zero theorem to list all possible…

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.\n\n21x³ - 79x² + 51x - 9 = 0; -4,6,1 by -300,300,100\n\nlist all possible rational roots.\n\n(type an integer or a simplified fraction. use a comma to separate answers as needed. type each answer only once.)

Answer

Explanation:

Step1: Identify (p) and (q)

For the polynomial (a_nx^n+\cdots+a_1x + a_0), here (a_n = 21) (leading - coefficient) and (a_0=- 9) (constant term). By the Rational Zero Theorem, if (x=\frac{p}{q}) is a rational root, then (p) is a factor of (a_0) and (q) is a factor of (a_n). The factors of (p=-9) (ignoring the sign) 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 (\frac{p}{q}) are: [ \begin{align*} \frac{p}{q}&=\frac{\pm1}{\pm1},\frac{\pm1}{\pm3},\frac{\pm1}{\pm7},\frac{\pm1}{\pm21},\frac{\pm3}{\pm1},\frac{\pm3}{\pm3},\frac{\pm3}{\pm7},\frac{\pm3}{\pm21},\frac{\pm9}{\pm1},\frac{\pm9}{\pm3},\frac{\pm9}{\pm7},\frac{\pm9}{\pm21}\ \end{align*} ] Simplify each fraction:

(\frac{\pm1}{\pm1}=\pm1)
(\frac{\pm1}{\pm3}=\pm\frac{1}{3})
(\frac{\pm1}{\pm7}=\pm\frac{1}{7})
(\frac{\pm1}{\pm21}=\pm\frac{1}{21})
(\frac{\pm3}{\pm1}=\pm3)
(\frac{\pm3}{\pm3}=\pm1) (already listed)
(\frac{\pm3}{\pm7}=\pm\frac{3}{7})
(\frac{\pm3}{\pm21}=\pm\frac{1}{7}) (already listed)
(\frac{\pm9}{\pm1}=\pm9)
(\frac{\pm9}{\pm3}=\pm3) (already listed)
(\frac{\pm9}{\pm7}=\pm\frac{9}{7})
(\frac{\pm9}{\pm21}=\pm\frac{3}{7}) (already listed)

Answer:

(-9,-3,-1,-\frac{3}{7},-\frac{1}{3},-\frac{1}{7},-\frac{1}{21},\frac{1}{21},\frac{1}{7},\frac{1}{3},\frac{3}{7},1,3,9)