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.

±1,±1/3,±1/7,±1/21,±3,±3/7,±9,±9/7

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

choose the correct graph below.

all graphs are -4,6,1 by -300,300,100.

Explanation:

Step1: Apply the Rational Zero Theorem

The Rational Zero Theorem states that if a polynomial \(P(x)=a_nx^n + a_{n - 1}x^{n-1}+\cdots+a_1x + a_0\) has integer coefficients, then the possible rational zeros are of the form \(\frac{p}{q}\), where \(p\) is a factor of the constant term \(a_0\) and \(q\) is a factor of the leading - coefficient \(a_n\).

For the polynomial \(P(x)=21x^{3}-79x^{2}+51x - 9\), \(a_n = 21\) and \(a_0=-9\).

The factors of \(a_0=-9\) (i.e., \(p\)) are \(\pm1,\pm3,\pm9\). The factors of \(a_n = 21\) (i.e., \(q\)) are \(\pm1,\pm3,\pm7,\pm21\).

Then the possible rational zeros \(\frac{p}{q}\) are:

$$
LATEXBLOCK0
$$

After removing duplicates, the possible rational roots are \(\pm1,\pm\frac{1}{3},\pm\frac{1}{7},\pm\frac{1}{21},\pm3,\pm\frac{3}{7},\pm9,\pm\frac{9}{7}\)

Step2: Test the possible rational roots

We can use synthetic division or direct substitution.

Let's test \(x = 3\):

$$P(3)=21\times3^{3}-79\times3^{2}+51\times3 - 9=21\times27-79\times9 + 153-9$$

$$=567-711+153 - 9=(567 + 153)-(711 + 9)=720 - 720=0$$

Since \(P(3)=0\), \(x = 3\) is a root.

We can then divide \(21x^{3}-79x^{2}+51x - 9\) by \((x - 3)\) using synthetic division:

3 |  21  -79   51  -9
   |     63  -48   9
   |----------------
     21  -16    3   0

The quotient is \(21x^{2}-16x + 3\).

We can solve \(21x^{2}-16x + 3=0\) using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), where \(a = 21\), \(b=-16\), \(c = 3\)

$$x=\frac{16\pm\sqrt{(-16)^{2}-4\times21\times3}}{2\times21}=\frac{16\pm\sqrt{256 - 252}}{42}=\frac{16\pm2}{42}$$

$$x_1=\frac{16 + 2}{42}=\frac{18}{42}=\frac{3}{7}$$

,

$$x_2=\frac{16- 2}{42}=\frac{14}{42}=\frac{1}{3}$$

Answer:

The actual rational roots of the equation \(21x^{3}-79x^{2}+51x - 9 = 0\) are \(x=\frac{1}{3},x=\frac{3}{7},x = 3\)