Question

use the rational zero test to list the possible rational zeros of f. verify that the zeros of f shown on the graph are co
f(x) = 9x⁵ - 27x⁴ - 10x³ + 30x² + x - 3

Explanation:

Step1: Identify constant & leading coeff

Constant term: $-3$, Leading coefficient: $9$

Step2: List factors of each

Factors of $-3$: $\pm1, \pm3$; Factors of $9$: $\pm1, \pm3, \pm9$

Step3: Apply Rational Zero Test

Possible rational zeros: $\pm1, \pm3, \pm\frac{1}{3}, \pm\frac{1}{9}$

Step4: Verify graph zeros (-1,1,3)

Check $x=-1$:

$f(-1)=9(-1)^5 -27(-1)^4 -10(-1)^3 +30(-1)^2 +(-1)-3$
$= -9 -27 +10 +30 -1 -3 = 0$

Check $x=1$:

$f(1)=9(1)^5 -27(1)^4 -10(1)^3 +30(1)^2 +1 -3$
$= 9 -27 -10 +30 +1 -3 = 0$

Check $x=3$:

$f(3)=9(3)^5 -27(3)^4 -10(3)^3 +30(3)^2 +3 -3$
$= 9(243) -27(81) -10(27) +30(9) +0$
$= 2187 - 2187 - 270 + 270 = 0$

Answer:

Possible rational zeros: $\pm1, \pm3, \pm\frac{1}{3}, \pm\frac{1}{9}$
Verified zeros from the graph: $x=-1$, $x=1$, $x=3$ (all are valid rational zeros of $f(x)$)