Question

find all rational roots of ( h(x) ).( h(x) = -3x^{4} + 2x^{3} + 10x^{2} - 11x + 2 )write your answer as a list of simplified values separated by commas, if there is more than one value.

Explanation:

Step1: Apply Rational Root Theorem

Possible rational roots: $\pm1, \pm2, \pm\frac{1}{3}, \pm\frac{2}{3}$

Step2: Test $x=1$

$h(1) = -3(1)^4 + 2(1)^3 + 10(1)^2 -11(1)+2 = -3+2+10-11+2=0$

Step3: Factor out $(x-1)$ via division

Use polynomial division or synthetic division:
$h(x)=(x-1)(-3x^3 -x^2 +9x -2)$

Step4: Test $x=2$ on cubic

$h(2) = -3(8)-4+18-2=-24-4+18-2=-12
eq0$; test $x=\frac{1}{3}$:
$-3(\frac{1}{3})^3 -(\frac{1}{3})^2 +9(\frac{1}{3}) -2 = -\frac{1}{9}-\frac{1}{9}+3-2=\frac{7}{9}
eq0$; test $x=-\frac{2}{3}$:
$-3(-\frac{8}{27}) -(\frac{4}{9}) +9(-\frac{2}{3}) -2 = \frac{8}{9}-\frac{4}{9}-6-2=\frac{4}{9}-8
eq0$; test $x=\frac{2}{3}$:
$-3(\frac{8}{27}) -(\frac{4}{9}) +9(\frac{2}{3}) -2 = -\frac{8}{9}-\frac{4}{9}+6-2=-\frac{12}{9}+4=\frac{8}{3}
eq0$; test $x=-1$:
$-3(-1)-1+(-9)-2=3-1-9-2=-9
eq0$; test $x=\frac{1}{3}$ again on cubic: no, test $x=2$ was wrong, test $x=\frac{1}{3}$ on original: $h(\frac{1}{3})=-3(\frac{1}{81})+2(\frac{1}{27})+10(\frac{1}{9})-11(\frac{1}{3})+2=-\frac{1}{27}+\frac{2}{27}+\frac{30}{27}-\frac{99}{27}+\frac{54}{27}=0$

Step5: Factor out $(3x-1)$ from cubic

$-3x^3 -x^2 +9x -2 = (3x-1)(-x^2 - \frac{2}{3}x + 2) = -(3x-1)(x^2 + \frac{2}{3}x - 2)$

Step6: Factor quadratic

$x^2 + \frac{2}{3}x -2 = 0 \implies 3x^2+2x-6=0$, discriminant $4+72=76$, roots irrational.

Step7: Identify rational roots

From steps 2 and 5, rational roots are $1, \frac{1}{3}$

Answer:

$1, \frac{1}{3}$