Answer:
The multiplicity of 3 is 2
Step-by-step explanation:
[tex]Use\:the\:rational\:root\:theorem[/tex]
[tex]a_0=9,\:\quad a_n=2[/tex]
[tex]\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:3,\:9,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\:2[/tex]
[tex]\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:3,\:9}{1,\:2}[/tex]
[tex]\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x-3[/tex]
[tex]=\left(x-3\right)\frac{2x^3-11x^2+12x+9}{x-3}[/tex]
since, [tex]\frac{2x^3-11x^2+12x+9}{x-3}=2x^2-5x-3[/tex]
[tex]\mathrm{Factor}\:2x^2-5x-3:\quad \left(2x+1\right)\left(x-3\right)[/tex]
[tex]=\left(x-3\right)\left(2x+1\right)\left(x-3\right)[/tex]
[tex]=\left(x-3\right)^2\left(2x+1\right)[/tex]
Therefore, the multiplicity of 3 is 2
