Answer: Option 'C' is correct.
Step-by-step explanation:
Since we have given that
[tex]\frac{\left(6x^4+15x^3+10x^2+10x+4\right)}{\left(3x^2+2\right)}[/tex]
Now, we will find the quotient by factoring the numerator:
[tex]\mathrm{Use\:the\:rational\:root\:theorem}\\a_0=4,\:\quad a_n=6\\\\\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:4,\:\quad \\\mathrm{The\:dividers\:of\:}a_n:\quad 1,\:2,\:3,\:6\\\\\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:4}{1,\:2,\:3,\:6}\\\\-\frac{2}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+2\\\\=\left(x+2\right)\frac{6x^4+15x^3+10x^2+10x+4}{x+2}\\\\=\frac{6x^4+15x^3+10x^2+10x+4}{x+2}=6x^3+3x^2+4x+2\\\\[/tex]
Now, we will factor it again:
[tex]=\left(6x^3+3x^2\right)+\left(4x+2\right)\\\\=2\left(2x+1\right)+3x^2\left(2x+1\right)\\\\=\left(2x+1\right)\left(3x^2+2\right)[/tex]
At last we get our factorised form :
[tex]=\left(x+2\right)\left(2x+1\right)\left(3x^2+2\right)\\\\=\frac{\left(x+2\right)\left(2x+1\right)\left(3x^2+2\right)}{3x^2+2}\\\\=\left(x+2\right)\left(2x+1\right)\\\\=2x^2+5x+2[/tex]
Hence, Option 'C' is correct.
