Answer: Option 'C' is correct.
Step-by-step explanation:
Since we have given that
![\frac{\left(6x^4+15x^3+10x^2+10x+4\right)}{\left(3x^2+2\right)}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cleft%286x%5E4%2B15x%5E3%2B10x%5E2%2B10x%2B4%5Cright%29%7D%7B%5Cleft%283x%5E2%2B2%5Cright%29%7D)
Now, we will find the quotient by factoring the numerator:
![\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\\\\](https://tex.z-dn.net/?f=%5Cmathrm%7BUse%5C%3Athe%5C%3Arational%5C%3Aroot%5C%3Atheorem%7D%5C%5Ca_0%3D4%2C%5C%3A%5Cquad%20a_n%3D6%5C%5C%5C%5C%5Cmathrm%7BThe%5C%3Adividers%5C%3Aof%5C%3A%7Da_0%3A%5Cquad%201%2C%5C%3A2%2C%5C%3A4%2C%5C%3A%5Cquad%20%5C%5C%5Cmathrm%7BThe%5C%3Adividers%5C%3Aof%5C%3A%7Da_n%3A%5Cquad%201%2C%5C%3A2%2C%5C%3A3%2C%5C%3A6%5C%5C%5C%5C%5Cmathrm%7BTherefore%2C%5C%3Acheck%5C%3Athe%5C%3Afollowing%5C%3Arational%5C%3Anumbers%3A%5Cquad%20%7D%5Cpm%20%5Cfrac%7B1%2C%5C%3A2%2C%5C%3A4%7D%7B1%2C%5C%3A2%2C%5C%3A3%2C%5C%3A6%7D%5C%5C%5C%5C-%5Cfrac%7B2%7D%7B1%7D%5Cmathrm%7B%5C%3Ais%5C%3Aa%5C%3Aroot%5C%3Aof%5C%3Athe%5C%3Aexpression%2C%5C%3Aso%5C%3Afactor%5C%3Aout%5C%3A%7Dx%2B2%5C%5C%5C%5C%3D%5Cleft%28x%2B2%5Cright%29%5Cfrac%7B6x%5E4%2B15x%5E3%2B10x%5E2%2B10x%2B4%7D%7Bx%2B2%7D%5C%5C%5C%5C%3D%5Cfrac%7B6x%5E4%2B15x%5E3%2B10x%5E2%2B10x%2B4%7D%7Bx%2B2%7D%3D6x%5E3%2B3x%5E2%2B4x%2B2%5C%5C%5C%5C)
Now, we will factor it again:
![=\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)](https://tex.z-dn.net/?f=%3D%5Cleft%286x%5E3%2B3x%5E2%5Cright%29%2B%5Cleft%284x%2B2%5Cright%29%5C%5C%5C%5C%3D2%5Cleft%282x%2B1%5Cright%29%2B3x%5E2%5Cleft%282x%2B1%5Cright%29%5C%5C%5C%5C%3D%5Cleft%282x%2B1%5Cright%29%5Cleft%283x%5E2%2B2%5Cright%29)
At last we get our factorised form :
![=\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](https://tex.z-dn.net/?f=%3D%5Cleft%28x%2B2%5Cright%29%5Cleft%282x%2B1%5Cright%29%5Cleft%283x%5E2%2B2%5Cright%29%5C%5C%5C%5C%3D%5Cfrac%7B%5Cleft%28x%2B2%5Cright%29%5Cleft%282x%2B1%5Cright%29%5Cleft%283x%5E2%2B2%5Cright%29%7D%7B3x%5E2%2B2%7D%5C%5C%5C%5C%3D%5Cleft%28x%2B2%5Cright%29%5Cleft%282x%2B1%5Cright%29%5C%5C%5C%5C%3D2x%5E2%2B5x%2B2)
Hence, Option 'C' is correct.
Answer:
y=28
Step-by-step explanation:
Hope this helps
The LCM of 6,2,and 22 is 66.
Happy studying ^-^
The answer should be C: 12