The quotient in polynomial form is C. x² - x ₊ 1.
<h3>What is synthetic division?</h3>
In algebra, synthetic division is a method for manually performing Euclidean division of polynomials, with less calculation than long division.
The given problem is-
-3 | 1 2 -2 3
Using synthetic division method,
-3 | 1 2 -2 3
<u>| -3 3 3 </u>
|1 -1 1 6
So, the remainder from the calculation is 6 .
And the quotient polynomial is x² - x ₊ 1.
Hence, the correct answer for the given question is C. x² - x ₊ 1
More about Synthetic division method :
brainly.com/question/12642065
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Step-by-step explanation:
How was the government able to make children go to residential school?
The value of f[ -4 ] and g°f[-2] are 
and 13 respectively.
<h3>What is the value of f[-4] and g°f[-2]?</h3>
Given the function;
- f[ -4 ] = ?
- g°f[ -2 ] = ?
For f[ -4 ], we substitute -4 for every variable x in the function.
For g°f[-2]
g°f[-2] is expressed as g(f(-2))
![g(\frac{3x-2}{x+1}) = (\frac{3x-2}{x+1}) + 5\\\\g(\frac{3x-2}{x+1}) = \frac{3x-2}{x+1} + \frac{5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) = \frac{3x-2+5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) = \frac{8x+3}{x+1}\\\\We\ substitute \ in \ [-2] \\\\g(\frac{3x-2}{x+1}) = \frac{8(-2)+3}{(-2)+1}\\\\g(\frac{3x-2}{x+1}) = \frac{-16+3}{-2+1}\\\\g(\frac{3x-2}{x+1}) = \frac{-13}{-1}\\\\g(\frac{3x-2}{x+1}) = 13](https://tex.z-dn.net/?f=g%28%5Cfrac%7B3x-2%7D%7Bx%2B1%7D%29%20%3D%20%20%28%5Cfrac%7B3x-2%7D%7Bx%2B1%7D%29%20%2B%205%5C%5C%5C%5Cg%28%5Cfrac%7B3x-2%7D%7Bx%2B1%7D%29%20%3D%20%20%5Cfrac%7B3x-2%7D%7Bx%2B1%7D%20%2B%20%5Cfrac%7B5%28x%2B1%29%7D%7Bx%2B1%7D%5C%5C%5C%5Cg%28%5Cfrac%7B3x-2%7D%7Bx%2B1%7D%29%20%3D%20%20%5Cfrac%7B3x-2%2B5%28x%2B1%29%7D%7Bx%2B1%7D%5C%5C%5C%5Cg%28%5Cfrac%7B3x-2%7D%7Bx%2B1%7D%29%20%3D%20%20%5Cfrac%7B8x%2B3%7D%7Bx%2B1%7D%5C%5C%5C%5CWe%5C%20substitute%20%5C%20in%20%5C%20%5B-2%5D%20%5C%5C%5C%5Cg%28%5Cfrac%7B3x-2%7D%7Bx%2B1%7D%29%20%3D%20%20%5Cfrac%7B8%28-2%29%2B3%7D%7B%28-2%29%2B1%7D%5C%5C%5C%5Cg%28%5Cfrac%7B3x-2%7D%7Bx%2B1%7D%29%20%3D%20%20%5Cfrac%7B-16%2B3%7D%7B-2%2B1%7D%5C%5C%5C%5Cg%28%5Cfrac%7B3x-2%7D%7Bx%2B1%7D%29%20%3D%20%20%5Cfrac%7B-13%7D%7B-1%7D%5C%5C%5C%5Cg%28%5Cfrac%7B3x-2%7D%7Bx%2B1%7D%29%20%3D%20%2013)
Therefore, the value of f[ -4 ] and g°f[-2] are and 13 respectively.
Learn more about composite functions here: brainly.com/question/20379727
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Answer:
Option a) 1/5 is most closely to 0.
Answer:
![\boxed{5 \cdot \sqrt{2} \cdot \sqrt[6]{5} }](https://tex.z-dn.net/?f=%5Cboxed%7B5%20%5Ccdot%20%5Csqrt%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B5%7D%20%7D)
Step-by-step explanation:
![\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B250%7D%20%5Ccdot%20%5Csqrt%7B%5Csqrt%5B3%5D%7B10%7D%20%7D)
![\sqrt{\sqrt[3]{10} } \implies (10^\frac{1}{3} )^\frac{1}{2} =10^\frac{1}{6} =\sqrt[6]{10}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Csqrt%5B3%5D%7B10%7D%20%7D%20%5Cimplies%20%2810%5E%5Cfrac%7B1%7D%7B3%7D%20%29%5E%5Cfrac%7B1%7D%7B2%7D%20%3D10%5E%5Cfrac%7B1%7D%7B6%7D%20%3D%5Csqrt%5B6%5D%7B10%7D)
![\therefore \sqrt{\sqrt[3]{10} }=\sqrt[6]{10}](https://tex.z-dn.net/?f=%5Ctherefore%20%5Csqrt%7B%5Csqrt%5B3%5D%7B10%7D%20%7D%3D%5Csqrt%5B6%5D%7B10%7D)
![\text{Solving }\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }](https://tex.z-dn.net/?f=%5Ctext%7BSolving%20%7D%5Csqrt%5B3%5D%7B250%7D%20%5Ccdot%20%5Csqrt%7B%5Csqrt%5B3%5D%7B10%7D%20%7D)
![\sqrt[3]{250}=\sqrt[3]{2\cdot 5^3}=5 \sqrt[3]{2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B250%7D%3D%5Csqrt%5B3%5D%7B2%5Ccdot%205%5E3%7D%3D5%20%20%5Csqrt%5B3%5D%7B2%7D)
Once
![\sqrt[6]{2} \cdot \sqrt[6]{5} = \sqrt[6]{10}](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B5%7D%20%3D%20%5Csqrt%5B6%5D%7B10%7D)
We have
![5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5}](https://tex.z-dn.net/?f=5%20%20%5Csqrt%5B3%5D%7B2%7D%20%5Ccdot%20%5Csqrt%5B6%5D%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B5%7D)
We can proceed considering the common base of exponentials
![\sqrt[3]{2} \cdot \sqrt[6]{2} = 2^{\frac{1}{3}} \cdot 2^{\frac{1}{6} } = 2^{\frac{3}{6} } = 2^{\frac{1}{2} }=\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B2%7D%20%20%3D%20%202%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%20%5Ccdot%20%202%5E%7B%5Cfrac%7B1%7D%7B6%7D%20%7D%20%20%3D%202%5E%7B%5Cfrac%7B3%7D%7B6%7D%20%7D%20%3D%202%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D%3D%5Csqrt%7B2%7D)
Therefore,
![5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5} = 5 \cdot \sqrt{2} \cdot \sqrt[6]{5}](https://tex.z-dn.net/?f=5%20%20%5Csqrt%5B3%5D%7B2%7D%20%5Ccdot%20%5Csqrt%5B6%5D%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B5%7D%20%3D%205%20%5Ccdot%20%5Csqrt%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B5%7D)
