Answer: Choice D
b greater-than 3 and StartFraction 2 over 15 EndFraction
In other words,
b > 3 & 2/15
or

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Explanation:
Let's convert the mixed number 2 & 3/5 into an improper fraction.
We'll use the rule
a & b/c = (a*c + b)/c
In this case, a = 2, b = 3, c = 5
So,
a & b/c = (a*c + b)/c
2 & 3/5 = (2*5 + 3)/5
2 & 3/5 = (10 + 3)/5
2 & 3/5 = 13/5
The inequality
is the same as 
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Let's multiply both sides by 15 to clear out the fractions

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Now isolate the variable b

Side note: Another way to go from 47/15 to 3 & 2/15 is to notice how
47/15 = 3 remainder 2
The 3 is the whole part while 2 helps form the fractional part. The denominator stays at 15 the whole time.
![\bf \begin{cases} x=1\implies &x-1=0\\ x=1\implies &x-1=0\\ x=-\frac{1}{2}\implies 2x=-1\implies &2x+1=0\\ x=2+i\implies &x-2-i=0\\ x=2-i\implies &x-2+i=0 \end{cases} \\\\\\ (x-1)(x-1)(2x+1)(x-2-i)(x-2+i)=\stackrel{original~polynomial}{0} \\\\\\ (x-1)^2(2x+1)~\stackrel{\textit{difference of squares}}{[(x-2)-(i)][(x-2)+(i)]}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Ax%3D1%5Cimplies%20%26x-1%3D0%5C%5C%0Ax%3D1%5Cimplies%20%26x-1%3D0%5C%5C%0Ax%3D-%5Cfrac%7B1%7D%7B2%7D%5Cimplies%202x%3D-1%5Cimplies%20%262x%2B1%3D0%5C%5C%0Ax%3D2%2Bi%5Cimplies%20%26x-2-i%3D0%5C%5C%0Ax%3D2-i%5Cimplies%20%26x-2%2Bi%3D0%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A%28x-1%29%28x-1%29%282x%2B1%29%28x-2-i%29%28x-2%2Bi%29%3D%5Cstackrel%7Boriginal~polynomial%7D%7B0%7D%0A%5C%5C%5C%5C%5C%5C%0A%28x-1%29%5E2%282x%2B1%29~%5Cstackrel%7B%5Ctextit%7Bdifference%20of%20squares%7D%7D%7B%5B%28x-2%29-%28i%29%5D%5B%28x-2%29%2B%28i%29%5D%7D)
![\bf (x^2-2x+1)(2x+1)~[(x-2)^2-(i)^2] \\\\\\ (x^2-2x+1)(2x+1)~[(x^2-4x+4)-(-1)] \\\\\\ (x^2-2x+1)(2x+1)~[(x^2-4x+4)+1] \\\\\\ (x^2-2x+1)(2x+1)~[x^2-4x+5] \\\\\\ (x^2-2x+1)(2x+1)(x^2-4x+5)](https://tex.z-dn.net/?f=%5Cbf%20%28x%5E2-2x%2B1%29%282x%2B1%29~%5B%28x-2%29%5E2-%28i%29%5E2%5D%0A%5C%5C%5C%5C%5C%5C%0A%28x%5E2-2x%2B1%29%282x%2B1%29~%5B%28x%5E2-4x%2B4%29-%28-1%29%5D%0A%5C%5C%5C%5C%5C%5C%0A%28x%5E2-2x%2B1%29%282x%2B1%29~%5B%28x%5E2-4x%2B4%29%2B1%5D%0A%5C%5C%5C%5C%5C%5C%0A%28x%5E2-2x%2B1%29%282x%2B1%29~%5Bx%5E2-4x%2B5%5D%0A%5C%5C%5C%5C%5C%5C%0A%28x%5E2-2x%2B1%29%282x%2B1%29%28x%5E2-4x%2B5%29)
of course, you can always use (x-1)(x-1)(2x+1)(x²-4x+5) as well.
The turning point is (3,3). The x to make the absolute value negative would be 2. The ordered pair would then be (0,3)
4³ ⋅ (100 ÷25) − 5^0
[parenthesis goes first, then exponents]
[any coefficient with zero exponent is equal to 1]
64 · 4 - 1
[multiplication, then subtraction]
256 - 1
<u>255</u>