![\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.
Answer:
F(-1) = 1
x = 3
Step-by-step explanation:
1) Assume that f(-1) = y1
So that point A (x1 = -1; y1) is a point lying in the graph representing the equation y = f(x)
As it can be seen in the figure, the graph representing that equation crosses point (-1; 1)
=> Point A has y1 = 1
=> f (-1) =1
2) Assume that f(x2) = -2
So that point B (x2; y2 = -2) is a point on the graphy representing the equation y = f(x)
As indicated in the figure, the graph crosses point (3; -2)
=> Point B has x2 = 3
=> When f(x) = -2, x = 3
The coordinate of the point is (6,-2)
<h3>How to determine the coordinate of the point?</h3>
The given parameters are:
A = (1,8)
B = (7,-4)
The location of the point (i.e 5/6) means that the ratio is:
m :n = 5 : (6 - 5)
m : n = 5 : 1
The coordinate of the point is then calculated as:

So, we have:

Evaluate

Evaluate the product
(x,y) = (6,-2)
Hence, the coordinate of the point is (6,-2)
Read more abut line segment ratio at:
brainly.com/question/12959377
#SPJ1
The most likely number of cats = 21/37 x 8 = 4.54 ≈ 5 cats
The most likely number of dogs = 16/37 x 8 = 3.46 ≈ 3 dogs
The probability of that arrangement happening = 1/8^2 = 1/64 = 0.00156