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Tpy6a [65]
2 years ago
8

The vertices of triangle WXY have coordinates W(3, 2), X(5, 5), and Y(7, 3). Triangle WXY was transformed to produce triangle Wꞌ

XꞌYꞌ, with coordinates Wꞌ(−2, −1) and Yꞌ(2, 0).
What are the coordinates of vertex Xꞌ?

Mathematics
1 answer:
dezoksy [38]2 years ago
8 0
We will do the analysis for each of the 3 vertices:
1st vertex:
W': We observe W'(x', y') and W(x, y)
x' = -2
x = 3
so, x plus some value will give the value of x'. We form the equation the following way:
x + Cx = x' (Cx will be the value used to MOVE the x to x' value)
When trying to solve the equation, we put values inside the equation:
3 + Cx = -2
The x coordinate is MOVED by some value Cx to become the x' coordinate (That is how we transform the triangle; by transforming each of its vertices by the same values for x and y coordinates, both respectfully).
Solving the equation, we get:
Cx = -2 - 3
Cx = -5.
The conclusion is: x coordinate is MOVED to the LEFT (because of the minus) by 5 units on axis.
Lets see for which amount y coordinate for this vertex is MOVED:
y'=-1
y=2
We obtain the equation again:
Add to y coordinate value some constant (Cy), to obtain the value of new (transformed coordinate) y'.
y+Cy=y'
2+Cy=-1
Cy=-1-2
Cy=-3
The conclusion is that y coordinate for this vertex is MOVED down on Y axis by 3 units on axis.
----------------------------------------------------------------------------------------------
We now observe in same manner the second vertex Y', and we expect to obtain the same constants as we obtained for the previous vertex, because that is the rule for the transformation being done, each vertex should MOVE by same amount of units on X and Y axis respectfully.
Hence, we now observe Y'(x', y') and Y(x, y):
x'=2
x=7
x+Cx=x' (some constant value added to x will produce the new value of x')
7+Cx=2
Cx=2-7
Cx=-5
So, the x coordinate value is MOVED to the left (meaning of minus) by 5 units, and the value of x' coordinate is obtained (transformation).
Next, we have for Y axis coordinates:
y'=0
y=3
y+Cy=y'
3+Cy=0
Cy=0-3
Cy=-3
So the conclusion is that y coordinate value is MOVED down by 3 units on Y axis, and in such way we obtained y' via transformation.
-----------------------------------------------------------------------------------------------
We do the same approach for the third vertex X':
X'(x',y') and X(x,y)
We have to find coordinates for vertex X'(x',y')
We have coordinates to use, for X(x,y)=X(5,5).
x'=?
y'=?
x=5
y=5
-------
1) x+Cx=x'
5+(-5)=x' (because for previous two vertices we obtained that the constant Cx is for both vertices -5, Cx=-5)
hence, x'=5-5
x'=0
2) y+Cy=y'
5+(-3)=y' (because for previous two vertices we obtained that the constant Cy is for both vertices -3, Cy=-3)
y'=5-3
y'=2
Hence, the third TRANSFORMED vertex X' is with coordinates X'(0,2).
I attached a .png image here, so you could see the transformed triangle in a coordinate system.

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x^3-2x^2+x-1 is one of the prime factors of the polynomial

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The question implies that we determine one of the prime factors of the polynomial.

The polynomial is given as:

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Expand the polynomial by adding 0's in the form of +a - a

x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^8 -2x^7 + 2x^7 - 4x^6 +x^6 + 2x^5 -2x^5- 3x^4 + 4x^4 + 2x^3 -6x^3+2x^3- x^2  -3x^2 +4x^2-2x+2x-1

Rearrange the terms

x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^8 -2x^7 + 2x^5 - 3x^4 + 2x^3 - x^2 + 2x^7 - 4x^6 + 4x^4 -6x^3+4x^2-2x+x^6-2x^5+2x^3-3x^2+2x-1

Factorize the expression

x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^2(x^6-2x^5+2x^3-3x^2+2x-1) + 2x(x^6-2x^5+2x^3-3x^2+2x-1) + 1(x^6-2x^5+2x^3-3x^2+2x-1)

Factor out x^6-2x^5+2x^3-3x^2+2x-1

x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x^2+2x + 1)(x^6-2x^5+2x^3-3x^2+2x-1)

Express x^2 + 2x + 1 as a perfect square

x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6-2x^5+2x^3-3x^2+2x-1)

Expand the polynomial by adding 0's in the form of +a - a

x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6- 2x^5+x^4-x^4-x^3 +x^3-2x^3-x^2 -2x^2 +x+x - 1)

Rearrange the terms

x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6- 2x^5+x^4-x^3-x^4-2x^3-x^2+x+x^3-2x^2 +x - 1)

Factorize the expression

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Factor out x^3-2x^2+x-1

x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^3 -x+1)(x^3-2x^2+x-1)

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This is the same as the option (c)

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