Simplify completely
(X^2+x-12/x^2-x-20)/(3x^2–24x+45/12 x^2-48-60)

(4 (x^4 - 20 x^2 - 12))/(3 x^2 (9 x^2 - 32 x - 144))

Step-by-step explanation:

Simplify the following:

(x^2 + x - x - 20 - 12/x^2)/((15 x^2)/4 + 3 x^2 - 24 x - 60 - 48)

Hint: | Put the fractions in x^2 + x - x - 20 - 12/x^2 over a common denominator.

Put each term in x^2 + x - x - 20 - 12/x^2 over the common denominator x^2: x^2 + x - x - 20 - 12/x^2 = x^4/x^2 + x^3/x^2 - x^3/x^2 - (20 x^2)/x^2 - 12/x^2:

(x^4/x^2 + x^3/x^2 - x^3/x^2 - (20 x^2)/x^2 - 12/x^2)/((45 x^2)/12 + 3 x^2 - 24 x - 60 - 48)

Hint: | Combine x^4/x^2 + x^3/x^2 - x^3/x^2 - (20 x^2)/x^2 - 12/x^2 into a single fraction.

x^4/x^2 + x^3/x^2 - x^3/x^2 - (20 x^2)/x^2 - 12/x^2 = (x^4 + x^3 - x^3 - 20 x^2 - 12)/x^2:

((x^4 + x^3 - x^3 - 20 x^2 - 12)/x^2)/((45 x^2)/12 + 3 x^2 - 24 x - 60 - 48)

Hint: | Group like terms in x^4 + x^3 - x^3 - 20 x^2 - 12.

Grouping like terms, x^4 + x^3 - x^3 - 20 x^2 - 12 = x^4 - 20 x^2 - 12 + (x^3 - x^3):

(x^4 - 20 x^2 - 12 + (x^3 - x^3))/(x^2 ((45 x^2)/12 + 3 x^2 - 24 x - 60 - 48))

Hint: | Look for the difference of two identical terms.

x^3 - x^3 = 0:

((x^4 - 20 x^2 - 12)/x^2)/((45 x^2)/12 + 3 x^2 - 24 x - 60 - 48)

Hint: | In (45 x^2)/12, the numbers 45 in the numerator and 12 in the denominator have gcd greater than one.

The gcd of 45 and 12 is 3, so (45 x^2)/12 = ((3×15) x^2)/(3×4) = 3/3×(15 x^2)/4 = (15 x^2)/4:

(x^4 - 20 x^2 - 12)/(x^2 (15 x^2/4 + 3 x^2 - 24 x - 60 - 48))

Hint: | Put the fractions in (15 x^2)/4 + 3 x^2 - 24 x - 60 - 48 over a common denominator.

Put each term in (15 x^2)/4 + 3 x^2 - 24 x - 60 - 48 over the common denominator 4: (15 x^2)/4 + 3 x^2 - 24 x - 60 - 48 = (15 x^2)/4 + (12 x^2)/4 - (96 x)/4 - 240/4 - 192/4:

(x^4 - 20 x^2 - 12)/(x^2 (15 x^2)/4 + (12 x^2)/4 - (96 x)/4 - 240/4 - 192/4)

Hint: | Combine (15 x^2)/4 + (12 x^2)/4 - (96 x)/4 - 240/4 - 192/4 into a single fraction.

(15 x^2)/4 + (12 x^2)/4 - (96 x)/4 - 240/4 - 192/4 = (15 x^2 + 12 x^2 - 96 x - 240 - 192)/4:

(x^4 - 20 x^2 - 12)/(x^2 (15 x^2 + 12 x^2 - 96 x - 240 - 192)/4)

Hint: | Group like terms in 15 x^2 + 12 x^2 - 96 x - 240 - 192.

Grouping like terms, 15 x^2 + 12 x^2 - 96 x - 240 - 192 = (12 x^2 + 15 x^2) - 96 x + (-192 - 240):

(x^4 - 20 x^2 - 12)/(x^2 ((12 x^2 + 15 x^2) - 96 x + (-192 - 240))/4)

Hint: | Add like terms in 12 x^2 + 15 x^2.

12 x^2 + 15 x^2 = 27 x^2:

(x^4 - 20 x^2 - 12)/(x^2 (27 x^2 - 96 x + (-192 - 240))/4)

Hint: | Evaluate -192 - 240.

-192 - 240 = -432:

(x^4 - 20 x^2 - 12)/(x^2 (27 x^2 - 96 x + -432)/4)

Hint: | Factor out the greatest common divisor of the coefficients of 27 x^2 - 96 x - 432.

Factor 3 out of 27 x^2 - 96 x - 432:

(x^4 - 20 x^2 - 12)/(x^2 (3 (9 x^2 - 32 x - 144))/4)

Hint: | Write ((x^4 - 20 x^2 - 12)/x^2)/((3 (9 x^2 - 32 x - 144))/4) as a single fraction.

Multiply the numerator by the reciprocal of the denominator, ((x^4 - 20 x^2 - 12)/x^2)/((3 (9 x^2 - 32 x - 144))/4) = (x^4 - 20 x^2 - 12)/x^2×4/(3 (9 x^2 - 32 x - 144)):

(4 (x^4 - 20 x^2 - 12))/(3 x^2 (9 x^2 - 32 x - 144))

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

the answer is c.

step-by-step explanation: