Please solve the problem in the image above thank you!

5a^4 b^2 + 32a^4 + 9b^2

Step-by-step explanation:

Step  1  :

Equation at the end of step  1  :

 ((32•(a^4))+(9•(b^2)))+(5a^4•b^2)

Step  2  :

Equation at the end of step  2  :

 ((32 • (a^4)) +  32b^2) +  5a^4 b^2

Step  3  :

Equation at the end of step  3  :

 (2^5a^4 +  3^2 b^2) +  5a^4 b^2

Step  4  :

Trying to factor a multi variable polynomial :

4.1    Factoring    5a^4 b^2 + 32a^4 + 9b^2  

the answer is c. 23/3

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'.