CAN YOU HELP ME ONE THESE 2 QUETIONS B4 I GO 2 SCHOOL?

Question

Please show your calculations and explain your solution in detail.

Factor the following polynomial: 25x4 - 49.

How would you simplify the following expression?

y4

------

y11

Answer 1

use difference of 2 square

so 5x^2 - 7 is ur answer

to do that u must find the square of each of the 2 terms

Answer 2

With experience, these answers will "jump off the page at you".

because 25 and 49 are square numbers, look at it this way:

(5x^2) ^2 - 7^2

To make this easier to recognize, we'll call 5x^2 "y":

y^2 - 7^2

The difference between 2 squares always factor to the sum and difference of their roots.

(y+7) (y-7) = Y*Y -7*7 + (7y -7y cancel each other out)

Then replace y...    (5x^2 + 7) (5x^2 - 7)

the rest is easy!

Multiplying powers = adding exponents.

Dividing powers = subtracting exponents

1 / y^11 = y^(-11), so y^4 * y^(-11) = y^(4-11) = y^(-7) or 1 / y^(7)

Answer 3

25x^4 - 49 = 25x^4 - 7^2  

= (5^2) [(x^2)^2)] - 7^2

=[5x^2]^2 - 7^2

this is in the form a^2 - b^2

so =(a-b) (a+b)  

a= 5x^2 , b =7

= (5x^2 - 7) (5x^2+7)  

2)  y^4/y^11 =  (y)^ (4-11) = y^ (-7) or  1/(y^7)   answer

good luck

Answer 4

y^4 / y^11

= y^(4 - 11)

= y^-7

or 1

   ----

   y^7

25^4 - 49

= (5x^2 + 7) ( 5x^2 - 7)

Thank you.

Answer 5

y4/y11= 1/y7

25x^4 - 49= (5x²-7)(5x²+7)

Answer 6

25(x^4) - 49 = [5(x^2) - 7][5(x^2) + 7]

y^4      

------ = 1/(y^7)

y^11