how do you do this problem 20y2- 5x2?

Question

the 2s are supposed to be raised

Answer 1

We have this:

20y^2 - 5x^2

To solve, we factor to simplify.

Factor out 5. Why? We factor out 5 because it is the BIGGEST number that evenly (without a remainder) can be divided by itself and 20. See it? I will divide 20 by 5 and then 5 by 5.

20y^2 - 5x^2 becomes 5(4y^2 - x^2).

Do you see the terms inside the parentheses?

We need to break this down a bit more.

The inside terms must be factored as well.

4y^2 - x^2 becomes (2y - x) (2y + x)

Final answer: 5 (2y-x) (2y+x)

Notice that our final answer is 3 factors. See it?

Guido

Answer 2

20y^2- 5x^2=5(4y^2-x^2) you didn't say, but I'm guessing that you want this to equal zero. If so
5(4y^2-x^2)=0
4y^2-x^2=0
4y^2=x^2
x=+/-2y

Answer 3

20y^2 - 5x^2 = 5(4y^2 - x^2) = 5(2y - x)(2y + x)

Answer 4

15y2x2 the twos are raised up like u said

Answer 5

well... you didn't set it equal to anything, but you can factor it as follows:
5(4y^2-x^2)
5(2y+x)(2y-X)

Answer 6

if u working ; on organic numbers or reals (the place u have finished variety gadget available) then that's 10; yet while u are engaged on limited contraptions like Z10; Z9... then that's diverse. So please specify the place are u working (wherein set)

Answer 7

5 (4y^2 - x^2)

= 5(2y + x)(2y - x)

Answer 8

assume that x=2

20y^2-5(2)^2=0
20y^2=5(2)^2
20y^2=20
y^2=20/20
y^2=1
y=1

20(1)^2=5(x)^2
20=5(x)^2
20/5=x^2
4=x^2
x=2


checking:
20y^2=5(x)^2
20(1)^2=5(2)^2
20(1)=5(4)
20=20

Answer 9

40y-10