Factor the following please help 32a^4-1250 b^4?

Answer 1

http://www.sparknotes.com/math/algebra2/factoring/section3.rhtml


From the above site,

1) Find if the coefficient of a^4 and b^4 is a perfect square.

32 is not a perfect square
1250 is not a perfect square
2) If not, try factoring them by finding its LCM so that it will result in a square number

32/2 = 16 is a perfect square because ( 4 x 4 or 4^2 = 16)
1250/2 = 625 is a perfect square because ( 25 x 25 or 25^2 = 625)

32a^4 - 1250b^4 =

2(16a^4 - 625b^4) ==> the terms in the parenthesis is a difference of two squares

2( 4a^2 + 25b^2) (4a^2 - 25b^2) =
The last term can still be factored because it is a difference of two squares

2(4a^2 + 25b^2) ( 2a + 5b) (2a - 5b) ===> the answer

Read the site because I cannot explain very well...

Answer 2

= 2(16a^4-625b^4) = 2(4a^2-25b^2)(4a^2+25b^2)
= 2(2a-5b)(2a+5b)(4a^2+25b^2)

Answer 3

= 2.(16a^(4) - 625b^(4))
= 2.(4a² - 25 b²).(4a² + 25b²)
= 2.(2a - 5b).(2a + 5b).(4a² + 25b²)

Answer 4

32a^4=1250b^4

2(16a^4-625b^4)
2(4a^2-25b^2)(4a^2+25b^2)
2(2a-5b)(2a+5b)(4a^2+25b^2)

Answer 5

32a^4-1250b^4
=2(16a^4-625b^4)
=2{(4a^2)^2-(25b^2)^2}
=2(4a^2+25b^2)(4a^2-25b^2)
=2(4a^2+25b^2){(2a)^2-(5b)^2}
=2(4a^2+25b^2)(2a+5b)(2a-5b)