625a^4 - 16b^4
2007-03-30 05:19:53 · 6 answers · asked by chrismath07 1 in Science & Mathematics ➔ Mathematics
difference of 2 squares
= (25a^2 - 4b^2) (25a^2 + 4b^2)
2007-03-30 05:22:34 · answer #1 · answered by Orinoco 7 · 3⤊ 1⤋
Check if there is a comon factor. As you can see the terms have no common factor.
To make it easier let a^2=t and b^4=p
our polynomial becomes:
625t^2-16p^2
Please notice that it's difference of 2 squares: (a-b)(a+b)=a^2-b^2
Our polynomial:
(25t-4p)(25t+4p)
Let substitute back (t and p):
(25a^2-4b^2)(25a^2+4b^2)
The first bracket is again the difference of 2 squares. There is nothing you can do with the first brackets:
(5a-2b)(5a+2b)(25a^a+4b^2) which is a complete factorization.
2007-03-30 12:35:59 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Difference of two squares
625a⁴ - 16b⁴
(25a² 4b²)(25a² + 4b²)
- - - - - - - - -s-
2007-03-30 12:37:47 · answer #3 · answered by SAMUEL D 7 · 0⤊ 1⤋
a^2 - b^2 = (a-b ) (a+b)
( 25a^2 ) ^2 - (4b^2 ) ^2
=(25a^2 - 4b^2) ( 25a^2 + 4b^2)
(25a^2 - 4b^2)
= ( 5a - 2b ) (5a + 2b)
hence , ans = ( 5a - 2b ) (5a + 2b) ( 25a^2 + 4b^2)
2007-03-30 12:31:18 · answer #4 · answered by priyadarshini u 2 · 0⤊ 0⤋
(5a+2b)(5a-2b)(25a^2+4b^2)
2007-03-30 12:30:35 · answer #5 · answered by Anonymous · 0⤊ 1⤋
(25a^2-4b^2)(25a^2+4b^2)
=(5a-2b)(5a+2b(25a^2+4b^2)
2007-03-30 12:27:39 · answer #6 · answered by ironduke8159 7 · 3⤊ 1⤋