How do I completely factor 64x^2-49??? Is it prime?
2007-04-10 05:36:55 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
64x^2 - 49 = (8x + 7)(8x - 7)
There's a trick to this. Any expression of the form a^2 - b^2 can always be factored as (a + b)(a - b). In this case, 64x^2 - 49 = (8x)^2 - 7^2, so we have a = 8x and b = 7, giving us a + b = 8x + 7 and a - b = 8x - 7.
2007-04-10 05:39:03 · answer #1 · answered by DavidK93 7 · 1⤊ 0⤋
the version of squares, a² - b², is factored as (a+b)(a-b). sixty 4 is a sq. (=8²); x² is a sq., of course; forty 9 is a sq. (=7²). as a result 64x² - forty 9 = (8x)² - 7² = (8x+7)(8x-7).
2016-10-21 13:01:50 · answer #2 · answered by ? 4 · 0⤊ 0⤋
This is an example of a 'difference of two squares'.
(a+b)(a-b) = a^2 + b^2
64x^2 + 49 is thus equal to (8x + 7)(bx - 7)
2007-04-10 05:46:01 · answer #3 · answered by Arch-RF d 1 · 0⤊ 0⤋
a difference of squares.
(8x+7) (8x-7)
learn to recognize this type of pattern, where there is no x term and the x2 and number term are perfect squares.
ex. x2-4 = (x+2) (x-2)
x2+9 = (x+3) (x-3)
2007-04-10 05:41:14 · answer #4 · answered by Anonymous · 0⤊ 0⤋
64x^2-49
=(8x)^2-7^2
=(8x-7)(8x+7)
2007-04-10 20:05:07 · answer #5 · answered by Toves 1 · 0⤊ 0⤋
(8x-7)(8x+7)
Mulyiply it out to get what you started with for your check that the answer is correct.
2007-04-10 05:56:03 · answer #6 · answered by TBU 2 · 0⤊ 1⤋
no
(8x)^2-7^2
=(8x-7)(8x+7)
2007-04-10 05:41:14 · answer #7 · answered by iyiogrenci 6 · 0⤊ 0⤋