2007-07-07 07:30:22 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Hi,
This is the difference of perfect squares. 4x² is really (2x) squared. 81 is 9². So both factors start with 2x and both end with 9. One binomial has a "-" and the other has a "+".
The factors are (2x - 9)(2x + 9).
I hope that helps!! :-)
2007-07-07 07:32:49 · answer #1 · answered by Pi R Squared 7 · 0⤊ 0⤋
The easiest way is to recognize that both 4x^2 and 81 are exact squares.
So, the original expression is the difference of 2 squares. Expressed as a^2-b^2= (a-b)*(a+b).
So for your original expression 4x^2-91=(2x-9)*(2x+9)
2007-07-07 07:41:23 · answer #2 · answered by dbear 2 · 0⤊ 0⤋
a squared - b squared = (a+b)(a-b)
This works as all of the terms are squares.
4x squared= 2x times 2x
81=9 times 9
4x^2-81=(2x+9)(2x-9)
2007-07-07 09:05:13 · answer #3 · answered by lulu 3 · 0⤊ 0⤋
write 4x^2-81 like : (2x)^2 -9^2
form: a^2-b^2
we know that:
a^2-b^2=(a+b)(a-b)
then,
(2x)^2-(9)^2 =(2x-9)(2x+9)
2007-07-07 08:01:43 · answer #4 · answered by Johnny 2 · 0⤊ 0⤋
(2x - 9).(2x + 9)
Check
2x.(2x + 9) - 9.(2x + 9)
4x² + 18x - 18x - 81
4x² - 81 as required.
2007-07-07 08:48:59 · answer #5 · answered by Como 7 · 0⤊ 0⤋
4x2 - 81 = (2x +9) * (2x - 9)
2007-07-11 06:00:57 · answer #6 · answered by Jun Agruda 7 · 2⤊ 0⤋
(2x-9)(2x+9) since a^2-b^2=(a-b)(a+b)
2007-07-07 07:36:48 · answer #7 · answered by chris 3 · 0⤊ 0⤋
(2x^2 + 9)(2x^2 - 9)
2007-07-07 07:37:48 · answer #8 · answered by Anonymous · 0⤊ 0⤋
(2x-9)(2x+9)
2007-07-10 23:21:28 · answer #9 · answered by ♫●GARV●♫ 6 · 0⤊ 0⤋