4x^(2) - 81
First remember the two numbers are both squared numbers.
4 = 2^(2) & 81 = 9^(2)
Hence
(2x)^(2) - 9^(2)
Next remember two squared terms with a negative(-) between them , will factor.
Hence
(2x - 9)(2x + 9)
Note the two different signs.
NB Two squared terms with a positive (+) between them does NOT factor. .
x2-81 = (x-9)(x+9)
well, the square root of 81 is 9, and seven squared is 49, so it wud be 40
81 x 81 = 6561
-9 squared = 81(-9)^2 = (-9) * (-9) = 81
812 = 81 x 81 = 6,561
(2x - 9)(2x - 9) or (2x - 9)2
There is a formula for factoring the "difference of squares." In this case, the answer is (9 - 4x)(9 + 4x)
(9 - 5u)(9 + 5u)
x2-81 = (x-9)(x+9)
I believe it is prime as you have written it. change the end to + 81 and it factors as (2x - 1)(x-9)(x+9)
9x2 - 6xy + y2 - 81 = (3x - y)2 - 81 = (3x - y - 9)*(3x - y + 9)
85
If that's 16x2 + 72x + 81, that factors to (4x + 9)(4x + 9) or (4x + 9)2
Type your answer here... 81
81 x 2 - 16 = 146
9x2 - 6xy + y2 - 81 = (3x - y)2 - 92 = (3x - y - 9)(3x - y + 9)
78 minus78