When factored it is: (x-9)(x+4)
. x + 4
sorry i am not sure but i dont think it has an answer If this equation is: x^2 - 36 = 5x, then: x^2 - 36 = 5x subtract 5x to both sides; x^2 - 5x - 36 = 0 Factor; (x + 4)(x - 9) = 0 x + 4 = 0 or x - 9 = 0 x = -4 or x = 9
(x - 2)(x - 3)
x2-5x+4 = (x-1)(x-4) when factord
It is (x - 9)*(x + 4)
When factored it is: (x-9)(x+4)
x2-5x-36 = (x-9(x+4) when factored
(x-6)(x+1)
(x - 7)(x + 2)
the question is to solve (x^2-5x+5)^(x^2-36)=1
(x + 4)(x + 1)
It is: (x2 + 5x) (x + 8)
There is a formula for the "difference of squares." In this case, the answer is (5x - 6)(5x + 6)
I assume x2 + 5x - 36 is the polynomial in the question. First, look for two factors of 36 that have a difference of 5, which would be 9 and 4. It would factor into (x + 9)(x - 4). To double check, multiplying them together results in x2 - 4x + 9x - 36 = x2 + 5x - 36. If the polynomial is 5x2 +7x +2
. x + 4
1st = x2nd = 5 - xx(5 - x) = -365x - x2 = -365x - x2 + 36 = -36 + 365x - x2 + 36 = 0-1(- x2 + 5x + 36 = 0)x2 - 5x - 36 = 0(x + 4)(x - 9) = 0(x + 4) = 0 or (x - 9) = 0x = -4 or x = 9Thus, the integers are -4 and 9