2x+5x-24 7x2-24
x^2 + 5x - 24 = (x - 3)(x + 8)
(x-3)(x+8)
To solve for x: x2 = 5x + 2 x2 - 5x = 2 x2 - 5x + (5/2)2 = 2 + (5/2)2 (x - 5/2)2 = 8/4 + 25/4 x - 5/2 = ± √(33/4) x = (5 ± √33) / 2
x2+5x+6 = (x+2)(x+3) when factored because 2 and 3 are factors of 6
2x+5x-24 7x2-24
x^2 + 5x - 24 = (x - 3)(x + 8)
(x-2)(x-3)
(x-3)(x+8)
To solve for x: x2 = 5x + 2 x2 - 5x = 2 x2 - 5x + (5/2)2 = 2 + (5/2)2 (x - 5/2)2 = 8/4 + 25/4 x - 5/2 = ± √(33/4) x = (5 ± √33) / 2
x2+5x+6 = (x+2)(x+3) when factored because 2 and 3 are factors of 6
the question is to solve (x^2-5x+5)^(x^2-36)=1
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
It is already in descending order and it factors to (x+2)(x-5)
Factor the polynomial x2 - 10x + 25. Enter each factor as a polynomial in descending order.
x2 + 5x - 120 can not be factored.
x2+10x+21 = (x+3)(x+7) when factored.....