x^3 - 121x
X2+11x+11 = 7x+9 X2+11x-7x+11-9 = 0 x2+4x+2 = 0 Solve as a quadratic equation by using the quadratic equation formula or by completing the square: x = -2 + or - the square root of 2
First, it is important to regroup, so I am going to rearrange this equation: (x4 - 7x2 - 18 - 3x3 + 27x I can now factor the first three terms and the last two terms: (x4 - 7x2 - 18) becomes (x2 - 9)(x2 + 2) -3x3 + 27x becomes -3x(x2 - 9); so the new equation looks like: (x2 - 9)(x2 + 2) - 3x(x2 - 9) From here, factor out what is common, in this case- x2- 9. Therefore, you will have (x2 - 9)(x2 + 2 - 3x), which can be rearranged to (x2 -9)(x2 - 3x + 2). Further factoring reveals (x + 3)(x - 3)(x - 1)(x - 2) as the final answer.
(x + 4) / (x3 - 11x + 20) = (x + 4) / (x2 + 4x - 5)(x + 4) = 1 / (x2 + 4x - 5) = 1 / (x + 5)(x - 1), where x ≠ -4
x+11x+30(x+5)(x+6)
This is the equation we want to factorize: x2 - 11x + 18 And there are two answers (the numbers are the same, but +/- signs are inverted): 1. (x - 2)(x - 9) = x2 - 9x - 2x -(-18) = x2 - 11x + 18 2. (-x + 2)(-x +9) = x2 -9x -2x + 18 = x2 - 11x + 18
x2 + 11x + 18 (x + 9)(x + 2) CHECK: x2 + 9x + 2x + 18 x2 + 11x + 18 SET EACH EQUAL TO ZERO: x + 9 = 0 x = -9 x + 2 = 0 x = -2 NOW YOU ARE DONE: Solution set: {-9, -2}
x2 -11x-12= (x-12)(x+1)
Assuming the question is written as: x2+11x-12 This would factor to: (x+12)(x-1)
21
x(6x - 11)
It doesn't factor evenly if it were x**2-11x+10: x**2-11x+10 = (x-1)(x-10)
Not possible to be specific since you have not provided the sign of the term in x, but there are two possible factorisations; (x + 3)(x + 8) = x2 + 11x + 24 and (x - 3)(x - 8) = x2 - 11x + 24
x2 - 11x + 24 = (x - 3)(x - 8)
To solve for x: x2 = 11x - 10 x2 - 11x = -10 x2 - 11x + 10 =0 (x - 1)(x - 10) = 0 x = {1, 10)
Can you improve the question? It's hard to tell what you mean.
(x + 9)(x + 2)