x2 = -11x - 10 x2 + 11x + 10 = 0 (x + 1)(x + 10) = 0
The LCM is the absolute value of x.
x2 + 11x + 24 You want the equation to be in the form (x + a)(x + b), where the two solutions are -a and -b.You know that...ab = 24...and...a + b = 11Note that 8 times 3 is 24, and 8 plus 3 is 11.x2 + 11x + 24 = (x + 3)(x + 8)The two answers are x = -3 and x = -8.
x2 -11x-12= (x-12)(x+1)
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 - 10 x2 + 11x + 10 = 0 (x + 1)(x + 10) = 0
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
It is: x^2 +11x +24 = (x+8)(x+3) when factored
x2 - 11x + 24 = (x - 3)(x - 8)
If you meant to write -x2 - 11x + 26, that factors to -(x - 2)(x + 13)
(x + 15)(x - 4)
To solve for x: x2 = 11x - 10 x2 - 11x = -10 x2 - 11x + 10 =0 (x - 1)(x - 10) = 0 x = {1, 10)
The LCM is the absolute value of x.
The factors of (x2 + 11x + 30) are (X + 5) and (x + 6)You want to look for two numbers, when multiplied equal 30, and when added equal 11. Start with factors of 30, and see what they add to: 5+6=11, and 5*6=30.
x2 = 11x - 10 ∴ x2 - 11x + 10 = 0 ∴ (x - 10)(x - 1) = 0 ∴ x ∈ {1, 10}
x2 + 11x + 24 You want the equation to be in the form (x + a)(x + b), where the two solutions are -a and -b.You know that...ab = 24...and...a + b = 11Note that 8 times 3 is 24, and 8 plus 3 is 11.x2 + 11x + 24 = (x + 3)(x + 8)The two answers are x = -3 and x = -8.
x^3 - 121x