This does not factor over the field of rational numbers. If does factor over the reals. Use the quadratic formula to find the roots for the related quadratic equation: x2-11x-10 = 0 Cal them r1 and r2. The factors are (x - r1) and (x - r2.)
x2 - 11x + 24 = (x - 3)(x - 8)
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
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 + 26 = -(x2 + 11x - 26) = -(x2 - 2x + 13x - 26) = -[ x(x - 2) + 13(x - 2) ] = -(x + 13)(x - 2) = (x + 13)(2 - x)
21
It doesn't factor evenly if it were x**2-11x+10: x**2-11x+10 = (x-1)(x-10)
To solve for x: x2 = 11x - 10 x2 - 11x = -10 x2 - 11x + 10 =0 (x - 1)(x - 10) = 0 x = {1, 10)
x2 = -11x - 10 x2 + 11x + 10 = 0 (x + 1)(x + 10) = 0
x2 = 11x - 10 ∴ x2 - 11x + 10 = 0 ∴ (x - 10)(x - 1) = 0 ∴ x ∈ {1, 10}
X2 -11x +10=(x - 1 )(x - 10)
x2 -11x-12= (x-12)(x+1)
Assuming the question is written as: x2+11x-12 This would factor to: (x+12)(x-1)
I assume you mean...,X2 + 11X + 10(X + 1)(X + 10)==============factored
This does not factor over the field of rational numbers. If does factor over the reals. Use the quadratic formula to find the roots for the related quadratic equation: x2-11x-10 = 0 Cal them r1 and r2. The factors are (x - r1) and (x - r2.)
x(6x - 11)
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