Unfortuanately, you can't factor this polynomial. To do so, you must find a pair of numbers that add up to the coefficient of the middle term, and multiply to equal the product of the coefficients of the first and last terms. In other words, we want a pair of numbers that add up to make -11, and multiply to make 21. Unfortunately there are no natural numbers that meet that condition.
x2 - 11x + 24 = (x - 3)(x - 8)
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.
It is: x^2 +11x +24 = (x+8)(x+3) when factored
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 + 30 = 0 (x + 5)(x + 6) = 0 so the roots are -5 and -6
Assuming the question is written as: x2+11x-12 This would factor to: (x+12)(x-1)
21
x2 - 11x + 24 = (x - 3)(x - 8)
x2 -11x-12= (x-12)(x+1)
x^3 - 121x
X2 -11x +10=(x - 1 )(x - 10)
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}
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.
x^2 + 11x + 6 has no rational zeros.
x(6x - 11)
x2+11x+11 = 7x+9 x2+11x-7x+11-9 = 0 x2+4x+2 = 0 The above quadratic equation can be solved by using the quadratic equation formula and it will have two solutions.
It doesn't factor evenly if it were x**2-11x+10: x**2-11x+10 = (x-1)(x-10)