x2+9x+18 find factors of 18 that add up to 9, in this case 6&3 (x+6)(x+3)
x2-9x = x(x-9)
Divide both sides of the equation by x: x2 = 9x x2 / x = 9x / x x = 9
x2 - y2 + 9x - 9y =(x2 + 9x) - (y2 + 9y) =x(x + 9) - y(y + 9)================================Another way to go after it:x2 - y2 + 9x - 9y =(x2 - y2) + 9x - 9y =(x + y) (x - y) + 9 (x - y) =(x + y + 9) (x - y)
When factoring it is fairly easy to check your answer. Say we want to factor x2+9x+14=0 (x+7)(x+2)=0 now if we want to check then we can just multiply these together and we should get the original polynomial... x2+2x+7x+14=0 x2+9x+14=0
x2 - 9x + 18 = (x - 3)(x - 6)
x^2-9x+18=(x-6)(x-3)
x2+9x+18 find factors of 18 that add up to 9, in this case 6&3 (x+6)(x+3)
x2-9x = x(x-9)
x2-9x+18 = (x-3)(x-6) when factorised
x2=9x-20 x2-9x+20=0 Factor: (x-4)(x-5)=0 x={4,5}
(x - 6)(x - 3)
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 - 9x = -8 ∴ x2 - 9x + 8 = 0 ∴ (x - 1)(x - 8) = 0 You can then solve it as well: ∴ x ∈ {1, 8}
I hope you mean x3+2x2-9x+18. Treat the equation like two equations: x3+2x2 and 9x+18. Factor out x2 from the first equation, so x2(x+2). Factor out 9 from the second equation, so 9(x+2). Now, the whole equation. x2(x+2)-9(x+2). Factor out (x+2), so (x+2)(x2-9). (x2-9) factors into (x+3)(x-3). So, (x+2)(x+3)(x-3) is the complete factorization. Hope this helps.
Step 1: Divide by 9x: 9x (x2 + 3x - 10) = 9x(x + 5)(x - 2)
(x - 9)(x2 + 9x + 81)