x^(3) - 12x^(2) + 35 x
Factor out 'x'
Hence
x(x^(2) - 12x + 35)
Factor the quadratic
x(x - 5)(x - 7) Done!!!!!
Factor out an X X(X^2 - 12X + 35) factor quadratic equation X(X - 7)(X - 5) ------------------- fully factored
x3 + 2x2 - 35x = x(x + 7)(x - 5)
35x + 115 = 360 35x = 360-115 35x = 245 x = 245/35 x = 7
42x+35x-35x=42x it's leave 85 so it's change into a negative 42x-85 is the answer
25x + 35x + 9 is 60x + 9, which factors to 3(20x + 3)
x3 - 12x2 + 35x = x(x2 - 12x + 35) = x(x - 5)(x - 7)
x3 - 12x2 + 35x = x (x2 - 12x + 35) = x (x - 7) (x - 5)
Factor out an X X(X^2 - 12X + 35) factor quadratic equation X(X - 7)(X - 5) ------------------- fully factored
x(x + 5)(x - 7)
it is below 555 and above 200 * * * * * x3 - 2x2 + 35x = x*(x2 - 2x - 35) = x*(x + 5)*(x - 7)
x3 + 2x2 - 35x = x(x + 7)(x - 5)
35x + 115 = 360 35x = 360-115 35x = 245 x = 245/35 x = 7
42x+35x-35x=42x it's leave 85 so it's change into a negative 42x-85 is the answer
25x + 35x + 9 is 60x + 9, which factors to 3(20x + 3)
5(x - 5)(x - 2)
35x + 10
Oh, what a happy little math problem we have here! To factor 35x + 63y, we can first look for the greatest common factor of the coefficients, which is 7. Then we can factor out the 7 to get 7(5x + 9y). And just like that, we've created a beautiful and simplified expression!