vaule of x 3x2 minus 30x minus 72 = 0
x^3 + x^2 - 30 x - 72 = 0 maybe, perhaps, possibly?
x^3 + x^2 - 30 x = 72
try x = -2
-8 + 4 + 60 = 56 not 72 so
try x = -1
-1 + 1 + 30 = 30
try x = 0
0 +0 + 0 = 0 , nope
try x = 2
8 + 4 - 60 = -48
headed wrong try x = -3
-27 + 9 + 90 = 72 !!! CARAMBA !!! x = -3 works
so we have one of the three roots
divide
(x^3 + x^2 - 30 x - 72) by (x+3)
wow now we have the quadratic to solve for the other two roots
x^2 -2x -24 = 0
(x-6)(x+4) = 0
x = 6 or x = 4 or x = -3
done
I am sure you know a better way, but that was fun :)
x^3 + x^2 - 30 x - 72 = 0
x^2(x+3) - 2x^2 - 30x - 72 = 0
x^2(x+3) - 2x(x+3) - 24x - 72 = 0
x^2(x+3) - 2x(x+3) - 24(x+3) = 0
(x^2-2x-24)(x+3) = 0
(x-6)(x+4)(x+3) = 0
x = 6 or x = -4 or x = -3
To find the value(s) of x in the equation 3x^2 - 30x - 72 = 0, we can use the quadratic formula. The quadratic formula is used to solve equations of the form ax^2 + bx + c = 0, where a, b, and c are constants.
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, we have a = 3, b = -30, and c = -72. Plugging these values into the quadratic formula, we get:
x = (-(-30) ± √((-30)^2 - 4(3)(-72))) / (2(3))
Simplifying further:
x = (30 ± √(900 + 864)) / 6
x = (30 ± √1764) / 6
x = (30 ± 42) / 6
We have two possible solutions:
1) When x = (30 + 42) / 6 = 72 / 6 = 12
2) When x = (30 - 42) / 6 = -12 / 6 = -2
Therefore, the values of x in the equation 3x^2 - 30x - 72 = 0 are 12 and -2.