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Identical expressions
- *x^ three - twelve *x^ two + thirty-six *x- thirty-two = zero
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Similar expressions
- *x^3+12*x^2+36*x-32=0
- *x^3-12*x^2-36*x-32=0
- *x^3-12*x^2+36*x+32=0
*x^3-12*x^2+36*x-32=0 equation
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The solution
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3 2 x - 12*x + 36*x - 32 = 0
$$\left(36 x + \left(x^{3} - 12 x^{2}\right)\right) - 32 = 0$$
Detail solution
Given the equation:
$$\left(36 x + \left(x^{3} - 12 x^{2}\right)\right) - 32 = 0$$
transform
$$\left(36 x + \left(\left(- 12 x^{2} + \left(x^{3} - 8\right)\right) + 48\right)\right) - 72 = 0$$
or
$$\left(36 x + \left(\left(- 12 x^{2} + \left(x^{3} - 2^{3}\right)\right) + 12 \cdot 2^{2}\right)\right) + \left(-36\right) 2 = 0$$
$$36 \left(x - 2\right) + \left(- 12 \left(x^{2} - 2^{2}\right) + \left(x^{3} - 2^{3}\right)\right) = 0$$
$$36 \left(x - 2\right) + \left(- 12 \left(x - 2\right) \left(x + 2\right) + \left(x - 2\right) \left(\left(x^{2} + 2 x\right) + 2^{2}\right)\right) = 0$$
Take common factor -2 + x from the equation
we get:
$$\left(x - 2\right) \left(\left(- 12 \left(x + 2\right) + \left(\left(x^{2} + 2 x\right) + 2^{2}\right)\right) + 36\right) = 0$$
or
$$\left(x - 2\right) \left(x^{2} - 10 x + 16\right) = 0$$
then:
$$x_{1} = 2$$
and also
we get the equation
$$x^{2} - 10 x + 16 = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{2} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{3} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -10$$
$$c = 16$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{2} = 8$$
$$x_{3} = 2$$
The final answer for x^3 - 12*x^2 + 36*x - 32 = 0:
$$x_{1} = 2$$
$$x_{2} = 8$$
$$x_{3} = 2$$
$$\left(36 x + \left(x^{3} - 12 x^{2}\right)\right) - 32 = 0$$
transform
$$\left(36 x + \left(\left(- 12 x^{2} + \left(x^{3} - 8\right)\right) + 48\right)\right) - 72 = 0$$
or
$$\left(36 x + \left(\left(- 12 x^{2} + \left(x^{3} - 2^{3}\right)\right) + 12 \cdot 2^{2}\right)\right) + \left(-36\right) 2 = 0$$
$$36 \left(x - 2\right) + \left(- 12 \left(x^{2} - 2^{2}\right) + \left(x^{3} - 2^{3}\right)\right) = 0$$
$$36 \left(x - 2\right) + \left(- 12 \left(x - 2\right) \left(x + 2\right) + \left(x - 2\right) \left(\left(x^{2} + 2 x\right) + 2^{2}\right)\right) = 0$$
Take common factor -2 + x from the equation
we get:
$$\left(x - 2\right) \left(\left(- 12 \left(x + 2\right) + \left(\left(x^{2} + 2 x\right) + 2^{2}\right)\right) + 36\right) = 0$$
or
$$\left(x - 2\right) \left(x^{2} - 10 x + 16\right) = 0$$
then:
$$x_{1} = 2$$
and also
we get the equation
$$x^{2} - 10 x + 16 = 0$$
This equation is of the form
a*x^2 + b*x + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{2} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{3} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -10$$
$$c = 16$$
, then
D = b^2 - 4 * a * c =
(-10)^2 - 4 * (1) * (16) = 36
Because D > 0, then the equation has two roots.
x2 = (-b + sqrt(D)) / (2*a)
x3 = (-b - sqrt(D)) / (2*a)
or
$$x_{2} = 8$$
$$x_{3} = 2$$
The final answer for x^3 - 12*x^2 + 36*x - 32 = 0:
$$x_{1} = 2$$
$$x_{2} = 8$$
$$x_{3} = 2$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -12$$
$$q = \frac{c}{a}$$
$$q = 36$$
$$v = \frac{d}{a}$$
$$v = -32$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 12$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 36$$
$$x_{1} x_{2} x_{3} = -32$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -12$$
$$q = \frac{c}{a}$$
$$q = 36$$
$$v = \frac{d}{a}$$
$$v = -32$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 12$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 36$$
$$x_{1} x_{2} x_{3} = -32$$
Sum and product of roots
[src]
sum
2 + 8
$$2 + 8$$
=
10
$$10$$
product
2*8
$$2 \cdot 8$$
=
16
$$16$$
16