Q&A for How to Solve a Cubic Equation

If you only have x³ in an equation, you can isolate it and find the cube root of both sides. This only works with really simple equations, though—factoring is the best way to solve more complex equations.

That equation has numerous answers because you've got three variables. To get one answer for three variables you need three equations. One possible answer would be x=1, y=-1, z=1 => (1)(-1)+1+1^3=1.

Solve the equation using the discriminant approach you will get three values of x. X=8, X=-7+(-1i)√71, X=-7+i√71. Easy from here, you pick the real value of x, that's 8 and your three numbers were 8, 10 and 12.

Yes, but it's highly impractical to memorize or even use: http://www.math.vanderbilt.edu/~schectex/courses/cubic/

In the ordinary sense, there is no such thing as the square root of a negative number. However, mathematicians have invented the "imaginary" number known as "i", which is defined as the square root of negative 1. The square root of -3 is equal to "i" multiplied by the square root of 3, or 1.73 i.

It is the constant term of the polynomial. This is because Minus Alpha x Minus Beta x Minus Gamma x Minus Delta = Plus Alpha Beta Gamma Delta.

By the Fundamental Theorem of Algebra, we have ax^3 + bx^2 + cx + d, which can be expressed as a(x-r)(x-s)(x-t). WLOG let the equation give r. Then, simply divide the cubic by (x-r) and we get a quadratic whose roots are the remaining two roots.

Multiply and collect: 6y3 + 4y2 - 5y = 2, therefore 21y = 2. The answer is y = 2 / 21.

Yes. It is possible. All you need to do is use the factoring approach, only first you must add 2 to both sides. After, you can factor it to (x) (x^2 + 4) = 2. Divide both sides by x to get x^2 + 4 = 2/x. Subtract 4 from both sides to get x^2 = 2/x - 4. Square root both sides to get x = ±√(2x - 4).

First simplify the equation to 4N = 2x. Then divide each side by 4 to get N = (2x)/(4).

Before trying advanced methods like the cubic formula, do a quick check for rational roots -- you might get lucky. Here the Rational Roots Theorem implies than any rational roots must be integer divisors of 60. A little trial and error then reveals 4^3 - 4 = 64-4 = 60, so A=4 is a solution. If you require all real and complex solutions, use the known solution to factor A^3 - A - 60 = (A-4)(A^2 + 4A + 15). The quadratic factor has no real roots, but its two complex solutions can be found via the quadratic formula.

Find the prime factors of 3 and -4 and divide each factor of -4 by each factor of 3, you will get a set of 12 numbers. you need to plug in these 12 numbers in your cubic equation, and whichever of them yields 0, is your answer. Here the answer is x= -1, -2 and (2/3).

Find factors of 1 (coefficient of x cube) and factors of -3 (constant term) then divide the factors of -3 by the factors of 1. You will get a set of 4 numbers. plug in these 4 numbers in your cubic equation and 3 of them will yield 0 which will be your solutions for x.

Differentiate the equation and equate the derivative to zero. Solve for the variable and you will get values of x at which the derivative is zero. Use those values to find the corresponding values of y and the points you obtain will be the points of inflection. Do the second derivative test, and the points at which the second derivative is positive, the function has local minima, and the points at which the second derivative is negative, the function has local maxima.

This is not a cubic. Anyway, always simplify noting that there is no simple term or factor that can be taken out first (e.g. p+3 etc). So 9 + 2p+6 -28 +4p +16p^2 = 0 -13 +6p +16p^2 = 0 16p^2 + 6p -13 = 0 Solve from here.

A quadratic equation can be expressed in the standard form ax² + bx + c = 0, where a does not equal zero. (This form can be thought of as +/- ax² +/- bx +/- c = 0.) Examples include 3x² + 2x + 5 = 0, and x² - x - 1 = 0.