Q&A for How to Factor a Cubic Polynomial

Then x is one of its factors. Factoring that out reduces the rest to a quadratic.

You just need to factor out the x. Let's say you're given: x^3+3x^2+2x=0. Then x(x^2+3x+2)=0. Roots: x -> 0; x^2+3x+1 -> (x+2)(x+1) -> -2, -1.

No, not if all coefficients are real. Complex roots always come in conjugate pairs and polynomials always have exactly as many roots as its degree, so a cubic might have 3 real roots, or 1 real root and 2 complex roots.

We divide the polynomial by the leading coefficient, then calculate the roots normally.

That would mean there are no variables (letters). Thus, it could be expressed as a single number.

No, the quadratic equation only contains a, b, and c terms. A cubic has a, b, c, and d terms. The quadratic can only be used on a quadratic equation of the form ax^2+bx+c.

Let the values which are factors of a constant term and check the polynomial for that values. You will get three factors.

It's a little hard to write out math on a computer, but here: x^2(x+6)+11x+6. Or you can take x out from three terms instead of the two: x(x^2+6x+11)+6. You can't solve a problem with a variable given only that.

This is an example of "the sum of cubes" (because x³ is the cube of x, and 27 is the cube of 3). The formula for factoring the sum of cubes is: a³ + b³ = (a + b)(a² - ab + b²). In this case, a is x, and b is 3, so use those values in the formula.

Your question is very abstract. There are several methods to find roots given a polynomial with a certain degree. The procedure for the degree 2 polynomial is not the same as the degree 4 (or biquadratic) polynomial. You can always factorize the given equation for roots -- you will get something in the form of (x +or- y).

The highest power of your variable in the equation tells you the number of factors that equation will have. So a cubic polynomial having a third degree power will have 3 factors. Of these, one will have to be real. The other two may be a complex conjugate pair, or will also both be real. When equating to zero to find roots, note that you can always assume more trivial roots of the equation by multiplying the entire equation with your variable.

There is always a constant term, but if it is zero, it is typically not written.

Not reliably, only if the roots are also relatively small such that their cubes are insufficient to overwhelm your very small coefficient.

A cubic equation must have a second degree term, unless it's equal to 0, in which case it won't be written out. If there is no second degree term, that means the coefficient of x^2 is 0.

If you factor the right common factors out, then yes. But when it has irrational roots, that's slightly different. In general, it works.

Given that your terms are ax^3, bx^2, and cx, you would find your GCF from each of those terms. Usually it would be just "x", but sometimes if all of the coefficients have a GCF, then that would be part of your GCF. Then separate the expression so it has an (x) value and an (ax^2+bx+c) value. Then solve the 1st degree function by dividing by its coefficient (linear), and the 2nd degree function algebraically using any quadratic way to solve it (completing the square, quadratic formula, factoring, etc.).

You can solve this equation by subtracting 512 from both sides to get (x^3 = -512) and then taking the cube root to find (x = -8).

You can't solve t^3 + t^2 - 1 because it's not an equation. This expression cannot be factored either.

To factor this polynomial, look for two numbers that multiply to 2 and add to -3, which are -1 and -2. Therefore, factor the polynomial as (x^2 - 1)(x - 2).

To find the zeros, set the polynomial equal to zero and solve. Start by using factor by grouping, factoring out a "z", and then proceed to solve from there.

This polynomial is not factorable because there are no common factors. Some polynomials cannot be factorized.

To solve this equation, subtract 512 from both sides to get x^3 = -512. Then, take the cube root of both sides to find x = -8.

To factor this polynomial, look for two numbers that multiply to \( x^3 \), which are \( x^2 \) and \( x \). Then, try to find two numbers that multiply to 1 and add to -12; however, there are none, indicating that it is not factorable. If the polynomial were \( x^3 - x - 12 \), you could factor it into \((x^2 - 4)(x + 3)\), which further factors into \((x - 2)(x + 2)(x + 3)\).

Add x to the equation, but remember to subtract x too. Then, if possible, factor by grouping.

That expression is not factorable with rational numbers.

A cubic polynomial with no constant would take the form ax³ + bx² + cx. In this case, you would factor the polynomial: x(ax² + bx + c).

This cannot be factored. Also, this isn't an equation and this can't be simplified any further.

This is actually already in its simplest form, meaning that it cannot be factorised.

That's fine. Sometimes non-linear roots just can't be simplified any further.

Not necessarily. Some cubic expessions such as (3x^2 - 6x^2 + x - 2) only have one *real* root (here's it's (x-2)), due to it's other root (it's this case, it's the square root of -1/3) being imaginary. Meanwhile other expressions such as (x^3 - a^3) only have two roots (in this case, they're (x-a) and (x^2 + ax + a^2)), as neither root can be simplified further.