Q&A for How to Factor a Cubic Polynomial

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What if the constant term is zero?

Orangejews

Community Answer

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

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What if the third degree polynomial does not have the constant term?

Community Answer

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.

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Can all three roots be imaginary?

Orangejews

Community Answer

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.

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Question

What if we have a coefficient to x-cubed?

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

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What if there is a polynomial of degree 0?

Donagan

Top Answerer

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

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Question

Can this be done using the quadratic formula since there are many difficult polynomials which have irrational roots?

Community Answer

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.

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How can I factor x^3+3x^2-16x+12=64?

Community Answer

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

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How do I solve x^3 + 6x^2 +11x +6?

Community Answer

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.

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How do I factorize x3 + 27?

Donagan

Top Answerer

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.

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How can I find roots?

Community Answer

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).

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How many factors does a cubic polynomial have?

Community Answer

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.

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What if there is no constant term in the cubic polynomial?

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

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Question

Suppose I know the first coefficient on X^3 is very small so the cubic is approximately a quadratic. Can this help me find the roots?

Community Answer

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

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What if the equation does not contain any second degree term?

Community Answer

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.

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Does factoring by grouping work on any cubic polynomial?

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

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How do I factor a cubic polynomial with less than 4 terms?

Community Answer

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.).

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Question

What if x^2 has a coefficient of zero?

Donagan

Top Answerer

The value of 0x² is zero. Zero multiplied by anything is zero.

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Do all cubic equations have 3 roots in the end?

Verin

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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.

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What if there is no linear term?

Verin

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

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How do I factorise (x^3 + 6x^2 - 7x - 6)?

Verin

Top Answerer

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

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Question

How do I factorize this equation 2x^3+3x^2-3x-2?

I_l1ke_gam3s

Community Answer

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

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What if the polynomial has no constant in it?

Community Answer

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).

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How do I factorize x³ - 64x - 14?

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That expression is not factorable with rational numbers.

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