Irreducible Quadratic Factors: Definition & Graphical Significance

Instructor: Laura Pennington

Laura received her Master's degree in Pure Mathematics from Michigan State University. She has 15 years of experience teaching collegiate mathematics at various institutions.

We'll define irreducible quadratic factors in the complete factorization of polynomials in this lesson. We'll also explain what this means for the graphs of polynomials with irreducible quadratic factors through examples.

Irreducible Quadratic Factors

Imagine that you are working on a top-secret team of analysts. Your latest mission is to save humanity from an asteroid that is heading towards the earth. While you and your team are assessing the situation, you come up with a plan that can divert the asteroid. The end step in this plan is to factor a polynomial completely into irreducible factors, where an irreducible factor is a polynomial that is not a constant and cannot be factored any further over the real numbers. You get right on it, and end up with the following:

x4 + 3x3 - 2x2 + 6x - 8 = (x - 1)(x + 4)(x2 + 2)

Hmmm…is this as far as you can take it? Survival of the human race depends on you, so we better investigate! You know that two of the factors, x - 1 and x + 4, are irreducible linear factors. This is because they are linear (have an exponent of 1) and have been factored as much as possible over the real numbers. Okay, but what about the third quadratic factor, x2 + 2?

When factoring a quadratic expression, we find the values of x that would make the expression equal to zero. We can call these values a and b, and we use them in the factorization (x - a)(x - b). In mathematical terms, we call a and b the zeros of the quadratic expression.

Consider our quadratic factor, x2 + 2, and let's find the zeros of this expression.


We get that the zeros are x = ± √(-2). We can't take the square root of a negative number using real numbers! Because x2 + 2 can't be factored any further over the real numbers, it is an irreducible quadratic factor. An irreducible quadratic factor is an irreducible factor that is quadratic, or has a highest exponent of 2.

When we're factoring a polynomial over the real numbers, and we get a quadratic factor that has no real numbers that make it equal zero, then the quadratic factor is irreducible, and we can't factor it any further over the real numbers. You quickly report your findings back to headquarters, and they are able to divert the asteroid. Phew! We're all safe thanks to you!

Graphical Significance

So what do these irreducible quadratic factors mean graphically? To answer this question, let's talk about the relationship between zeros of a polynomial and the x-intercepts on the graph of that polynomial. You see, the zeros of a polynomial are the values of x that make the polynomial equal to 0. These values correspond to the x-intercepts on the graph of that polynomial, where x-intercepts are where the graph crosses the x-axis.

When it comes to irreducible quadratic factors, there can't be any x-intercepts corresponding to this factor, since there are no real zeros. In other words, if we have an irreducible quadratic factor, f(x), then the graph will have no x-intercepts if we graph y = f(x).

For instance, consider our asteroid polynomial. If we graph the irreducible factor x2 + 2 (not the whole polynomial), we see the graph doesn't cross the x-axis.


Furthermore, this tells us that the graph of the entire polynomial (x - 1)(x + 4)(x2 + 2) won't have any x-intercepts that correspond to this factor. However, it will still have x-intercepts corresponding to the other two factors at x = 1 and x = -4, so we know the graph will have just two x-intercepts.


Just as we expected, there are exactly two x-intercepts that correspond with the irreducible linear factors, but no x-intercepts that correspond to the irreducible quadratic factor.

Another Example

Let's look at another example to really solidify our understanding of this concept. Consider the factorization of the following polynomial:

  • x5 - 2x4 + 4x3 - 8x2 + 3x - 6 = (x2 + 1)(x2 + 3)(x - 2)

We see that the factorization contains two irreducible quadratic factors, x2 + 1 and x2 + 3, because if we try to find the zeros of these two factors, we end up with non-real numbers, ± √(-1) and ± √(-3).


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