Factoring Quadratics: The Weird Case

3x⁶ + 7x³ + 2

3(x³)² + 7(x³) + 2

At first glance, this does not appear to be a quadratic — and, in technical terms, it isn't. But this expression is quadratic in form, meaning that it can be restated as a quadratic, it follows the same patterns, and it can be factored by using the same techniques.

This polynomial has three terms (so it's a trinomial). Taking a closer look at those exponents, I see that the power on the leading term is 4, and the power on the middle term is 2, which is half of 4.

In a regular quadratic, I would have the powers 2 and 1, where 1 is half of 2. (And, in either case, the third term is just a number.) So this polynomial follows the quadratic pattern; it is a quadratic "in x²", rather than "in x". In fact, I can rewrite the expression as:

My leading coefficient is just (the "understood") 1, so this is a case of simple quadratic factoring. I'll draw my parentheses as usual, but I'll put an x² at the beginning of each factor:

Factors of −8 that add to −2 are −4 and +2, so I get:

Because this didn't start out as a regular quadratic, but instead started out as a polynomial in quadratic form, I still need to check to see if there is any further factoring possible. In this case, my first factor is a difference of squares, x² − 4, and that is something I can factor.

This gives me a final answer of:

y² − 2y − 8

(y − 4)(y + 2)

(x² − 4)(x² + 2)

x⁴ − 2x² − 8

(x²)² − 2(x²) − 8

y² − 2y − 8

(y − 4)(y + 2)

(x² − 4)(x² + 2)

a(variable stuff)² + b(same stuff) + c

This polynomial has three terms, and the degree of the middle term, being 3, is half of the degree of the leading term, being 6. So this is a quadratic "in x³"; in fact, it's even a simple quadratic to factor.

I need factors of +5 that add to +6, so I'll use +1 and +5 to split the middle term's coefficient:

I didn't start with an actual quadratic, so I might not actually be done with this factorization. I need to check to see if there is more that I can do. So I look at the two factors I've got. I see that the second of these factors is a sum of cubes, which is something that I can factor further:

Putting this all together, my complete answer is:

This expression isn't even a polynomial, since polynomials are required to have whole-number exponents. However, this expression does have three terms, and the degree on the middle term is half of the degree on the leading term; and the third term is just a number. So this is in quadratic form; it's a quadratic in x^1/3. And, because the leading coefficient is just 1, this is actually a simple case of factoring.

I need factors of −6 that add to −1. I'll use −3 and +2 to split the middle term's coefficient. This gives me:

There's nothing more that I can do with this, so I'm done. My answer is:

This expression isn't entirely a polynomial, because of the one-half power on the middle term's variable. But the expression has three terms, and the power on the middle term is half of the power on the leading term (being the understood 1), so this is of quadratic form; it's a quadratic in x^1/2, and it's a simple case of factoring.

Because the one-half power is the same as the square root, my answer, in radical form (that is, when written with the square-root symbol), is:

This has only two terms, and nothing comes out of both. So they've thrown me a curveball; this is actually just a difference of squares. I will factor in the usual way:

Checking, I see that neither of these factors is a difference of squares, nor a sum or difference of cubes, so there's nothing more I can do. My answer is:

This expression is a polynomial, and it does have three terms, though they're big lumpy ones. The power on the middle term (namely, 2) is half of the power on the leading term (namely, 4), so this is quadratic in form; it's a quadratic in (x − 3)². I will factor as usual....

Well, actually, because I'm lazy and this particular expression has such big lumpy terms, I'll do the substitution thing. But I won't bother with a variable; I'll just use a shape.

I'm not done yet, because I've got a squared binomial in there that I can multiply out, and then simplify inside each of the two resulting parentheticals:

Now I need to check to see if either of these simple-case quadratics factors further.

In the first parenthetical, I need factors of +13 that add to −6; the only factor pair for 13 is 1 and 13, so the first quadratic doesn't factor. In the second parenthetical, I need factors of +7 that add to −6; the only factor pair for 7 is 1 and 7. But I'm multiplying to a "plus", so the factors have to have the same sign; neither −1 − 7 nor +1 + 7 equals −6, so this quadratic doesn't factor, either.

My final answer then is:

x⁴ − 12x³ + 56x² − 120x + 91