# Why are higher-degree polynomial equations more difficult to solve?

I am confused about the significance of the powers on equations. For example, in $ax = b$, intuitively $b$ is a value $x$ multiplied $a$ times. In $ax + b = c$, $c$ is a value $x$ multiplied $a$ times added to by $b$. In $ax^2 + bx + c = d$, $d$ is a value $x$ multiplied by itself and by another value $a$ added to by the $x$ multiplied by a value $b$ and that added to by another value $c$. This and so on.

Even in the simple equation $x^2 = y$, $x$ is intuitively $\sqrt{y}$ and so on for arbitrary powers and roots. For the equation $x + x = y$, $x$ is placed doubly and $y$ is equal to that doubled x, so $x$ would be $y$ halved. But when a particular case like $x^2 + x = y$ arises and so on with all orders and polynomials of those orders ("higher order equations"), it cannot be solved so easily and the whole process begins to seem much more foreign and contrived.

Why am I no longer able to intuitively solve for the unknown in such an equation when an exponent greater than one is used? Why doesn't the seamless reversing of operations to solve extend to higher order equations?

• The only reason you think that $x^2=y$ is easier to solve for $x$ than $x^2+x=y$ is that we happen to have invented a notation $\sqrt {y}$ that means "the solution to $x^2=y$". If you want to invent a notation that means "the solution to $x^2+x=y$", then that equation will be equally easy to solve. – WillO Jun 16 '14 at 17:51
• In your example you can actually add $\frac14$ to both sides of the equation, collapse LHS into $\left(x+\frac12\right)^2$, and then it's obvious how to proceed. You can think of adding $\frac14$ as reversing its subtraction from the complete square. – Ruslan Jun 17 '14 at 15:44

## 2 Answers

I think the difference you are picking up on has to do with the structure of the equation itself. In some equations, the unknown $x$ is acted on serially through a sequence of nested operations; solving the equation amounts to "unwinding" those operations one by one. For example, $\frac{(3x+5)^2-19}{2}=10$ can be thought of as "take $x$, triple it, add $5$, square the result, subtract $19$, and divide by $2$; the result is $10$. You can diagram this as a series of functions like so: So to solve it, you reverse those steps: Start with $10$, multiply it by $2$, add $19$, take the square root, subtract $5$, divide by $3$.

But in other equations -- even fairly simple ones -- have a different kind of structure. For example, $x^2+5x=10$ looks like this: Notice that this structure resists any attempt to solve it by "unwinding", precisely because of the fork in the diagram. The $x$ flows through more than one path, which makes it impossible to trace the result backwards to its source.

Some equations are presented in a form that at first appears to have multiple-paths (e.g. $3x + 5x = 16$) but we can rearrange them to a single-path structure (for example, via the distributive property / combining like terms). But higher-degree polynomial equations typically have terms that cannot be combined, and this is what makes them resistant to the kind of intuitive solution you are asking about.

Update: In the comments below, the OP asks:

This makes sense, but then how are general solutions for higher order equations like the quadratic formula derived algebraically/symbolically? I am familiar with the geometric proof by completing the square, but do higher order equations suddenly require the vantage of geometry to solve? Or is there a pure/direct algebraic derivation that can be thought to extend from the basis of "unwinding" or the like?

Although completing the square is historically geometric in origin, it can be understood in a purely algebraic way as a method of restructuring a function so that it has a "serial" structure, enabling it to be solved via unwinding. Let's take the example of $x^2+5x=10$, already diagrammed above. In completing the square, you first add $(\frac{5}{2})^2=\frac{25}{4}$ to both sides of the equation, obtaining $x^2 + 5x + \frac{25}{4} = \frac{65}{4}$. Then you recognize the left-hand side as a perfect square trinomial, so the equation can be written $(x + \frac{5}{2})^2 = \frac{65}{4}$. This equation, if represented diagrammatically, would have a simple serial structure: Start with $x$, add $\frac{5}{2}$, square it, and end up with $\frac{65}{4}$. It can then be solved by unwinding: Start with $\frac{65}{4}$, take the square root(s) (keeping in mind that there are two square roots, one positive and one negative), and subtract $\frac{5}{2}$.

Of course, this is not the only method that can be used to tackle quadratics. Consider the slightly different example of $x^2 + 5x = 24$. If we rearrange this as $x^2 + 5x - 24 = 0$ and factor the LHS, we get $(x-3)(x+8)=0$. This equation can be diagrammed as follows: At first glance this looks to be no better than the original equation; it has a fork in it, and seems resistant to unwinding. But! There is this property of real numbers, the "zero product property", which says that if two numbers multiply to be zero, then one of them must be zero. And that allows you to split the diagram into two separate diagrammatic cases: And each of those can be tackled via a very simple unwinding method.

In short, most of the techniques that are taught (at least at the high school level) for solving polynomial equations can be understood as "methods for re-structuring equations so that unwinding techniques can be employed". (I'm not claiming that they are taught in those terms, or that they should be, but merely that they can be thought of that way.)

Unfortunately this only gets you so far. Once you get to 5th degree polynomials, it is a famous result that there may be solutions that cannot be expressed by a combination of "simple" operations (see here). That means, among other things, that there is no way to restructure a general 5th degree polynomial so as to enable a solution via unwinding techniques.

• Hello, thank you for the great answer! This makes sense, but then how are general solutions for higher order equations like the quadratic formula derived algebraically/symbolically? I am familiar with the geometric proof by completing the square, but do higher order equations suddenly require the vantage of geometry to solve? Or is there a pure/direct algebraic derivation that can be thought to extend from the basis of "unwinding" or the like? – Mathemert Jun 17 '14 at 2:13
• @Mathemert See my update in the answer. – mweiss Jun 17 '14 at 3:21
• Thank you for the elaboration; that helps very much! Could you clear up though, how b/2a is derived algebraically / symbolically? Or can it only be derived geometrically, and then if that is true, when an equation becomes a higher order, does that mean it must be studied geometrically in order to find a solution? – Mathemert Jun 17 '14 at 3:41
• @Mathemert here is a good algebraic derivation: purplemath.com/modules/sqrvertx2.htm – stackErr Jun 17 '14 at 10:16
• I would say it's just pattern-matching. We have an expression of the form $a^2x^2 + bx$. We want to add some square $c^2$ to this so that the result $a^2x^2 + bx + c^2 = (ax+c)^2$. If we expand the RHS we find that it is $a^2x^2 + 2acx + c^2$. Comparing those two expressions we find that they will match precisely if we set $2ac = b$, i.e. $c = b/2a$. – mweiss Jun 17 '14 at 11:33

Here's how I see it. Equations involving only multiplication are easy to solve, for instance $$b^{18}x^{20012} = a^{42},$$ and similarly equations involving only addition are easy to solve. Multiplication and addition of real numbers, taken individually, have a lot of structure, and this structure helps with calculations.

Unfortunately, the only thing connecting multiplication and addition of real numbers is the rather weak distributive law, and the consequences of this weak coupling are felt in many areas of mathematics, whenever you see problems involving addition and multiplication acting in concert. In addition to solving algebraic equations, some examples include:

• Infinitude of primes of the form $x^2+1$. Looking at tables of numbers, it is obvious that there are infinitely many primes of this form, and in fact it is almost certain that much stronger results are true. But we don't currently have any idea how to systematically attack this problem to find a proof -- the difficulty, again, is that primes are a purely multiplicative concept, while $x^2+1$ mixes in some addition.

• Other famous conjectures involving primes, such as the Goldbach and Twin Prime conjectures, again involve mixing primes with addition.

• The Collatz Conjecture. Consider the recurrence $$f(n) = \begin{cases}n/2, & n\equiv 0 \mod 2\\3n+1, & n\equiv 1 \mod 2.\end{cases}$$ Does repeatedly applying $f$ to an integer always eventually reach $1$? Probably, but nobody knows for sure. Again, notice the conspicuous mixture of addition and multiplication in the recurrence.

Over many years we have developed a very solid understanding of how to deal with linear equations, and slowly, we are also building a complete theory of quadratic and higher-order forms. Maybe, someday, we'll see the whole picture clearly.

EDIT: Here's what I mean by weak. The structure of the integers is completely determined by the multiplication and addition tables; these tables have to satisfy certain axioms such as associativity of addition and multiplication, commutativity, distributivity, etc, collectively called the ring axioms (see http://en.wikipedia.org/wiki/Ring_axioms#Definition).

Let's say we keep the multiplication table but completely forget the addition table. How much of the table can we reconstruct, knowing only the multiplication table, and the ring axioms? Notice that our main weapon here is the distributive law, since it is the only axiom that couples multiplication and addition. If the distributive law were very strong, the structure of the multiplication table would completely determine the addition table. But it turns out this is not the case -- here, for instance, is a non-standard addition table

$$\begin{array}{c|cccccc} + & 1 & 2 & 3 & 4 & 5 & 6\\ \hline 1 & 3 & 17 & 7 & 11 & 6 & 13\\ 2 & 17 & 6 & 19 & 34 & 23 & 14\\ 3 & 7 & 19 & 9 & 29 & 13 & 51\\ 4 & 11 & 34 & 29 & 12 & 31 & 38\\ 5 & 6 & 23 & 13 & 31 & 15 & 37\\ 6 & 13 & 14 & 51 & 38 & 37 & 18 \end{array}$$ and you can check that it obeys the ring axioms: for instance $2\cdot(3+1) = 14 = 2\cdot 3 + 2\cdot 1$ and $(2+2)\cdot 3 = 18 = 2\cdot 3+2\cdot 3.$

In fact you can construct many such "nonstandard" addition tables, and it is a good exercise to work out 1) how I built the one above, and 2) what all of the possible valid (in the sense of obeying the ring axioms with standard multiplication) addition tables are.

• Thank you for the answer! But would you mind clarifying some things? What do you mean "weak" distributive property? Isn't multiplication very closely tied to addition, in fact defined in terms of addition as multiplication is just repeated addition? Why does this seemingly simple/direct relationship yield such a complication? And equations such as ax + b include both operations but aren't so complicated; why is this, and what distinguishes the $x^n + x^{n-1} + ... x$ form? – Mathemert Jun 17 '14 at 2:03
• Hm, according to the table, $(3+1) = 2$, so $2*(3+1) = 4 \neq (2*3) + (2*1) = 8$, so it is not distributing. Where am I going wrong? How do you construct these tables (generally) such that they abide by the ring axioms? And, it would be helpful if you could explicitly flesh out the connection between the weakness of the distributive law and the complications that arise from higher degree polynomials (eg, $x^2 + x$). – Mathemert Jun 17 '14 at 5:51
• @Mathemert $2\cdot 3 + 2\cdot 1 = 6 + 2 = 4.$ – user7530 Jun 17 '14 at 6:04
• Ah, I see, I didn't perform addition according to the table. How do you construct these tables (generally) such that they abide by the ring axioms? – Mathemert Jun 17 '14 at 6:07
• Oh, are you intending for the exercise to be left to the reader (me)? – Mathemert Jun 17 '14 at 6:45