Factoring higher order polynomials
by Tim Isbell, October 2015
This material is a crib sheet, built on a third order example, of how to factor equations that are higher order than a quadratic. It is also available at mc.isbellonline.org.
Important theorems and tools… with example
Fundamental Theorem of Algebra
Here’s another useful tool: Lower & Upper Bound Theorem
A shortcut for our example problem
General Strategy for factoring high ordered polynomials
Important theorems and tools… with example
Fundamental Theorem of Algebra
Roots (or zeros) of a function are the values of x where the function equals zero, meaning these are the x-intercepts of the graph of the function. Roots are found by factoring the polynomial. For example, to find the zeros of the function f(x) = x2 +3x +2, we set f(x) = 0 and factor the quadratic into the factors: (x+1)*(x+2). At any points where f(x) = 0, one of these two factors must equal zero. So to find the values of x that make f(x) = 0, we set each of these factors equal to zero, and find that if either x = -1 or x = -2 then f(x) = 0. So the roots (or zeros) of f(x) are x = -1, -2.
The Fundamental Theorem of Algebra says that the number of roots of a polynomial function is equal to the highest order term in the function. So a function with the highest order term of x5 will have 5 roots. Each root may have only a real part, or they may have a real and an imaginary part (which is called a complex number).
Complex Conjugates Theorem
Whenever a polynomial function has complex roots, they always come in pairs, called complex conjugates. So if (3 + 2i) is a root, then you can be certain that (3 - 2i) is also a root.
Similarly, square-root roots also always come in pairs. So if 
is a root, so is 
.
Rational Root Theorem (& Factor Theorem)
This theorem finds all possible rational (ie. real) roots of a polynomial function with integer coefficients.
Consider the function f(x) = qx3 + bx2 + cx + p, where q is the coefficient of the highest order term and p is the constant. In this case, all the possible zeros are:
all the factors of p, divided by 
all the factors of q.
For example, if f(x) = 4x3 - 2x2 - 24x - 18. Since the factors of p are 
9, 18 and the factors of q are 
the list of possible p/q’s is very long: 
Note that there 12 possible positive rational roots and another 12 possible negative rational roots!
Fortunately, from the Fundamental Theorem of Algebra we already know that our function has a maximum of 3 roots. These may all be real, or two of them may be complex. To search for the real roots, we can divide each possible root into f(x) in order to locate the ones with a remainder of zero, but that’s very time consuming. We can divide faster with synthetic division, but that will still take a long time.
We need to find ways to shorten the list of possibilities.
Descartes’ Rule of Signs
Here’s one way. Zeros (or roots) are where f(x) crosses the x axis. This rule says:
- The maximum number of positive real zeros in f(x) is the number of sign changes in f(x).
- The maximum number of negative real zeros in f(x) is the number of sign changes in f(-x).
For our example, f(x) = 4x3 - 2x2 - 24x - 18, the sign changes only once (between the first and second terms). So, at most, there is one positive real root. So as soon as we find one real positive root, there’s no need to evaluate any more positive p/q’s.
Also for our example, f(-x) = -4x3 - 2x2 + 24x - 18, the sign changes twice (between the second and third terms, and again between the third and fourth terms). So, at most, there are two negative real roots in f(x). So as soon as we find two negative real roots, we can stop looking for more.
Polynomial Remainder Theorem
This is a quick way to check to see if a simple linear expression of the form (x - a) is a factor of a high order polynomial. It is quick if a is an integer (for instance, x - 2) but not so quick if a is any other real number.
If P(x)/(x - a) has a remainder, its remainder will be P(a).
So it follows that if P(a) = 0, then we know that (x - a) is a factor of P(x). So a fast way to search for zeros is to start evaluating P(a) for different values of a, and when we find one with a remainder of 0 then that is a real root. This is sometimes called the Factor Theorem.
It’s time to start at the beginning of our list of possible zeros (the p/q’s that we found from the Rational Root Theorem), and see if we can find a root.
Our p/q list looks like: The order doesn’t matter, but integers are quicker to evaluate than fractions so let’s start plugging through the integers in our p/q list.
f(1) = +4 -2 -24 -18 = -38. Nope, f(1) is not zero, so x = 1 is not a zero. And (x-1) is not a factor of f(x).
f(-1) = -4 -2 + 24 -18 = 0. YES, f(-1) is zero, so x = -1 is a zero. And (x+1) is a factor of f(x)! Now we know one of 2 possible negative roots. Let’s continue on, looking for another root.
f(2) = 32 - 8 - 48 - 18 = -42. Nope.
f(-2) = -32 - 8 + 48 - 18 = -10. Nope.
f(3) = 108 - 18 - 72 - 18 = 0. YES, (x = 3) is a positive root. Now we’ve found one negative and one positive root, and we know the remaining root is negative. So as we move further through our list we can skip testing any more positive p/q’s. In other words, we can ignore all the p/q’s that are positive.
f(-3) = -108 -18 +72 -18 = -72. Nope.
f(-6), f(-9), f(-18) are also non zero. So… Nope.
Now it’s time to start working our way through the fractional p/q’s! It’s tedious, but when we do it this is what we find:
f(-½), f(-¼) are also non zero. Nope again.
f(-3/2) = 4(-3/2)3 -2(-3/2)2 -24(-3/2) -18 = 0. Finally! We’ve found the third root.
So the factors of f(x) = 4x3 - 2x2 - 24x - 18 = (x-3)(x+1)(x+3/2). Or, said another way, the roots (or zeros) of f(x) are x = -3/2, -1, and +3.
Here’s another useful tool: Lower & Upper Bound Theorem
We didn’t use this tool to find the zeros for the f(x) in the above example, but we could have. This tool gives us another way to trim off the extremes of the p/q list. We do this by arbitrarily choosing a value of x=k, and then use this tool to see if there are any zeros beyond it. Here’s how it works:
If you have a polynomial with real coefficients and a positive leading coefficient, then ...
Finding the Upper Bound. Use synthetic division to divide f(x) by (x - k), where k > 0. Note, this means you make k any number you want, to see if there are any zeros above that k. So look at your list of p/q’s and choose a k that cuts off some of the high ones. Then use synthetic division to divide f(x) by (x-k). If all the signs in the bottom row of the synthetic division are positive, then x=k is an upper bound (no roots are to the right of x=k).
Finding the Lower Bound. Use synthetic division to divide f(x) by (x - k), where k < 0, and the signs in the bottom row of the synthetic division alternate, then x=-k is a lower bound (no roots are to the left of x=-k).
Note! If there is the integer zero in the bottom row of these synthetic divisions, consider that one either positive or negative, as needed.
A caution: this tool is not a scalpel. Suppose you apply this to (x-1)3 = x3-3x2+3x-1. You know the highest root is x=1. But when you use synthetic division to check if x=2 is an upper bound the bottom row of numbers are not all positive. When you decide to check if x=3 is a positive bound, the second number in the bottom row is a zero and you can’t go on. Finally, if you check if x=4 is a positive bound you find that all the bottom row numbers are positive, confirming that there are now roots above x=4. That’s helpful, but not as precise as you might like.
A shortcut for our example problem
After finding the first root, x = -1, we knew there were only 2 roots left, one positive and one negative. They could be a complex pair, or both real. Knowing that x = -1 is a root means that if we divide f(x) by by (x + 1), using regular long division or synthetic division, we’ll get a quotient without a remainder! That’s important, because doing this division reduces our f(x) from the original third order (a cubic) to the second order (a quadratic), which we will call g(x):
g(x) = 4x2 - 6x - 18 = 0
It’s just a quadratic and we have several simple ways to factor these. First, we try to factor by inspection. Because there are several factors of 4 and 18, this is a bit hard to factor intuitively. So instead we can “complete the square.” Probably an even smarter strategy is to just shove g(x) into the quadratic formula. This quickly generates the remaining two roots, whether they are real or complex.
Using the quadratic formula on g(x) identifies the last two roots as x = +3 and x = -3/2.
So instead of just methodically plodding our way through a long list of p/q’s, each time we find a root we need to use this knowledge to simplify the polynomial so that it is increasingly easy to find the remaining roots.
Graph the function
Use Desmos (a browser-based graphing calculator) or something similar to check your work. Or… before you even start, graph the function so you know where you’re going. Obviously, you can’t do this on a test. But in real life, graphing calculators do a lot of this work for us. So get familiar with graphing calculators. They also help us gain an intuitive feel for many of these concepts, and especially how the constants in a function affect its graph. Check out this Desmos graph of our example function (see below). Notice that it confirms our three real roots (places where the graph of the function intersects with the x-axis).
General Strategy for factoring high ordered polynomials
- Use the Fundamental Theorem of Algebra to find how many roots are buried in your function. (The highest ordered exponent is also the number of roots of the equation. Some of the roots may be complex, meaning they have an imaginary part.)
- Remember: Complex Conjugates Theorem. (Imaginary roots ALWAYS come in pairs)
- Use the Rational Roots Theorem to make a list of all possible roots.
- Especially if the equation has a lot of potential rational roots, use Descartes Rule of Signs to find out the maximum number of possible, positive, rational roots. (Rational roots are always real roots, because rational numbers are a subset of real numbers. These real roots are places where the graph of the function crosses the x-axis).
- Use the Polynomial Remainder Theorem. (This theorem will quickly identify some rational roots, start working through the possible roots found in step 3 above. Keep going until you only have 2 left. Now go to step 6.)
- Use synthetic division to divide f(x) by one root after another. You’ll be left with just a quadratic function from those last 2 roots you didn’t identify. Now go to step 7.
- Solve the remaining quadratic function by any means you can: inspection, complete the square, or quadratic formula.
That should get you to a clean list of roots for any polynomial.
If you have a graphing calculator (such as Desmos), graph the equation to see if your roots are correct.
If factoring a high ordered polynomial is a homework assignment and your teacher allows it, use the graphing calculator before you even start. Then you know where the path leads.