Section 6.3: Polynomial Division and Factoring

In the previous section, we discussed a technique for sketching polynomials that depended on our ability to find all the factors/roots of a polynomial.  This is a very difficult problem, in general, and it is complicated by the fact that the answer depends on the number system you use: rational, real, or complex numbers.  Using the complex numbers is very convenient, because every polynomial factors completely into complex linear factors; unfortunately, actually finding all the factors is quite challenging.  On the other extreme, using only rational numbers, we have a straightforward and moderately efficient technique for finding all possible linear factors.  That is because, we can quickly convert the problem to that of factoring over the integers!  However, not all integer/rational polynomials factor completely into linear factors.  Working over the real numbers is somewhere in between these two extremes.

Although we cannot give explain all these difficult issues in detail (that would take several more courses in advanced mathematics!), we can discuss some specific, elementary ideas and techniques that can be quite useful in certain cases.  Specifically, we will discuss:

• a general strategy for factoring,
• how to divide polynomials, and 
• how to narrow the search for potential factors/roots over the rational numbers.  

Along the way, we will discover an important formula which shows exactly why roots and linear factors are so closely related.  We will end with some brief observations about factoring over the real and complex numbers and "prime" polynomials.

Polynomial Division

In order to think about factoring polynomials, it is helpful to consider a similar problem, that of factoring integers.  The Fundamental Theorem of Arithmetic says that every integer can be factored into "smallest pieces"; for example, we can write 360 = 2³3²5.  The numbers 2, 3, and 5 are called "prime" numbers, because they cannot be factored any more into smaller integers.  The usual strategy for factoring an integer, n, is to:

1. Look through the list of primes, {2, 3, 5, 7, 11, 13, ... } for a number, p, that is likely to be a factor.
2. Check that that p does in fact divide evenly into n, and by computing the quotient, q = n/p.
3. Repeat the process on q, until we are left with a prime quotient.

For example, to factor 360:

• We notice that 360 is even, so it should be divisible by 2.
• Dividing gives 360/2 = 180.
• Since this is even, we divide by 2 again: 180/2 = 90.
• Again, we can divide by 2: 90/2 = 45.
• Since 45 ends in 5, we know that it is divisible by 5, so we do it: 45/5 = 9.
• We know that 9 is a multiple of 3: 9/3 = 3.
• Since 3 is prime, we are done, and we can conclude that 360 = 2·2·2·5·3·3.

Notice how we used some common divisibility rules (i.e., "even numbers are divisible by 2", "numbers ending in 5 are divisible by 5", etc.) in order to narrow our search to primes that are likely to be factors.

In the same way, using advanced algebra, one can show that that every polynomial (in whatever number system we choose!) can be factored into "prime" (i.e., unfactorable) polynomials.  We can use exactly the same strategy to try and factor polynomials, once we know:

• What are the "prime" polynomials?
• How do we narrow our search to likely factors?
• How do we divide polynomials?

We will address these questions in reverse order.  That is, we will begin by demonstrating how to divide polynomials.

Polynomial division is very similar to ordinary integer division.  Consider how one divides 23 into 4770:

1. We compare the most significant digits of the divisor (i.e., 23) and the dividend (i.e., 4770) to estimate the most significant digit of the quotient.  Since 2 goes into 4 twice, we guess the first digit of our divisor to be 2.

2. We then multiply that digit by 23 to obtain 2·23 = 46, write that below the most significant digits of the dividend, and subtract.

3. We keep carrying down less significant digits of the dividend, putting 0's in the quotient, until we obtain a number that is bigger than the divisor, and we repeat steps 1) and 2).

4. When we are left with a remainder that is smaller than the divisor, we stop.  

We can summarize the results of this division by the equation 4770 = 23·207 + 9, or in words, "23 divides 207 times into 4770, with a remainder of 9".

Compare this with the process of dividing x² + 1 into x⁵ - 2x⁴ + 3x + 7:

With integer division, we automatically put the most significant digits on the left and insert 0's where appropriate; with polynomials we must explicitly line up the terms, from most significant (i.e., highest powers) to least significant, inserting 0 terms where necessary to keep like terms aligned.   As before, this calculation can be summarized by the equation:

x⁵ - 2x⁴ + 3x + 7 = (x² + 1)(x³ - 2x² - x + 2) + (4x + 5), 

or in words, "x² + 1 divides x³ - 2x² - x + 2 times into x⁵ - 2x⁴ + 3x + 7, with a remainder of 4x + 5".  In general, when we divide p by d and get a quotient, q, with a remainder, r, we can summarize our calculation by the equation p = d· q + r.

In some ways, polynomial division is even easier than integer division, because, when we are estimating the next term in the quotient, we can simply take the quotient of the most significant terms in the divisor and dividend (e.g., the second term in the quotient is simply -2x⁴/x² = -2x²).  You should notice that this means that it is possible to get fractional coefficients in the quotient, such as when dividing 2x³ + 6x² - 3x - 1 by 2x - 1:

For example, at the second step, since 7x²/(2x)= 7/2x, we take the second term in the quotient to be 7/2x.

We end this section with a simple consequence of this division algorithm which makes clear the connection between roots and linear factors.  Notice how, if we divide a polynomial, p, by a linear function x - b, we will obtain a quotient, q, and a remainder which, being smaller than a 1st degree polynomial, must be a constant, c.  These must all be related by the equation, p(x) = (x - b)q(x) + c.  We can can solve for the remainder, c, if we simply plug in the corresponding root b for x:

p(b) = (b - b)q(b) + c = 0·q(b) + c = c.

This leads to the general formula for polynomials:

p(x) = (x - b)q(x) + p(b)

This means that:

The linear polynomial x - b divides evenly into (i.e., is a factor of) p if and only if p(b) = 0, that is, b is a root of p.

More generally, we can use the same reasoning to show that:

The linear polynomial ax - b  is a factor of p if and only if b/a is a root of p.

This is sometimes called the Root-Factor Theorem.  Therefore, the problem of finding all the roots/horizontal intercepts of a polynomial is precisely equivalent to determining all linear factors of the polynomial.

Make sure that you can divide polynomials by completing the following exercises.

Likely Factors of Rational Polynomials and Rational Roots

In order to generalize our factoring strategy to polynomials, we still need to discuss:

• How can we efficiently limit our search for likely factors of a polynomial?
• What are the "prime" polynomials?

As we mentioned in the introduction, the answers to these questions depends on the number system we are using.  In this section, we will focus our attention exclusively on rational polynomials and their corresponding rational roots.  We first show, with an example, how to convert the problem to that of factoring over the integers.

When we are looking at a polynomial with rational coefficients, such as 4x³ - 11.6x² - 2x + 2.4, it is more convenient to convert this to a problem over the integers by taking out a common factor.  First, write each coefficient over a least common denominator, then factor out the greatest common factor of the numerator and the common denominator:

4x³ - 11.6x² - 2x + 2.4 = 20/5x³ - 58/5x² - 10/5x + 12/5 = 2/5(10x³ - 29x² - 5x + 6)

By factoring out 2/5, we can now shift our focus to the process of factoring the remaining polynomial, p(x) = 10x³ - 29x² - 5x + 6, which has integer coefficients with no common factor.  Since we are only interested in rational roots, that is,  roots of of the form b/a where a and b are integers, and their corresponding linear factors, ax - b, we have converted the original problem to one of factoring over the integers.  

Now we show how to quickly narrow down the likely possible factors to a relatively short list.  First, observe that if ax - b is a factor of p, by long division we know that the quotient, q(x) = cx² + dx + e, will also have integer coefficients.  Since this would imply the equation 10x³ - 29x² - 5x + 6 = (ax - b)(cx² + dx + e), multiplying out gives 10x³ - 29x² - 5x + 6 = acx³ + ... + be; in particular, a must be a factor of 10 and b must be factor of 6.  In general, we can say that:

If a 1st degree polynomial, ax - b, with integer coefficients is a factor of another polynomial, p(x) = a_nxⁿ + a_{n - 1}x^{n - 1} + ... + a₀ with integer coefficients, then a is a factor of the coefficient of the highest degree term, a_n, while b is a factor of the constant term, a₀.

This is known as the Rational Root Theorem, since the Root-Factor Theorem implies that rational roots correspond to integral, linear (i.e., 1st degree) factors.  While it still leaves a large number of possibilities to try, this "divisibility rule" narrows down the likely factors/roots a bit.  For example, in the case of p, the only integral, linear factors we need to examine are:

x ± 1, x ± 2, x ± 3, x ± 6, 2x ± 1, 2x ± 3, 5x ± 1, 5x ± 2, 5x ± 3, 5x ± 6, 10x ± 1, and 10x ± 3; 

that is the only possible rational roots are:

x = ±1, x = ±2, x = ±3, x = ±6, x = ±0.5, x = ±1.5, x = ±0.2, 
x = ±0.4, x = ±0.6, x = ±1.2, x = ±0.1, and x = ±0.3.  

You may have observed that we did not bother two include possibilities such as 2x ± 2, 2x ± 6, or 10x ± 2, since these are simply multiples of other possibilities in the list, that is, they are not "prime".  They do not lead to any new roots anyway.  

Notice that we now have a partial answer to both of our original questions, in the case of integer/rational polynomials:

• Some likely factors of a_nxⁿ + a_{n - 1}x^{n - 1} + ... + a₀, where the a_i's are all integers, are linear polynomials ax - b, where a is a factor of a_n and b is a factor of a₀.  

• A linear polynomial, ax - b, where a and b have no common factor, cannot be factored over the integers, that is, it is "prime".  

Now the process of factoring reduces to searching through the list of possible factors to find those that actual factors.  We do this in two ways:

1. We successively test various possibilities, by simply plugging the corresponding root in to p to see if we get 0.

2. We use this information to eliminate possible factors and narrow our search until we have accounted for all the possibilities. 

For example, plugging in the largest and smallest roots and 0 gives:

Note: Since we know that:

• from far out, it looks like 10x³ and 
• it has a vertical intercept of 6,

we could have guessed (without even plugging in) that we would get a large, negative value at -6, a large positive value at 6, and a positive value at 0.  

We can use this information to speed up our search for a root/factor, by exploiting an important observation (which is proven in Calculus):

Whenever the outputs of a polynomial changes from a negative to positive value, or vice versa, there must be a root between the corresponding inputs.

For example, since the graph goes from negative to positive between -6 and 0, there must be a negative root between these two numbers.  This means that we can focus our attention on the remaining possible negative roots x = -3, -2, -1.5, -1.2, -1, -0.6, -0.5, -0.4, -0.3, -0.2, and -0.1 (i.e., we've narrowed our search to less than half of the original possibilities).  Plugging in the middle possibility, x = -1, gives -28.  Again, since the graph is negative here and positive at 0, we can focus our attention on the possible roots between these two values, x = -0.6, -0.5, -0.4, -0.3, -0.2, and -0.1 (again, cutting the number of possibilities in half).  Again, we plug in a value in the middle to find that p(-0.4) = 2.72.  Therefore, the actual root must be either x = -0.6 or -0.5.  Notice: Be selecting the middle possibility, we are guaranteed to cut our search down by at least half each time.

Since p(-0.5) = 0, this is a root of p.  By the Root-Factor Theorem, this means that 2x + 1 is a factor.  Following the same procedure as for factoring integers, we can confidently divide out 2x + 1 and repeat the process on the quotient, which by long division:

works out to be 5x² - 17x + 6.  

Applying the Rational Root Theorem to this smaller polynomial, we see that the only possible remaining integral factors are:

x - 1, x ± 2, x ± 3, 5x ± 1, 5x - 2, 5x ± 3, and 5x ± 6; 

that is, we can eliminate the possible factors of 2x ± 1, 2x ± 3, 10x ± 1, and 10x ± 3 (as well as, x ± 6, x + 1, and 5x + 2, since ±6, -1, and -0.4 were not roots).  Eliminating the corresponding possible roots, we are left with the following list of potential roots:

x = -3, -2, -1.2, -0.6, -0.2, 0.2, 0.4, 0.6, 1, 1.2, 2, and 3.  

Proceeding as before, we plug in the values on the ends and at 0:

and we happen to find that 3 is a root.  Therefore, x - 3 is another factor, and dividing out:

gives the remaining factor of 5x - 2.  This means that we have completely factored p as p(x) = (2x + 1)(x - 3)(5x - 2).  

Reflecting back on what we did to factor p, we obtain a general strategy for finding rational roots (i.e., linear factors with integer coefficients) of polynomials with integer coefficients:

Note: When you get down to a quadratic polynomial, you may use the quadratic formula to locate the final two roots; for example, with 5x² - 17x + 6 we could simply have noticed that the roots are , so that it must factor as (x - 3)(5x - 2).

Practice applying these ideas for factoring over the rationals/integers by completing the following Exercises.  

Factoring Over the Real and Complex Numbers and Prime Polynomials

While the example in the previous section factored completely into integral, linear factors, we can easily find polynomials that will not.  For example, we can use the Rational Root Theorem to show that p(x) = x² - x - 1, q(x) = x² + x + 1, and r(x) = x⁴ - 2 have no integral, linear factors!  The Rational Root Theorem says that the only possible rational roots of p, q, or r are x = ±1, with the additional two possibilities of  x = ±2 for r.  Since none of these possibilities evaluate to 0, these polynomials have no 1st degree factors with integer coefficients.  In particular, this immediately implies that the first two are "prime"; with a tiny bit more algebra, we can show that r is "prime", as well, as long as we restrict ourselves to the integer/rational number system.

However, the quadratic formula shows that p does have two real roots, namely, .  By the Root-Factor Theorem, this means that it should factor as .  Multiplying out:

we can see that this is a valid factorization.  

Likewise, we can see that  are two real roots of r.  This means that we can divide r by :

so that we have the factorization .

In general, by working over a larger number system, which may include more of the actual roots of the polynomial, we can hope to factor any polynomial more completely.  The factorizations of p and r are pretty typical of those over the real numbers, in the sense that:

Every polynomial with real coefficients factors completely into 1st and 2nd degree polynomials with real coefficients.

Unfortunately, while this statement is true, practically speaking it is quite hard to actually compute the factors, in general.  However, it does tell us about the "prime" polynomials, if we are working in the real number system:

The only real, prime polynomials are of the form x - a or x² + bx + c, with b² - 4c < 0.

Note: The condition b² - 4c < 0 guarantees that the quadratic formula does not give us any real roots.

For example, this theorem implies that is a "prime" polynomial over the real numbers.  We can also see this directly from its graph; it is simply the graph of x² shifted up, so cannot cross the horizontal axis, that is, it has no real roots.  Therefore, by the Root-Factor Theorem, it cannot factor any more into real linear factors.  Likewise, we can see from its graph:

that q has no horizontal intercepts, so it cannot factor any more over the real numbers.

Although, in general, complex roots/factors are just about as difficult to compute as real factorizations, by working with the complex numbers, we are finally guaranteed to be able to factor every polynomial completely into linear factors:

Every polynomial with complex coefficients factors completely into 1st degree polynomials with complex coefficients.  In particular, the complex, prime polynomials are those of the form x - z.  

Note: This property of the complex number system is so important as to be called the Fundamental Theorem of Algebra. 

For example, using complex numbers, we can factor q and r completely into linear factors.  That is because, by the quadratic formula, x² + x + 1 and each have two complex roots, namely  and , respectively.  This means that we may factor q and r as:

and: 

, 

respectively.

Practice factoring over the integers, reals, and complex numbers by completing the following Exercises.  

Go to Sketching Rational Functions.

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Send questions or comments to jwicks@northpark.edu

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