5.4 Roots of Polynomial Functions

Learning Objectives

By the end of this section, you will be able to:

• perform polynomial long division
• factor polynomials using inspection and polynomial division
• apply the Rational Root Theorem to find roots
• factor higher-degree polynomials and find roots using a combination of techniques

When we're at a loss to factor a quadratic function, we can always fall back on the quadratic formula, but we're not so lucky in higher degrees. There's a closed formula for cubics but it sucks to use, quartics get truly demented, and everything is terrible after that. So really, we just want to figure out how to factor as best we can to break higher-degree polynomials down into linear and quadratic factors. A tool we need in our toolbox for this (and other math courses) is called polynomial long division.

Polynomial Long Division

We want to divide a (dividend) polynomial \( p(x)\) by another polynomial divisor \( d(x) \), resulting in some kind of quotient polynomial \(q(x)\) and possibly a remainder polynomial \(r(x)\). Forget polys for now and think about numbers. If I divide 37 by 5, I can write two (true) equations

\[ \frac{p}{d} = q + \frac{r}{d} \quad \rightarrow \quad \frac{37}{5} = 7 + \frac{2}{5} \notag \]
\[ p = qd + r \quad \rightarrow \quad 37 = 7(5) + 2 \notag \]

Make sure you agree with me about those statements. The same statements are true when I divide polynomials instead of numbers.

\[ \frac{ p(x)}{d(x)} = q(x) + \frac{ r(x)}{d(x)} \quad \text{ and } \quad p(x) = q(x) \cdot d(x) + r(x) \notag \]

Each of the ingredients in the above equations may be polynomials of some sort, whether constants, linear functions, quadratics, etc. The only rule is that the degree of \(r(x)\) must be smaller than the degree of \(d(x)\).

So given a division of \(p(x)\) by \(d(x)\), how do we figure out \(q(x)\) and \(r(x)\)? Here's a summary in English:

1. Set up a long division bar with \(p(x)\) underneath and the divisor \(d(x)\) out front.
2. Compare the first terms of \(p(x) \) and \(d(x)\) and ask yourself, "What do I need to multiply \(d(x)\) by in order to make these first terms match exactly?" Write what is needed above the long division bar.
3. Actually perform the multiplication through \(d(x)\), writing the result underneath the \( p(x)\). Subtract down the relevant columns, resulting in a remainder. Bring down the next term in the dividend \(p(x)\).
4. Compare the first terms of the remainder and the \(d(x)\), and repeat Steps 2-3.
5. Continue until you run out of terms to bring down.
6. The terms you wrote above the long division bar give you the quotient \( q(x)\), and the final remainder gives you \(r(x)\).

It's much easier to learn this via example, though.

Example \(\PageIndex{1}\)

Divide \( p(x) = 2x^2 + 12x - 4 \) by \(d(x) = x-2 \). Write the result in the form \( p(x) = q(x) \cdot d(x) + r(x) \), and in the form \(\frac{ p(x)}{d(x)} = q(x) + \frac{ r(x)}{d(x)}\).

Solution

1. Set up the problem:

\begin{array}{r}
\phantom{)} \\
x-2{\overline{\smash{\big)}\,2x^2+12x-4\phantom{)}}}\\
\notag
\end{array}

2. Compare the first terms: \(p(x)\) has \(2x^2\) and \(d(x) \) has \( x\). If I multiply \(x\) by \(2x\), it will match the \(2x^2\). Write \(2x\) above the bar, lined up properly.

\begin{array}{r}
2x \phantom{-10} \phantom{)} \\
x-2{\overline{\smash{\big)}\,2x^2+12x-4\phantom{)}}}\\
\notag
\end{array}

3. Perform the multiplication \( (2x)(x-2)\) and write it lined up below \(p(x)\), and subtract down the appropriate columns. Bring down the next term (shown in red).

\begin{array}{r}
2x \phantom{-10}\phantom{)} \\
x-2{\overline{\smash{\big)}\,2x^2+12x-4\phantom{)}}}\\
\underline{-~\phantom{(}(2x^2-4x)\phantom{-b)}}\\
0+16x\textcolor{red}{-4}\phantom{)}\\
\notag
\end{array}

4. Compare the first terms \(x\) and \(16x\). To match, we must multiply the \(x\) by \(16\). Write that above, multiply it through \(d(x)\), and write the result underneath. Subtract down the columns.

\begin{array}{r}
2x+16\phantom{)} \\
x-2{\overline{\smash{\big)}\,2x^2+12x-4\phantom{)}}}\\
\underline{-~\phantom{(}(2x^2-4x)\phantom{-b)}}\\
0+16x-4\phantom{)}\\
\underline{-~\phantom{()}(16x-32)}\\
0+28\phantom{)}
\notag
\end{array}

5. There is no other term to bring down, and our remainder is just a constant, so we stop here. The quotient is \(q(x) = 2x+16 \) and the remainder is \(r(x) = 28\). Our final answer is

\[ 2x^2 + 12x - 4 = (2x+16)(x-2) + 28 \quad \text{ or } \quad \frac{ 2x^2 + 12x - 4 }{x-2} = 2x+16 + \frac{28}{x-2} \notag \]

Exercise \(\PageIndex{1}\)

Divide \( p(x) = 3x^2 + 9x + 12 \) by \(d(x) = x+1 \). Write the result in the form \( p(x) = q(x) \cdot d(x) + r(x) \), and in the form \(\frac{ p(x)}{d(x)} = q(x) + \frac{ r(x)}{d(x)}\).

Answer

\begin{array}{r}
3x+6\phantom{)} \\
x+1{\overline{\smash{\big)}\,3x^2+9x+12\phantom{)}}}\\
\underline{-~\phantom{(}(3x^2+3x)\phantom{-b)}}\\
0+6x+12\phantom{)}\\
\underline{-~\phantom{()}(6x+6)}\\
0+6\phantom{)}
\notag
\end{array}

We have

\[ 3x^2 + 9x + 12 = (3x+6)(x+1) + 6 \quad \text{ or } \quad \frac{3x^2 + 9x + 12}{x+1} = 3x+6 + \frac{6}{x+1} \notag \]

ALERT: One important thing to watch out for is when dividends are missing certain powers of \(x\) terms! Before performing polynomial long division, insert placeholders with zero coefficients as needed into your \(p(x)\). For example:

\[ x^3 - 1 \quad \rightarrow \quad x^3 \textcolor{red}{+0x^2 + 0x}-1 \notag \]

Otherwise you will run into trouble and nonsense when you try to line your columns up and subtract correctly.

There's no reason you can only divide by using linear polynomials as \(d(x)\). Let's step it up with a more complex example and exercise.

Example \(\PageIndex{2}\)

Divide \(p(x) = x^4 - 1\) by \( d(x) = x^2 - 1\). Write the result in the form \( p(x) = q(x) \cdot d(x) + r(x) \), and in the form \(\frac{ p(x)}{d(x)} = q(x) + \frac{ r(x)}{d(x)}\).

Solution

First, I insert placeholders: \( p(x) = x^4 + 0x^3 + 0x^2 + 0x - 1\). I write this below the long division sign and put \(d(x)\) out front.

\begin{array}{r}
\phantom{)} \\
x^2-1{\overline{\smash{\big)}\,x^4 + 0x^3 + 0x^2 + 0x - 1\phantom{)}}}\\
\notag
\end{array}

Comparing the first terms, I need to multiply \(x^2\) by \(x^2\) to make it match the \(x^4\) so that they will kill each other off when I subtract. Place that up above, and perform the multiplication, writing it underneath in the appropriate columns. Subtract down the columns and bring down the next term(s).

\begin{array}{r}
x^2\phantom{+0x^2+1}\phantom{)} \\
x^2-1{\overline{\smash{\big)}\,x^4 + 0x^3 + 0x^2 + 0x - 1\phantom{)}}}\\
\underline{-~\phantom{(}(x^4\qquad \quad -x^2)\phantom{ + 0x - 1b)}}\\
0+0+x^2 \textcolor{red}{+0x-1}\phantom{)}\\
\notag
\end{array}

Now we compare the first terms and see that \(x^2\) already matches \(x^2\). When this happens, you don't need to multiply by anything but 1, so put a 1 above the division bar. We perform the multiplication and write it underneath, lined up correctly. Subtract again.

\begin{array}{r}
x^2 \quad \quad+1\phantom{)} \\
x^2-1{\overline{\smash{\big)}\,x^4 + 0x^3 + 0x^2 + 0x - 1\phantom{)}}}\\
\underline{-~\phantom{(}(x^4\qquad \quad -x^2)\phantom{ + 0x - 1b)}}\\
0+0+x^2 +0x-1\phantom{)}\\
\underline{-~\phantom{()}(x^2 \quad \quad -1)}\\
0\phantom{)}
\notag
\end{array}

Notice that on our last subtraction, everything wiped out, leaving a remainder of \( r(x) = 0\)! This means that the divisor \(d(x)\) divides evenly into \(p(x)\). We write our final answer:

\[ x^4 - 1 = (x^2 +1)(x^2 -1) \quad \text{ or } \quad \frac{x^4 - 1}{x^2-1} = x^2+1 \notag \]

Exercise \(\PageIndex{2}\)

Divide \(p(x) = x^3 - 27\) by \( d(x) =x^2 + 3x + 9 \). Write the result in the form \( p(x) = q(x) \cdot d(x) + r(x) \), and in the form \(\frac{ p(x)}{d(x)} = q(x) + \frac{ r(x)}{d(x)}\).

Answer

\begin{array}{r}
x-3\phantom{)} \\
x^2 + 3x + 9{\overline{\smash{\big)}\,x^3+0x^2+0x-27\phantom{)}}}\\
\underline{-~\phantom{(}(x^3+3x^2+9x)\phantom{-b)}}\\
0-3x^2-9x-27\phantom{)}\\
\underline{-~\phantom{()}(-3x^2-9x-27)}\\
0\phantom{)}
\notag
\end{array}

\[ x^3 - 27 = (x-3)(x^2+3x+9) \quad \text{ or } \quad \frac{ x^3-27}{x^2+3x+9} = x-3 \notag \]

Say I'm dividing \(p(x)\) by a certain \( d(x) = x-c\). Then I can write the result as

\[ p(x) = q(x) \cdot (x-c) + r \notag\]

like we've been doing. Here, \(r\) is a constant number, because I'm dividing by a degree-one linear factor \(x - c\). What would happen if I evaluate \( p(x)\) at \(x = c\)? Let's plug it in.

\[ p(c) = q(c) \cdot (\textcolor{red}{c-c}) + r = \textcolor{red}{0} + r = r \notag \]

This gives me an interesting fun fact: the function value \( p(c)\) is exactly the remainder \(r\) that you get when dividing \(p(x)\) by \( x-c\)! Now, on the flip side, say I'm in the situation of the above exercise, where the remainder turned out to be zero. If that ever happens, you know for a fact that \( p(c) = 0\), telling you that \(c\) is indeed a root.

Rational Root Theorem

We're going to pick up another tool to use with polynomial long division when factoring to find roots. Here's the official statement:

Rational Root Theorem

If a polynomial \( f(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 \) has integer coefficients \(a_n\) and \(a_0\) (neither being zero), then every rational root of \(f\) is of the form \( \frac{p}{q} \), where \(p\) is a factor of \(a_0\), and \(q\) is a factor of \(a_n\).

In other words, to find the list of possible rational number roots of \(f\), make two lists,

• all the integer factors \(\pm p_1, \pm p_2, \pm p_3 ... \) of the constant term \(a_0\),
• all the integer factors \(\pm q_1, \pm q_2, \pm q_3 ... \) of the leading coefficient \(a_n\),

and then list all possible combinations \(\pm \dfrac{p_i}{q_j}\) where the numerator is from the first list and the denominator is from the second list. These are the only rational numbers that could be roots of \(f\), and you can check which are actually roots by evaluating \(f\) to see if you get zero there.

Example \(\PageIndex{3}\)

1. List all rational numbers that could be roots of \(f(x) = 6x^3 + 23x^2 - 6x - 8\). Test the options until you find two that are in fact roots. You can use a calculator this time to speed things up.

2. Which of the following polynomials might have a root at \(x = \dfrac{5}{4} \)?

• \( f(x) = 12x^4 + x^2 + 10 \)
• \( g(x) = 6x^3 - 2x^2 - 5 \)
• \( h(x) = 20x^5 + 15x^3 - 17x^2 + x - 20 \)

Solution

1. We first make the two lists of factors:

• \( a_0 = -8 \) has factors \( \pm 1, \pm 2, \pm 4, \pm 8 \)
• \( a_n = 6 \) has factors \( \pm 1, \pm 2, \pm 3, \pm 6 \)

Now the possible combinations of numerators and denominators are, after simplification and without writing duplicates,

\[ \pm 1, \pm 2, \pm 4, \pm 8, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}, \pm \frac{8}{3}, \pm \frac{1}{6} \notag \]

Generally, pick the easiest ones to test out first, especially if you're not using a calculator. Plug the options into the polynomial and see which ones turn out to give 0. The ones that do work are \(x = -4, -\frac{1}{2}, \frac{2}{3} \).

2. We are looking for polynomials that have 5 as a factor of their constant term and \(4\) as a factor of their leading coefficient. This is true of \(f\) because 5 divides 10 and 4 divides 12. Since 4 doesn't divide 6, \(g\) is disqualified, but \(h\) works.

Exercise \(\PageIndex{3}\)

1. List all rational numbers that could be roots of \(f(x) = 2x^3 + 5x^2 - 4x - 3\). Test the options until you find two that are in fact roots.

2. Which of the following polynomials might have a root at \(x = \dfrac{2}{3} \)?

• \( f(x) = 12x^4 + x^2 + 10 \)
• \( g(x) = 9x^3 - x - 12 \)
• \( h(x) = 20x^5 + 15x^3 - 17x^2 + x - 20 \)

Answer

1. Options: \( \pm 1, \pm 3, \pm \dfrac{1}{2}, \pm \dfrac{3}{2} \). The ones that are actually roots are \(x = 1, -3, -\dfrac{1}{2} \).

2. The ones with compatible \(a_0\) and \(a_n\) factors are \(f\) and \(g\).

Factoring Higher-Degree Polynomial Functions

Armed with these tools, we are now equipped to factor polynomials, which will also tell us their roots. It's a simple process:

To factor a polynomial function \(f\) and find its real roots:

Step 1: Find a root \(c\) by inspection (look at the polynomial and making an educated guess at a number easy to plug in, like \( \pm 1\) or \( \pm 2\), or use the Rational Root Theorem, and test options). One you find \(c\), you know \( (x-c)\) must appear in the polynomial's factorization, like \( f(x) = (x-c)( \text{something else}) \).

Step 2: Divide \(f\) by \( (x-c)\) using polynomial division. The quotient \(q(x)\) is the \( (\text{something else})\) factor.

Step 3: If the \(q(x)\) is not a linear factor or an irreducible quadratic, repeat the process on \(q(x)\) to further factor it. So on and so forth, until every factor is linear of the form \( (x-c)\) or an irreducible quadratic.

Step 4: Determine the real roots using the linear factors.

Example \(\PageIndex{4}\)

Factor \( f(x) = 2x^3 - x^2 - 2x + 1 \) and find all real roots.

Solution

First solution: We first check if something easy to plug in will result in \(f\) giving 0. The easiest thing to plug in first is 1: \( f(1) = 2 - 1 - 2 + 1 = 0 \). Yep! So \(x = 1\) is a root, and we know that \(f\) must factor as \( f(x) = (x-1)(\text{something}) \). To determine the other factor, we divide by \( (x-1)\):

\begin{array}{r}
\textcolor{red}{2x^2+x-1}\phantom{)} \\
x-1{\overline{\smash{\big)}\,2x^3-x^2-2x+1\phantom{)}}}\\
\underline{-~\phantom{(}(2x^3-2x^2)\phantom{-2x+)}}\\
0+x^2-2x\phantom{+1)}\\
\underline{-~\phantom{()}(x^2-x)\phantom{+1)}}\\
0+-x +1 \phantom{)}\\
\underline{-~\phantom{()}(-x+1)}\\
0 \phantom{)}
\notag
\end{array}

So now we have \( f(x) = (x-1)(2x^2 + x -1) \), and we can factor the quadratic further using any preferred method to get a final answer of \(f(x) = (x-1)(2x-1)(x+1)\). The real roots are \( x = -1, \frac{1}{2}, 1\).

Second solution: Since this function's \(a_0\) and \(a_n\) are very small numbers with limited factor options, it would also be quick to just make the rational roots list and test all the options to see if we get three roots. Remember that since \(f\) is degree-3, it can't possibly have more than three distinct real roots! If we find them all we'll be done. Using the Rational Root Theorem, we have a short list of possibilities: \( \pm 1, \pm \frac{1}{2} \). We already saw \(f(1) = 0,\) and \(f(-1) = -2 -1+2+1 = 0 \), so that's good too. Then \( f\left( \frac{1}{2}\right) = \frac{1}{4} - \frac{1}{4} - 1 + 1 = 0 \) is also easy to compute. We stop there, having found three real roots!

Third solution: This function is a great candidate for factoring by grouping. We start with \( f(x) = x^2(2x-1) - (2x-1) \) to get \( f(x) = (x^2-1)(2x-1)\). Then we recognize the difference of squares and factor completely as \( f(x) = (x+1)(x-1)(2x-1) \).

Try it yourself below, and then get more practice in the exercises section!

Exercise \(\PageIndex{4}\)

Factor \( f(x) = 3x^4+4x^3-5x^2-2x \) and find all real roots.

Answer

First factor \( x(3x^3+4x^2-5x-2) \) and focus on the cubic factor. Notice \(x = 1\) is a root, so divide by \( (x-1) \) to get to \( x(x-1)(3x^2 +7x + 2) = x(x-1)(3x+1)(x+2) \). The roots are \( x = 0, 1, -\frac{1}{3}, -2\).