# A.16: Roots of Polynomials

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$

( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\id}{\mathrm{id}}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\kernel}{\mathrm{null}\,}$$

$$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$

$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$

$$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$

$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$

$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vectorC}[1]{\textbf{#1}}$$

$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$

$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$

$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

Being able to factor polynomials is a very important part of many of the computations in this course. Related to this is the process of finding roots (or zeros) of polynomials. That is, given a polynomial $$P(x)\text{,}$$ find all numbers $$r$$ so that $$P(r)=0\text{.}$$

In the case of a quadratic $$P(x)=ax^2+bx+c\text{,}$$ we can use the formula

\begin{align*} x &= \frac{-b \pm \sqrt{b^2-4ac}}{2a} \end{align*}

The corresponding formulas for cubics and quartics 1 are extremely cumbersome, and no such formula exists for polynomials of degree 5 and higher 2.

Despite this there are many tricks 3 for finding roots of polynomials that work well in some situations but not all. Here we describe approaches that will help you find integer and rational roots of polynomials that will work well on exams, quizzes and homework assignments.

Consider the quadratic equation $$x^2 - 5x + 6=0\text{.}$$ We could 4 solve this using the quadratic formula

\begin{align*} x &= \frac{5 \pm \sqrt{25-4\times1\times6}}{2} = \frac{5 \pm 1}{2} = 2,3. \end{align*}

Hence $$x^2 - 5x + 6$$ has roots $$x = 2,3$$ and so it factors as $$(x - 3)(x - 2)\text{.}$$ Notice 5 that the numbers $$2$$ and $$3$$ divide the constant term of the polynomial, $$6\text{.}$$ This happens in general and forms the basis of our first trick.

##### Lemma A.16.1 A very useful trick

If $$r$$ or $$-r$$ is an integer root of a polynomial $$P(x)=a_nx^n+\ \cdots\ +a_1x+a_0$$ with integer coefficients, then $$r$$ is a factor of the constant term $$a_0\text{.}$$

Proof

If $$r$$ is a root of the polynomial we know that $$P(r)=0\text{.}$$ Hence

\begin{align*} a_n \cdot r^n+\ \cdots\ +a_1\cdot r+a_0&=0 \end{align*}

If we isolate $$a_0$$ in this expression we get

\begin{align*} a_0 &=-\big[a_n r^n+\ \cdots\ +a_1r\big] \end{align*}

We can see that $$r$$ divides every term on the right-hand side. This means that the right-hand side is an integer times $$r\text{.}$$ Thus the left-hand side, being $$a_0\text{,}$$ is an integer times $$r\text{,}$$ as required. The argument for when $$-r$$ is a root is almost identical.

Let us put this observation to work.

##### Example A.16.2 Integer roots of $$x^3-x^2+2$$

Find the integer roots of $$P(x)=x^3-x^2+2\text{.}$$

Solution:

• The constant term in this polynomial is $$2\text{.}$$
• The only divisors of $$2$$ are $$1,2\text{.}$$ So the only candidates for integer roots are $$\pm 1, \pm 2\text{.}$$
• Trying each in turn

\begin{align*} P(1)&=2 & P(-1)&=0\\ P(2)&=6 & P(-2) &=-10 \end{align*}

• Thus the only integer root is $$-1\text{.}$$
##### Example A.16.3 Integer roots of $$3x^3+8x^2-5x-6$$

Find the integer roots of $$P(x)= 3x^3+8x^2-5x-6\text{.}$$

Solution:

• The constant term is $$-6\text{.}$$
• The divisors of $$6$$ are $$1,2,3,6\text{.}$$ So the only candidates for integer roots are $$\pm1, \pm 2, \pm 3, \pm 6\text{.}$$
• We try each in turn (it is tedious but not difficult):

\begin{align*} P(1) &= 0 & P(-1) &= 4\\ P(2) &= 40 & P(-2) &= 12\\ P(3) &= 132 & P(-3) &= 0\\ P(6) &= 900 & P(-6) &= -336 \end{align*}

• Thus the only integer roots are $$1$$ and $$-3\text{.}$$

We can generalise this approach in order to find rational roots. Consider the polynomial $$6x^2-x-2\text{.}$$ We can find its zeros using the quadratic formula:

\begin{align*} x &= \frac{1 \pm \sqrt{1 + 48}}{12} = \frac{1\pm 7}{12} = -\frac{1}{2}, \frac{2}{3}. \end{align*}

Notice now that the numerators, 1 and 2, both divide the constant term of the polynomial (being 2). Similarly, the denominators, 2 and 3, both divide the coefficient of the highest power of $$x$$ (being 6). This is quite general.

##### Lemma A.16.4 Another nice trick

If $$b/d$$ or $$-b/d$$ is a rational root in lowest terms (i.e. $$b$$ and $$d$$ are integers with no common factors) of a polynomial $$Q(x) = a_nx^n+\ \cdots\ +a_1x+a_0$$ with integer coefficients, then the numerator $$b$$ is a factor of the constant term $$a_0$$ and the denominator $$d$$ is a factor of $$a_n\text{.}$$

Proof

Since $$\frac{b}{d}$$ is a root of $$P(x)$$ we know that

\begin{align*} a_n(b/d)^n+\ \cdots\ +a_1(b/d)+a_0 &=0 \end{align*}

Multiply this equation through by $$d^n$$ to get

\begin{align*} a_n b^n+\ \cdots\ +a_1 b d^{n-1}+a_0d^n &=0 \end{align*}

Move terms around to isolate $$a_0 d^n\text{:}$$

\begin{align*} a_0d^n &= - \big[ a_n b^n+\ \cdots\ +a_1 b d^{n-1} \big] \end{align*}

Now every term on the right-hand side is some integer times $$b\text{.}$$ Thus the left-hand side must also be an integer times $$b\text{.}$$ We know that $$d$$ does not contain any factors of $$b\text{,}$$ hence $$a_0$$ must be some integer times $$b$$ (as required).

Similarly we can isolate the term $$a_n b^n\text{:}$$

\begin{align*} a_n b^n &= - \big[ a_{n-1} b^{n-1}d+\ \cdots\ +a_1 b d^{n-1} + a_0 d^n \big] \end{align*}

Now every term on the right-hand side is some integer times $$d\text{.}$$ Thus the left-hand side must also be an integer times $$d\text{.}$$ We know that $$b$$ does not contain any factors of $$d\text{,}$$ hence $$a_n$$ must be some integer times $$d$$ (as required).

The argument when $$-\frac{b}{d}$$ is a root is nearly identical.

We should put this to work:

##### Example A.16.5 Rational roots of $$2x^2-x-3$$

$$P(x)=2x^2-x-3\text{.}$$

Solution:

• The constant term in this polynomial is $$3=1\times 3$$ and the coefficient of the highest power of $$x$$ is $$2=1\times 2\text{.}$$
• Thus the only candidates for integer roots are $$\pm 1,\ \pm 3\text{.}$$
• By our newest trick, the only candidates for fractional roots are $$\pm \frac{1}{2},\ \pm\frac{3}{2}\text{.}$$
• We try each in turn 6

\begin{align*} P(1)&=-2 & P(-1)&=0\\ P(3)&=12 & P(-3)&=18\\ P\left(\tfrac{1}{2}\right) &= -3 & P\left(-\tfrac{1}{2}\right) &= -2\\ P\left(\tfrac{3}{2}\right) &= 0 & P\left(-\tfrac{3}{2}\right) &= 3 \end{align*}

so the roots are $$-1$$ and $$\frac{3}{2}\text{.}$$

The tricks above help us to find integer and rational roots of polynomials. With a little extra work we can extend those methods to help us factor polynomials. Say we have a polynomial $$P(x)$$ of degree $$p$$ and have established that $$r$$ is one of its roots. That is, we know $$P(r)=0\text{.}$$ Then we can factor $$(x-r)$$ out from $$P(x)$$ — it is always possible to find a polynomial $$Q(x)$$ of degree $$p-1$$ so that

\begin{gather*} P(x) = (x-r)Q(x) \end{gather*}

In sufficiently simple cases, you can probably do this factoring by inspection. For example, $$P(x)=x^2-4$$ has $$r=2$$ as a root because $$P(2)=2^2-4=0\text{.}$$ In this case, $$P(x)=(x-2)(x+2)$$ so that $$Q(x)=(x+2)\text{.}$$ As another example, $$P(x)=x^2-2x-3$$ has $$r=-1$$ as a root because $$P(-1)=(-1)^2-2(-1)-3=1+2-3=0\text{.}$$ In this case, $$P(x)=(x+1)(x-3)$$ so that $$Q(x)=(x-3)\text{.}$$

For higher degree polynomials we need to use something more systematic — long divison.

##### Lemma A.16.6 Long Division

Once you have found a root $$r$$ of a polynomial, even if you cannot factor $$(x-r)$$ out of the polynomial by inspection, you can find $$Q(x)$$ by dividing $$P(x)$$ by $$x-r\text{,}$$ using the long division algorithm you learned 7 This is a standard part of most highschool mathematics curricula, but perhaps not all. You should revise this carefully. in school, but with $$10$$ replaced by $$x\text{.}$$

##### Example A.16.7 Roots of $$x^3-x^2+2$$

Factor $$P(x)=x^3-x^2+2\text{.}$$

Solution:

• We can go hunting for integer roots of the polynomial by looking at the divisors of the constant term. This tells us to try $$x=\pm1, \pm2\text{.}$$
• A quick computation shows that $$P(-1)=0$$ while $$P(1),P(-2),P(2) \neq 0\text{.}$$ Hence $$x=-1$$ is a root of the polynomial and so $$x+1$$ must be a factor.
• So we divide $$\frac{x^3-x^2+2}{x+1}\text{.}$$ The first term, $$x^2\text{,}$$ in the quotient is chosen so that when you multiply it by the denominator, $$x^2(x+1)=x^3+x^2\text{,}$$ the leading term, $$x^3\text{,}$$ matches the leading term in the numerator, $$x^3-x^2+2\text{,}$$ exactly.
• When you subtract $$x^2(x+1)=x^3+x^2$$ from the numerator $$x^3-x^2+2$$ you get the remainder $$-2x^2+2\text{.}$$ Just like in public school, the $$2$$ is not normally “brought down” until it is actually needed.
• The next term, $$-2x\text{,}$$ in the quotient is chosen so that when you multiply it by the denominator, $$-2x(x+1)=-2x^2-2x\text{,}$$ the leading term $$-2x^2$$ matches the leading term in the remainder exactly.

And so on.

• Note that we finally end up with a remainder $$0\text{.}$$ A nonzero remainder would have signalled a computational error, since we know that the denominator $$x-(-1)$$ must divide the numerator $$x^3-x^2+2$$ exactly.
• We conclude that

\begin{gather*} (x+1)(x^2-2x+2)=x^3-x^2+2 \end{gather*}

To check this, just multiply out the left hand side explicitly.
• Applying the high school quadratic root formula $$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ to $$x^2-2x+2$$ tells us that it has no real roots and that we cannot factor it further 8.

We finish by describing an alternative to long division. The approach is roughly equivalent, but is perhaps more straightforward at the expense of requiring more algebra.

##### Example A.16.8 Roots of $$x^3-x^2+2$$ again

Factor $$P(x)=x^3-x^2+2\text{,}$$ again.

Solution: Let us do this again but avoid long division.

• From the previous example, we know that $$\frac{x^3-x^2+2}{x+1}$$ must be a polynomial (since $$-1$$ is a root of the numerator) of degree 2. So write

\begin{gather*} \frac{x^3-x^2+2}{x+1}=ax^2+bx+c \end{gather*}

for some, as yet unknown, coefficients $$a,\ b$$ and $$c\text{.}$$
• Cross multiplying and simplifying gives us

\begin{align*} x^3-x^2+2&=(ax^2+bx+c)(x+1)\\ &=ax^3+(a+b)x^2+(b+c)x+c \end{align*}

• Now matching coefficients of the various powers of $$x$$ on the left and right hand sides

\begin{align*} &\text{coefficient of $x^3$:}\qquad&a&=1\\ &\text{coefficient of $x^2$:}&a+b&=-1\\ &\text{coefficient of $x^1$:}& b+c&=0\\ &\text{coefficient of $x^0$:}& c&=2 \end{align*}

• This gives us a system of equations that we can solve quite directly. Indeed it tells us immediately that that $$a=1$$ and $$c=2\text{.}$$ Subbing $$a=1$$ into $$a+b=-1$$ tells us that $$1+b=-1$$ and hence $$b=-2\text{.}$$
• Thus

\begin{align*} x^3-x^2+2 &= (x+1)(x^2-2x+2). \end{align*}

This page titled A.16: Roots of Polynomials is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.