Zeros of polynomials

To find the zeros of a polynomial ${P}$, we solve ${P(x)=0}$. For degree ${1}$ and ${2}$, we have formulas. For higher degrees, this page gives a plan: guess a zero from a short list, divide it out, and repeat. We also see why a polynomial of degree ${n}$ has exactly ${n}$ zeros, if we count complex zeros.

The rational zeros theorem

A rational number is a fraction of two integers, such as ${\dfrac{3}{2}}$ or ${-5}$. The next theorem gives a short list of the only rational numbers that can be zeros:

Rational zeros theorem: Let ${P(x)=a_nx^n+\cdots+a_1x+a_0}$ have integer coefficients. If ${\dfrac{p}{q}}$ is a rational zero in lowest terms, then

${p}$ is a factor of the constant term ${a_0}$, and

${q}$ is a factor of the leading coefficient ${a_n}$.

The theorem does not say that these numbers are zeros. It only says that no other rational number can be one. We still have to test each candidate.

Optional

Why it is true. Put ${x=\dfrac{p}{q}}$ into ${P(x)=0}$, and multiply both sides by ${q^n}$:

$a_np^n+a_{n-1}p^{n-1}q+\cdots+a_1pq^{n-1}+a_0q^n=0$

Every term except the last one has a factor ${p}$. So ${p}$ is also a factor of the last term, ${a_0q^n}$. Since ${p}$ and ${q}$ have no common factor, ${p}$ must be a factor of ${a_0}$. In the same way, every term except the first has a factor ${q}$. So ${q}$ is a factor of ${a_n}$.

To find all zeros of a polynomial with integer coefficients:

  1. List the possible rational zeros ${\dfrac{p}{q}}$.
  2. Test them, one at a time, with synthetic division. A remainder of ${0}$ means you found a zero.
  3. Use the quotient for the next step. It has a lower degree. Repeat until the quotient is quadratic.
  4. Solve the quadratic by factoring or with the quadratic formula.

Example 1: Find all zeros of ${P(x)=2x^3+x^2-13x+6}$.

Solution:

The constant term is ${6}$. Its factors are ${\pm1}$, ${\pm2}$, ${\pm3}$, and ${\pm6}$. These are the choices for ${p}$.

The leading coefficient is ${2}$. Its factors are ${\pm1}$ and ${\pm2}$. These are the choices for ${q}$.

So the possible rational zeros ${\dfrac{p}{q}}$ are

$\pm1,\ \pm2,\ \pm3,\ \pm6,\ \pm\dfrac{1}{2},\ \pm\dfrac{3}{2}$.

Try ${x=1}$ first: ${P(1)=2+1-13+6=-4}$. Not a zero.

Try ${x=2}$ with synthetic division:

${2}$${2}$${1}$${−13}$${6}$
${4}$${10}$${−6}$
${2}$${5}$${−3}$${0}$

The remainder is ${0}$, so ${2}$ is a zero. The quotient is ${2x^2+5x-3}$. Factor it:

${2x^2+5x-3=(2x-1)(x+3)}$

So ${P(x)=(x-2)(2x-1)(x+3)}$, and the zeros are ${2}$, ${\dfrac{1}{2}}$, and ${-3}$. All three were on our list.

Example 2: Find all zeros of ${P(x)=x^4-5x^3-5x^2+23x+10}$.

Solution:

The leading coefficient is ${1}$, so ${q=\pm1}$. The possible rational zeros are the factors of ${10}$:

${\pm1,\ \pm2,\ \pm5,\ \pm10}$

Test them:

${P(1)=1-5-5+23+10=24}$

${P(-1)=1+5-5-23+10=-12}$

${P(2)=16-40-20+46+10=12}$

${P(-2)=16+40-20-46+10=0}$

So ${-2}$ is a zero. Divide by ${x+2}$, so ${c=-2}$:

${−2}$${1}$${−5}$${−5}$${23}$${10}$
${−2}$${14}$${−18}$${−10}$
${1}$${−7}$${9}$${5}$${0}$

The quotient is ${Q(x)=x^3-7x^2+9x+5}$. Its possible rational zeros are ${\pm1}$ and ${\pm5}$. We already know ${\pm1}$ are not zeros of ${P}$, so they are not zeros of ${Q}$ either. Try ${5}$:

${5}$${1}$${−7}$${9}$${5}$
${5}$${−10}$${−5}$
${1}$${−2}$${−1}$${0}$

So ${5}$ is a zero, and the new quotient is ${x^2-2x-1}$. This does not factor nicely, so use the quadratic formula:

$\begin{align*}&x\\&=\dfrac{2\pm\sqrt{4+4}}{2}\\&=\dfrac{2\pm 2\sqrt{2}}{2}\\&=1\pm\sqrt{2}\end{align*}$

The zeros are ${-2}$, ${5}$, ${1+\sqrt{2}}$, and ${1-\sqrt{2}}$. The last two are irrational, so they were not on the list. We found them only after dividing out the rational zeros.

Tip: A graph of ${P}$ on a graphing calculator shows roughly where the real zeros are. Test the candidates near those points first. That saves many tries.

The fundamental theorem of algebra

Some polynomials have no real zeros. For example, ${x^2+1=0}$ has no real solution. But it does have the complex solutions ${i}$ and ${-i}$ (see complex numbers). In fact, every polynomial has a complex zero:

Fundamental theorem of algebra: Every polynomial of degree ${n\ge 1}$ has at least one complex zero.

Remember that real numbers are complex numbers too, with imaginary part ${0}$. So "complex zero" includes real zeros.

This theorem has a strong consequence. Say ${c_1}$ is a zero of ${P}$. By the factor theorem, ${P(x)=(x-c_1)Q_1(x)}$. The quotient ${Q_1}$ has degree ${n-1}$, so it has a zero ${c_2}$ too. Repeat this ${n}$ times. Each step takes away one factor:

${P(x)=a(x-c_1)(x-c_2)\cdots(x-c_n)}$

Here ${a}$ is the leading coefficient, and ${c_1,\ldots,c_n}$ are complex numbers. Some of them can be equal. So:

A polynomial of degree ${n}$ has exactly ${n}$ zeros, if a zero of multiplicity ${m}$ is counted ${m}$ times.

Example 3: Factor ${P(x)=x^3-3x^2+4x-12}$ completely, and find all its zeros.

Solution:

Factor by grouping:

$\begin{align*}&P(x)\\&=x^2(x-3)+4(x-3)\\&=(x-3)(x^2+4)\end{align*}$

The factor ${x^2+4}$ has no real zeros. But ${x^2=-4}$ gives ${x=\pm 2i}$. So ${x^2+4=(x-2i)(x+2i)}$, and

${P(x)=(x-3)(x-2i)(x+2i)}$.

The zeros are ${3}$, ${2i}$, and ${-2i}$. That is ${3}$ zeros for degree ${3}$.

Example 4: Find all zeros of ${P(x)=x^4+4x^2}$, and give their multiplicities.

Solution:

Take out the common factor ${x^2}$. Then factor ${x^2+4}$ as in Example 3:

$\begin{align*}&P(x)\\&=x^2(x^2+4)\\&=x^2(x-2i)(x+2i)\end{align*}$

The zero ${0}$ has multiplicity ${2}$. The zeros ${2i}$ and ${-2i}$ each have multiplicity ${1}$. Counted this way, there are ${2+1+1=4}$ zeros, which matches the degree ${4}$.

Complex zeros come in pairs

In Examples 3 and 4, the complex zeros came as ${2i}$ and ${-2i}$. These are conjugates: ${a+bi}$ and ${a-bi}$. This always happens when the coefficients are real:

Conjugate zeros theorem: If ${P}$ has real coefficients and ${a+bi}$ is a zero of ${P}$, then its conjugate ${a-bi}$ is also a zero.

So a polynomial with real coefficients has an even number of non-real zeros. A polynomial of odd degree must therefore have at least one real zero.

The product of a conjugate pair of factors has real coefficients. Use ${(A-B)(A+B)=A^2-B^2}$ with ${A=x-a}$ and ${B=bi}$:

$\begin{align*}&\big(x-(a+bi)\big)\big(x-(a-bi)\big)\\&=(x-a)^2-(bi)^2\\&=(x-a)^2+b^2\end{align*}$

Example 5: Find a polynomial ${P}$ of degree ${3}$ with real coefficients and leading coefficient ${1}$. Its zeros include ${2}$ and ${1+i}$.

Solution:

The coefficients are real, so the conjugate ${1-i}$ is also a zero. That gives three zeros: ${2}$, ${1+i}$, and ${1-i}$. Multiply the conjugate pair first, with ${a=1}$ and ${b=1}$:

$\begin{align*}&\big(x-(1+i)\big)\big(x-(1-i)\big)\\&=(x-1)^2+1\\&=x^2-2x+2\end{align*}$

Then multiply by ${x-2}$:

$\begin{align*}&P(x)\\&=(x-2)(x^2-2x+2)\\&=x^3-2x^2+2x-2x^2+4x-4\\&=x^3-4x^2+6x-4\end{align*}$

Check: ${P(2)=8-16+12-4=0}$.

Example 6: The number ${i}$ is a zero of ${P(x)=x^4-2x^3+6x^2-2x+5}$. Find all the zeros.

Solution:

The coefficients are real, so ${-i}$ is also a zero. So both ${x-i}$ and ${x+i}$ are factors, and so is their product:

${(x-i)(x+i)=x^2+1}$

Divide ${P(x)}$ by ${x^2+1}$ with long division (see dividing polynomials). The remainder is ${0}$, and the quotient is ${x^2-2x+5}$. So

${P(x)=(x^2+1)(x^2-2x+5)}$.

Check: Multiply out the right side:

$\begin{align*}&x^4-2x^3+5x^2+x^2-2x+5\\&=x^4-2x^3+6x^2-2x+5\end{align*}$

Solve ${x^2-2x+5=0}$ with the quadratic formula:

$\begin{align*}&x\\&=\dfrac{2\pm\sqrt{4-20}}{2}\\&=\dfrac{2\pm\sqrt{-16}}{2}\\&=\dfrac{2\pm 4i}{2}=1\pm 2i\end{align*}$

The four zeros are ${i}$, ${-i}$, ${1+2i}$, and ${1-2i}$. They form two conjugate pairs.

Summary

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